FUNKCJA WYKŁADNICZA – rozwiązanie zadań

Zadanie 1

1. $7^{x - 4} = ({\sqrt{7})}^{2 - 3x}$

$$7^{x - 4} = \lbrack{\left( 7 \right)^{\frac{1}{2}}\rbrack}^{2 - 3x}$$

$$7^{x - 4} = 7^{1 - \frac{3}{2}x}$$

funkcja roznowartosciowa

$$x - 4 = 1 - \frac{3}{2}x$$

$$x + \frac{3}{2}x = 5$$

$$2\frac{1}{2}x = 5$$

$$\frac{5}{2}x = 5$$

$$\frac{1}{2}x = 1$$

x = 2

1. ${(\frac{1}{8}\sqrt{2})}^{x} = 8*4^{3 - 2x}$

$$\left( \frac{1}{2^{3}}*2^{\frac{1}{2}} \right)^{x} = 2^{3}*2^{6 - 4x}$$

$${(\frac{2^{\frac{1}{2}}}{2^{3}})}^{x} = 2^{3}*2^{6 - 4x}$$

$${\lbrack({2)}^{\frac{1}{2} - 3}\rbrack}^{x} = 2^{9 - 4x}$$

$$\left( 2 \right)^{- \frac{5}{2}x} = 2^{9 - 4x}$$

funkcja roznowartosciowa

$$- \frac{5}{2}x = 9 - 4x$$

$$- \frac{5}{2}x + 4x = 9$$

$$- 2\frac{1}{2}x + 4x = 9$$

$$1\frac{1}{2}x = 9$$

$$\frac{3}{2}x = 9$$

$$\frac{1}{2}x = 3$$

x = 6

1. ${(\frac{1}{8}*4^{x})}^{x} = 2^{3x - 4}$

(2⁻³*2^2x)^x = 2^3x − 4

(2^{−3 + 2x})^x = 2^3x − 4

funkcja roznowartosciowa

−3x + 2x² = 3x − 4

2x² − 6x + 4 = 0

=36 − 32 = 4

$$\sqrt{} = 2$$

$$x_{1} = \frac{6 - 2}{4} = 1$$

$$x_{2} = \frac{6 + 2}{4} = 2$$

1. 7^5x − 7^5x − 1 = 6

7^5x − 7^5x * 7⁻¹ = 6

7^5x = t

t > 0

$$t - \frac{1}{7}t = 6$$

$$\frac{6}{7}t = 6$$

$$\frac{1}{7}t = 1$$

t = 7

7^5x = 7

funkcja roznowartosciowa

5x = 1

$$x = \frac{1}{5}$$

1. 2 * 16^x − 17 * 4^x + 8 = 0

2 * 4^2x − 17 * 4^x + 8 = 0

t = 4^x

t > 0

2t² − 17t + 8 = 0

=289 − 64 = 225

$$\sqrt{} = 15$$

$$x_{1} = \frac{17 - 15}{4} = \frac{1}{2}$$

$$4^{x} = \frac{1}{2}$$

4^x = 2⁻¹

funkcja roznowartosciowa

2^2x = 2⁻¹

2x = −1

$$x = - \frac{1}{2}$$

$$x_{2} = \frac{17 + 15}{4} = \frac{32}{4} = 8$$

4^x = 8

2^2x = 2³

funkcja roznowartosciowa

2x = 3

$$x = \frac{3}{2}$$

1. 4^x − 9 * 2^x + 8 = 0

2^2x − 9 * 2^x + 8 = 0

2^x = t

t > 0

t² − 9t + 8 = 0

=49

$$\sqrt{} = 7$$

x₁ = 1

x₂ = 8

funkcja roznowartosciowa

2^x = 1    ∨    2^x = 8

2^x = 2⁰    ∨    2^x = 2³

x = 0    ∨    x = 3

1. 8^x − 3 = 9^x − 3

$$\frac{8^{x - 3}}{9^{x - 3}} = 1$$

$$({\frac{8}{9})}^{x - 3} = 1$$

$$\left( \frac{8}{9} \right)^{x - 3} = ({\frac{8}{9})}^{0}$$

funkcja roznowartosciowa

x − 3 = 0

x = 3

1. 15^2x + 4 = 3^3x * 5^4x − 4

3^2x + 4 * 5^2x + 4 = 3^3x * 5^4x − 4      |  :  3^2x + 4 * 5^2x + 4

$$1 = \frac{3^{3x}*5^{4x - 4}}{3^{2x + 4}*5^{2x + 4}}$$

1 = 3^{3x − 2x − 4} * 5^{4x − 4 − 2x − 4}

1 = 3^x − 4 * 5^2x − 8

1 = 3^x − 4 * 5^2(x−4)

1 = 3^x − 4 * 25^x − 4

(3*25)⁰ = (3 * 25)^x − 4

funkcja roznowartosciowa

x − 4 = 0

x = 4

1. 6^2x + 4 = 3^3x * 2^x + 8

3^2x + 4 * 2^2x + 4 = 3^3x * 2^x + 8      |  :  3^2x + 4 * 2^2x + 4

$$1 = \frac{3^{3x}*2^{x + 8}}{3^{2x + 4}*2^{2x + 4}}$$

1 = 3^x − 4 * 2^−x + 4

1 = 3^x − 4 * 2^−1(x−4)

$$1 = 3^{x - 4}*\frac{1}{2}^{(x - 4)}$$

$$(3{*\frac{1}{2})}^{0} = ({3*\frac{1}{2})}^{x - 4}$$

funkcja roznowartosciowa

x − 4 = 0

x = 4

Zadanie 2

1. ${\sqrt{2 + \sqrt{3}}}^{x} + {\sqrt{2 - \sqrt{3}}}^{x} = 4$

$${\sqrt{\frac{\left( 2 + \sqrt{3} \right)\left( 2 - \sqrt{3} \right)}{(2 - \sqrt{3)}}}}^{x} + {\sqrt{2 - \sqrt{3}}}^{x} = 4$$

$${\sqrt{\frac{4 - 3}{2 - \sqrt{3}}}}^{x} + {\sqrt{2 - \sqrt{3}}}^{x} = 4$$

$${\sqrt{\frac{1}{2 - \sqrt{3}}}}^{x} + {\sqrt{2 - \sqrt{3}}}^{x} = 4$$

$$\frac{1}{{\sqrt{2 - \sqrt{3}}}^{x}} + {\sqrt{2 - \sqrt{3}}}^{x} = 4$$

$${\sqrt{2 - \sqrt{3}}}^{x} = t$$

t > 0

$$\frac{1}{t} + t = 4\ \ |*t$$

1 + t² − 4t = 0

t² + 4t + 1 = 0

=16 − 4 = 12

$$\sqrt{} = 2\sqrt{3}$$

$$t_{1} = \frac{4 - 2\sqrt{3}}{2} = 2 - 2\sqrt{3}$$

$${\sqrt{2 - \sqrt{3}}}^{x} = 2 - \sqrt{3}$$

$$\left( 2 - \sqrt{3} \right)^{\frac{1}{2}x} = \left( 2 - \sqrt{3} \right)^{1}$$

funkcja roznowartosciowa

$$\frac{1}{2}x = 1$$

x = 2

$$t_{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3}$$

$${\sqrt{2 - \sqrt{3}}}^{x} = 2 + \sqrt{3}$$

$$\left( 2 - \sqrt{3} \right)^{\frac{1}{2}x} = \left( 2 + \sqrt{3} \right)^{1}$$

$$({2 - \sqrt{3})}^{\frac{1}{2}x} = \frac{\left( 2 + \sqrt{3} \right)(2 - \sqrt{3}}{(2 - \sqrt{3})}$$

$$\left( 2 - \sqrt{3} \right)^{\frac{1}{2}x} = \frac{1}{2 - \sqrt{3}}$$

$$\left( 2 - \sqrt{3} \right)^{\frac{1}{2}x} = \left( 2 - \sqrt{3} \right)^{- 1}$$

$$\frac{1}{2}x = - 1$$

x = −2

1. $\left( \sqrt{3 - 2\sqrt{2}} \right)^{x} + \left( \sqrt{3 + 2\sqrt{2}} \right)^{x} = 6$

$${\sqrt{\frac{1}{3 + 2\sqrt{2}}}}^{x} + {\sqrt{3 + 2\sqrt{2}}}^{x} = 6$$

$$\frac{1}{{\sqrt{3 + 2\sqrt{2}}}^{x}} + {\sqrt{3 + 2\sqrt{2}}}^{x} = 6$$

$$\left( \sqrt{3 + 2\sqrt{2}} \right)^{x} = t$$

t > 0

$$\frac{1}{t} + t = 6\ \ |*t$$

1 + t² − 6t = 0

t² − 6t + 1 = 0

=36 − 4 = 32

$$\sqrt{} = 4\sqrt{2}$$

$$t_{1} = \frac{6 - 4\sqrt{2}}{2} = 3 - 2\sqrt{2}$$

$${(\sqrt{3 + 2\sqrt{2}})}^{x} = 3 - 2\sqrt{2}$$

$${(3 + 2\sqrt{2})}^{\frac{1}{2}x} = (3 - 2\sqrt{2})$$

funkcja roznowartosciowa

$$\frac{1}{2}x = 1$$

x = 2

$$t_{2} = \frac{6 + 4\sqrt{2}}{2} = 3 + 2\sqrt{2}$$

$$\left( \sqrt{3 + 2\sqrt{2}} \right)^{x} = 3 + 2\sqrt{2}$$

$$\left( 3 + 2\sqrt{2} \right)^{\frac{1}{2}x}{= \left( 3 + 2\sqrt{2} \right)}^{1}$$

$$\left( 3 + 2\sqrt{2} \right)^{\frac{1}{2}x} = \frac{(3 + 2\sqrt{2)}(3 - 2\sqrt{2})}{3 + 2\sqrt{2}}$$

$${(3 + 2\sqrt{2})}^{\frac{1}{2}x} = \frac{1}{3 + 2\sqrt{2}}$$

$${(3 + 2\sqrt{2)}}^{\frac{1}{2}x} = (3{+ 2\sqrt{2})}^{- 1}$$

$$\frac{1}{2}x = - 1$$

x = −2

1. ${(\frac{1}{3})}^{x} < ({\frac{1}{3})}^{2}$

$$\frac{1}{3} \in \left( 0,1 \right)$$

funkcja malejaca

$${(\frac{1}{3})}^{x} < ({\frac{1}{3})}^{2}$$

x > 2

x ∈ (2,+∞)

1. 3^x + 2 + 7^x < 4 * 7^x − 1 + 34 * 3^x − 1

3^x * 3² − 34 * 3^x − 1 < 4 * 7^x * 7¹ − 7^x

$$9*3^{x} - \frac{34}{3}*3^{x} < \frac{4}{7}*7^{x} - 7^{x}$$

$$3^{x}\left( 9 - \frac{34}{3} \right) < 7^{x}\left( \frac{4}{7} - 1 \right)$$

$$3^{x}\left( - \frac{7}{3} \right) < 7^{x}\left( - \frac{3}{7} \right)\ \ \ |:7^{x}$$

$$\frac{3^{x}\left( - \frac{7}{3} \right)}{7^{x}} = \left( - \frac{3}{7} \right)\ \ \ \ |*( - \frac{3}{7})$$

$$\left( \frac{3}{7} \right)^{x} > \frac{9}{49}$$

$$\left( \frac{3}{7} \right)^{x} > \left( \frac{3}{7} \right)^{2}$$

funkcja malejaca

x < 2

x ∈ (−∞,2)

Zadanie 3

1. 4^x − 3 * 2^x − m = 0

2^2x − 3 * 2^x − m = 0

t = 2^x

t > 0

t² − 3t − m = 0

$$\frac{a = 1}{\begin{matrix} b = - 3 \\ c = - m \\ \end{matrix}}$$

1   a ≠ 0

1 ≠ 0 dla m ∈ R

2    > 0

b² − 4ac > 0

9 + 4m > 0

4m > −9

$$m > - \frac{9}{4}$$

3 t₁ * t₂ > 0

$$\frac{c}{a} > 0$$

$$\frac{- m}{1} > 0$$

−m > 0

m < 0

4 t₁ + t₂ > 0

$$\frac{- b}{a} > 0$$

$$\frac{3}{1} > 0$$

3 > 0

m ∈ R

zestawienie zalozen

$$m \in ( - \frac{9}{4},0)$$

1. 16^x − 4^x + 1 − m² + 1 = 0

4^2x − 4^x * 4¹ − m² + 1 = 0

4^x = t

t > 0

t² − 4t − m² + 1 = 0

1) funkcja kwadratowa

1   a ≠ 0

1 ≠ 0 dla m ∈ R

2    = 0

b² − 4(−m²+1) = 0

16 + 4m² − 4 = 0

4m² + 12 = 0

m² + 3 = 0

m² = −3

brak rozwiazania

$$3\ \ \ \frac{- b}{2a} > 0$$

2) funkcja kwadratowa

1   a ≠ 0

1 ≠ 0

2    > 0

b² − 4ac > 0

12 + 4m² > 0

m ∈ R

3   t₁ * t₂ < 0

$$\frac{c}{a} < 0$$

$$\frac{- m^{2} + 1}{1} < 0$$

−m² + 1 < 0

1 − m² < 0

(1−m)(1+m) < 0

m = 1     ∨    m = −1

m ∈ (−∞,−1) ∪ (1,+∞)

3) funkcja liniowa

a = 0

1 ≠ 0 dla m ∈ R

Zadanie 4

1. $\left\{ \begin{matrix} 2^{x}*3^{y} = 12 \\ 3^{x}*2^{y} = 18 \\ \end{matrix} \right.\ $

$$\frac{2^{x}*3^{y}}{3^{x}*2^{y}} = \frac{12}{18}$$

$$\left( \frac{2}{3} \right)^{x}*\left( \frac{3}{2} \right)^{y} = \frac{2}{3}$$

$$\left( \frac{2}{3} \right)^{x}*\left( \frac{2}{3} \right)^{- y} = \frac{2}{3}$$

$$\left( \frac{2}{3} \right)^{x - y}{= \left( \frac{2}{3} \right)}^{1}$$

funkcja roznowartosciowa

x − y = 1

x = y + 1

2^y + 1 * 3^y = 12

2 * 2^y * 3^y = 12   |  : 2

2^y * 3^y = 6

(2*3)^y = (2*3)¹

funkcja roznowartosciowa

$$\left\{ \begin{matrix} x = 2 \\ y = 1 \\ \end{matrix} \right.\ $$

1. $\left\{ \begin{matrix} 5*5^{x - y} = \sqrt{25^{2y + 1}} \\ 8^{2x + 1} = 32*2^{4y - 1} \\ \end{matrix} \right.\ $

$$\left\{ \begin{matrix} {5*5}^{x - y} = {(25^{2y + 1})}^{\frac{1}{2}} \\ 8^{2x + 1} = 32*2^{4y - 1} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} 5*5^{x - y} = {(5^{4y + 2})}^{\frac{1}{2}} \\ 2^{6x + 3} = 32*2^{4y - 1} \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} 5*5^{x}*5^{- y} = 5^{2y}*5 \\ 2^{6x}*2^{3} = 2^{4}*2^{4x - 1} \\ \end{matrix} \right.\ $$

funkcja roznowartosciowa

$$\left\{ \begin{matrix} 1 + x - y = 2y + 1 \\ 6x + 3 = 4 + 4y \\ \end{matrix} \right.\ $$

$$\left\{ \begin{matrix} x - 3y = 0 \\ 6x - 4y = 1 \\ \end{matrix} \right.\ $$

x = 3y

6 * 3y − 4y = 1

18y − 4y = 1

14y = 1

$$y = \frac{1}{14}$$

$$x = \frac{3}{14}$$

$$\left\{ \begin{matrix} x = \frac{3}{14} \\ y = \frac{1}{14} \\ \end{matrix} \right.\ $$

Zadanie 5

1. 16^x + 4^x + 2 − 36 = 0

4^2x + 16 * 4^x − 36 = 0

zalozenie:

t = 4^x

t > 0

t² + 16t − 36 = 0

=256 + 144 = 400

$$\sqrt{} = 20$$

$$t_{1} = \frac{- 16 - 20}{2} = - 18$$

t₁ nie spelnia zalozenia

$$t_{2} = \frac{- 16 + 20}{2} = 2$$

4^x = 2

funkcja roznowartosciowa

2^2x = 2¹

2x = 1

$$x = \frac{1}{2}$$

2. 5^x − 1 − 5 * 2^x = 5^x − 2 + 5 * 2^x − 2

5^x − 1 − 5^x − 2 = 5 * 2^x − 2 + 5 * 2^x

$$\frac{1}{5}*5^{x} - \frac{1}{25}*5^{x} = \frac{5}{4}*2^{x} + {5*2}^{x}$$

$$5^{x}\left( \frac{1}{5} - \frac{1}{25} \right) = 2^{x}\left( \frac{5}{4} + 5 \right)$$

$$5^{x}\left( \frac{4}{25} \right) = 2^{x}\left( \frac{25}{4} \right)$$

$$\frac{5^{x}}{2^{x}} = \frac{25}{4}*\frac{4}{25}$$

$$\left( \frac{5}{2} \right)^{x} = 1$$

$${(\frac{5}{2})}^{x} = ({\frac{5}{2})}^{0}$$

x = 0

1. $2^{\sqrt{x}} = \sqrt{16^{\sqrt{x}}} - 2$

$$2^{\sqrt{x}} = 2^{2\sqrt{x}} - 2$$

$$- 2^{2\sqrt{x}} + 2^{\sqrt{x}} - 2 = 0$$

$$2^{2\sqrt{x}} - 2^{\sqrt{x}} + 2 = 0$$

zalozenie:

$$2^{\sqrt{x}} = t$$

t > 0

t² − t − 2 = 0

=1 + 8 = 9

$$\sqrt{} = 3$$

$$t_{1} = \frac{1 - 3}{2} = - 1$$

t₁ nie spelnia zalozenia

$$t_{2} = \frac{1 + 3}{2} = 2$$

$$2^{\sqrt{x}} = 2$$

$$2^{\sqrt{x}} = 2^{1}$$

funkcja roznowartosciowa

$$\sqrt{x} = 1$$

x = 1

1. $\left( \sqrt{4 - \sqrt{15}} \right)^{x} + \left( \sqrt{4 + \sqrt{15}} \right)^{x} = 8$

$${\sqrt{\frac{\left( 4 - \sqrt{15} \right)\left( 4 + \sqrt{15} \right)}{\left( 4 + \sqrt{15} \right)}}}^{x} + {\sqrt{4 + \sqrt{15}}}^{x} = 8$$

$${\sqrt{\frac{16 - 15}{4 + \sqrt{15}}}}^{x} + {\sqrt{4 + \sqrt{15}}}^{x} = 8$$

$${\sqrt{\frac{1}{4 + \sqrt{15}}}}^{x} + {\sqrt{4 + \sqrt{15}}}^{x} = 8$$

$$\frac{1}{{\sqrt{4 + \sqrt{15}}}^{x}} + {\sqrt{4 + \sqrt{15}}}^{x} = 8$$

zalozenie:

$${\sqrt{4 + \sqrt{15}}}^{x} = t$$

t > 0

$$\frac{1}{t} + t - 8 = 0$$

t² − 8t + 1 = 0

=64 − 4 = 60

$$\sqrt{} = 2\sqrt{15}$$

$$t_{1} = \frac{8 + 2\sqrt{15}}{2} = 4 + \sqrt{15}$$

$${\sqrt{4 + \sqrt{15}}}^{x} = 4 + \sqrt{15}$$

$$\left( 4 + \sqrt{15} \right)^{\frac{1}{2}x} = ({4 + \sqrt{15})}^{1}$$

funkcja roznowartosciowa

$$\frac{1}{2}x = 1$$

x = 2

$$t_{2} = \frac{8 - 2\sqrt{15}}{2} = 4 - \sqrt{15}$$

$${\sqrt{4 + \sqrt{15}}}^{x} = 4 - \sqrt{15}$$

$${(4 + \sqrt{15})}^{\frac{1}{2}x} = 4 - \sqrt{15}$$

$${(4 + \sqrt{15})}^{\frac{1}{2}x} = \frac{\left( 4 - \sqrt{15} \right)(4 + \sqrt{15})}{4 + \sqrt{15}}$$

$$\left( 4 + \sqrt{15} \right)^{\frac{1}{2}x} = \frac{1}{4 + \sqrt{15}}$$

$$\left( 4 + \sqrt{15} \right)^{\frac{1}{2}x} = ({4 + \sqrt{15})}^{- 1}$$

funkcja roznowartosciowa

$$\frac{1}{2}x = - 1$$

x = −2

2. $5*25^{\frac{1}{x}} + 3*10^{\frac{1}{x}} = 2*4^{\frac{1}{x}}$

$$5*25^{\frac{1}{x}} + 3*10^{\frac{1}{x}} = 2*4^{\frac{1}{x}}\ \ \ \ \ |:\ 4^{\frac{1}{x}}$$

x ≠ 0

$$5*5^{\frac{2}{x}} + 3*2^{\frac{1}{x}}*5^{\frac{1}{x}} - 2*2^{\frac{2}{x}} = 0$$

$$5*5^{\frac{1}{x}}*5^{\frac{1}{x}} + 3*2^{\frac{1}{x}}*5^{\frac{1}{x}} - 2*2^{\frac{1}{x}}*2^{\frac{1}{x}} = 0$$

$$3*5^{\frac{1}{x}}*5^{\frac{1}{x}} + 2*5^{\frac{1}{x}}*5^{\frac{1}{x}} + 3*2^{\frac{1}{x}}*5^{x} - 2*2^{\frac{1}{x}}*2^{\frac{1}{x}} = 0$$

$$\frac{5*25^{\frac{1}{x}}}{4^{\frac{1}{x}}} + \frac{3*2^{\frac{1}{x}}*5^{\frac{1}{x}}}{4^{\frac{1}{x}}} = 2$$

$$5*\left( \left( \frac{5}{2} \right)^{2} \right)^{\frac{1}{x}} + \ \frac{3*2^{\frac{1}{x}}*5^{\frac{1}{x}}}{2^{\frac{2}{x}}} = 2$$

$$5*\left( \frac{5}{2} \right)^{\frac{2}{x}} + 3*\frac{5^{\frac{1}{x}}}{2^{\frac{1}{x}}} = 2$$

$$5*\left( \frac{5}{2} \right)^{\frac{2}{x}} + 3*\left( \frac{5}{2} \right)^{\frac{1}{x}} - 2 = 0$$

$$\left( \frac{5}{2} \right)^{\frac{1}{x}} = t$$

t > 0

=9 + 40 = 49

$$\sqrt{} = 7$$

$$t_{1} = \frac{- 3 - 7}{10} = - 1$$

t₁ nie spelnia zalozenia

$$t_{2} = \frac{- 3 + 7}{10} = \frac{2}{5}$$

$$\left( \frac{5}{2} \right)^{\frac{1}{x}} = ({\frac{5}{2})}^{- 1}$$

funkcja roznowartosciowa

$$\frac{1}{x} = - 1$$

−x = 1

x = −1

1. $4^{\left| x - 1 \right|} + 2^{\sqrt{x^{2} - 2x + 1}} - 2 = 0$

$$2^{2\left| x - 1 \right|} + 2^{\sqrt{({x - 1)}^{2}}} - 2 = 0$$

2^2|x−1| + 2^|x−1| − 2 = 0

zalozenie:

2^|x−1| = t

t > 0

t² + t − 2 = 0

=1 + 8 = 9

$$\sqrt{} = 3$$

$$t_{1} = \frac{- 1 - 3}{2} = \frac{- 4}{2} = - 2$$

t₁ nie spelnia zalozenia

$$t_{2} = \frac{- 1 + 3}{2} = 1$$

2^|x−1| = 1

2^|x−1| = 2⁰

funkcja roznowartosciowa

x − 1 = 0

x = 1

Zadanie 6

$$\sqrt[3]{5\sqrt{2} + 7} - \sqrt[3]{5\sqrt{2} - 7}$$

(a+b)³ = a³ + 3a²b + 3ab² + b³

(a − b)³ = a³ − 3a²b + 3ab² + b³

$${(\sqrt{2} + 1)}^{3} = \sqrt{2^{3}} + 3*{(\sqrt{2})}^{2}*1 + 3*\sqrt{2}*1 = 2\sqrt{2} + 6 + 3\sqrt{2} + 1 = 7 + 5\sqrt{2}$$

$${(\sqrt{2} - 1)}^{3} = \sqrt{2^{3}} - 3*{(\sqrt{2})}^{2}*1 + 3\sqrt{2}*1 - 1 = \underset{\sqrt{8}}{} - 6 + 3\sqrt{2} - 1 = 5\sqrt{2} - 7$$

$$\sqrt[3]{5\sqrt{2} + 7} - \sqrt[3]{5\sqrt{2} - 7} = \left( \sqrt{2} + 1 \right) - \left( \sqrt{2} - 1 \right) = \sqrt{2} + 1 - \sqrt{2} + 1 = \sqrt{2} - \sqrt{2} + 1 + 1 = 2$$

2 jest liczba calkowita

Zadanie 7

(5 − x)^{x3 − 4x2 + x + 6} = 1

1   5 − x > 0

−x > −5

x < 5

(5−x)^{x3 − 4x2 + x + 6}=(5−x)⁰

funkcja roznowartosciowa

x³ − 4x² + x + 6 = 0

(x−2)(x²−2x−3) = 0

x = 2

x² − 2x − 3 = 0

=4 + 12 = 16

$$\sqrt{} = 4$$

$$x_{1} = \frac{2 - 4}{2} = - 1$$

$$x_{2} = \frac{2 + 4}{2} = 3$$

2   5 − x = 1

−x = −4

x = 4

Odpowiedz :    x ∈ {−1, 2, 3, 4}