$a = 16\sqrt[3]{4} = 2^{4}*2^{\frac{2}{3}} = 2^{2}$
Egret
x2 − 3x − 4 = 0 =9 − 4 * 1 * (−4) = 9 + 16 = 25
$x_{1} = \frac{3 - 5}{2} = \frac{- 2}{2} = - 1\ \ \ \ \ \ \ \ x_{1} = \frac{3 + 5}{2} = \frac{8}{2} = 4$ x ∈ (−1,4)
25log5 2 = 22log5 2 = 22 = 4
an = −n2 + 16 x ∈ {1,2,3}
gfg
$x = \frac{3}{\sqrt{5} - 2}*\frac{\sqrt{5} + 2}{\sqrt{5} + 2} = \frac{3\sqrt{5} + 6}{5 - 4} = 3\sqrt{5} + 6$ $y = \frac{12}{\sqrt{5} - 1}*\frac{\sqrt{5} + 1}{\sqrt{5} + 1} + 1 = \frac{12\sqrt{5} + 12}{5 - 1} = 3\sqrt{5} + 3$ z < y < x
S2n = [(2n)+(2n+2)]n = n(4n+2) = 4n2 + 2n
$\left( \frac{a}{2} \right)^{2} + \left( 2a \right)^{2} = x^{2} = > x^{2} = 4a^{2} + \frac{1}{4}a^{2} = 4\frac{1}{4}a^{2} = \frac{17}{4}a^{2} = > x = \frac{\sqrt{17}}{2}a$ $\sin \propto = \frac{2a}{x} = \frac{2a\text{\ \ \ }}{\frac{\sqrt{17}}{2}a} = 2*\frac{2}{\sqrt{17}} = \frac{4\sqrt{17}}{17}$
x2 + 4 ≠ 0 = >x2 ≠ −4 sprzeczność x ∈ R
Liczba przeciwna to liczba z przeciwnym znakiem
y = (x−p) + q = >y = f(x−0) − 10 = f(x) − 10
A= (2,3,5) $\overset{}{A}$={3} $\overset{}{\Omega}$= {6} P=$\text{\ \ }\frac{3}{6}$
$x = \sqrt{5 + 12} = \sqrt{25 + 144} = 13$ $\cos \propto = \frac{5}{x} = \frac{5}{13}$
W = x3 − 2 x2 − 4 x + 8 = x2(x−2) − 2(x−2) = (x2−2)(x−2) = (x+2)(x−2)2
|−8| + |−4| = 12 −4 + 6 = 2
=16 − 4 * 5 * 2 = −24 $p = - \frac{- 4}{2*2} = 1\ \ \ $
$2\ x - 9\ y + 6 = 0 = > \ y = \frac{2\ x + 6}{9} = \frac{2}{9}x + \frac{2}{3}$ $k\bot l < = > a_{k}*a_{l} = - 1 = > a_{k} = \frac{- 1}{a_{l}} = \frac{- 1}{\frac{2}{9}} = - 1*\frac{9}{2} = - \frac{9}{2}$
Iloraz ma nad sobą n-tą potęgę
S = (a,b) $a = - 6\ \ \ b = 0\ \ r = \sqrt{4} = 2$
$R = \frac{1}{2}d = > d = 2*2\sqrt{3} = 4\sqrt{3}$ $d = a\sqrt{2} = > a = \frac{4\sqrt{3}}{\sqrt{2}} = \frac{4\sqrt{6}}{2} = 2\sqrt{6}$
8 * 7 − 8 * 6 = 8 * 1 = 8
$\frac{\left| 0A \right|}{\left| 0A \right| + \left| \text{AB} \right|} = \frac{\left| 0D \right|}{\left| 0C \right|} = > \left| 0D \right| = \frac{\left| 0C \right|*\left| 0A \right|}{\left| 0A \right| + \left| \text{AB} \right|} = \frac{48*6}{16} = \frac{288}{16} = 18$
$\left\{ \begin{matrix} a_{2} = 7 \\ a_{6} = 17 \\ \end{matrix} \right.\ $ − $\left\{ \begin{matrix} a_{1} + 5r = 17 \\ a_{1} + r = 7 \\ \end{matrix} \right.\ = > 4r = 10 = > r = \frac{5}{2}$ $a_{1} + \frac{5}{2} = 7 = > a_{1} = 7 - \frac{5}{2} = \frac{14 - 5}{2} = \frac{9}{2}$
$\frac{6*174 + 2x}{8} = 174,5 = > 2x = 1396 - 1044 = 352 = > x = \frac{352}{2} = 176$
Mjuyk