\begin{enumerate}[label=\alph*)] \item $\frac{x^2-1}{3}+\frac{x^2+1}{2}=3,5$\newline $\frac{2\left(x^2-1\right)}{2\cdot 3}+\frac{3\left(x^2+1\right)}{3\cdot 2}=3,5$\newline $\frac{2 x^2-2}{6}+\frac{3 x^2+3}{6}=3,5$\newline $\frac{2 x^2-2+3 x^2+3}{6}=3,5$.\newline $\frac{5 x^2+1}{6}=3,5$.\newline $5 x^2+1=3,5 \cdot 6$\newline $5 x^2+1=21$\newline $5 x^2=21-1$\newline $5 x^2=20$\newline $x^2=4$\newline $x=\pm 2$. \item $\frac{x^2-1}{2}+\frac{x^2+5}{6}=\frac{1}{2}$\newline $\frac{3\left(x^2-1\right)}{3\cdot 2}+\frac{x^2+5}{6}=\frac{1}{2}$\newline $\frac{3 x^2-3}{6}+\frac{x^2+5}{6}=\frac{1}{2}$\newline $ \begin{aligned} &\frac{3 x^2-3+x^2+5}{6}=\frac{1}{2} \\ &\frac{4 x^2+2}{6}=\frac{1}{2} \\ &2\left(4 x^2+2\right)=1\cdot 6 \\ &8 x^2+4=6 \\ &8 x^2=6-4 \\ &8 x^2=2 \\ &x^2=\frac{1}{4} \\ &x=\pm \frac{1}{2} \end{aligned} $ \item $ \begin{aligned} &x^2+2 x=\frac{2 x+9 x}{6} \\ &6\left(x^2+2 x\right)=2 x^2+9 x \\ &6 x^2+12 x=2 x^2+9 x \\ &6 x^2-2 x^2=9 x-12 x \\ &4 x^2=-3 x \\ &4 x^2+3 x=0 \\ &x(4 x+3)=0 \end{aligned} $\newline $\left\{\begin{array}{l}x=0 \\ 4 x+3=0\end{array}\left\{\begin{array}{l}x=0 \\ 4 x=-3\end{array}\left\{\begin{array}{l}x=0 \\ x=-\frac{3}{4}\end{array}\right.\right.\right.$ \item $ \begin{aligned} &\left(x-\frac{1}{3}\right)\left(x+\frac{1}{3}\right)=\frac{8}{9}\\ &x^2-\left(\frac{1}{3}\right)^2=\frac{8}{9}\\ &x^2-\frac{1}{9}=\frac{8}{9}\\ &x^2=\frac{8}{9}+\frac{1}{9}\\ &x^2=1\\ &x=\pm 1 \text {. } \end{aligned} $

Homepage › Solution manuals › Найма Гахраманова › Математика - 8 › Задание 10, стр 49

Математика - 8

Задание 10, стр 49

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a): $\frac{x^{2} - 1}{3} + \frac{x^{2} + 1}{2} = 3, 5$
$\frac{2 (x^{2} - 1)}{2 \cdot 3} + \frac{3 (x^{2} + 1)}{3 \cdot 2} = 3, 5$
$\frac{2 x^{2} - 2}{6} + \frac{3 x^{2} + 3}{6} = 3, 5$
$\frac{2 x^{2} - 2 + 3 x^{2} + 3}{6} = 3, 5$ .
$\frac{5 x^{2} + 1}{6} = 3, 5$ .
$5 x^{2} + 1 = 3, 5 \cdot 6$
$5 x^{2} + 1 = 21$
$5 x^{2} = 21 - 1$
$5 x^{2} = 20$
$x^{2} = 4$
$x = \pm 2$ .
b): $\frac{x^{2} - 1}{2} + \frac{x^{2} + 5}{6} = \frac{1}{2}$
$\frac{3 (x^{2} - 1)}{3 \cdot 2} + \frac{x^{2} + 5}{6} = \frac{1}{2}$
$\frac{3 x^{2} - 3}{6} + \frac{x^{2} + 5}{6} = \frac{1}{2}$

$\begin{aligned} \frac{3 x^{2} - 3 + x^{2} + 5}{6} = \frac{1}{2} \\ \frac{4 x^{2} + 2}{6} = \frac{1}{2} \\ 2 (4 x^{2} + 2) = 1 \cdot 6 \\ 8 x^{2} + 4 = 6 \\ 8 x^{2} = 6 - 4 \\ 8 x^{2} = 2 \\ x^{2} = \frac{1}{4} \\ x = \pm \frac{1}{2} \end{aligned}$
c): $\begin{aligned} x^{2} + 2 x = \frac{2 x + 9 x}{6} \\ 6 (x^{2} + 2 x) = 2 x^{2} + 9 x \\ 6 x^{2} + 12 x = 2 x^{2} + 9 x \\ 6 x^{2} - 2 x^{2} = 9 x - 12 x \\ 4 x^{2} = - 3 x \\ 4 x^{2} + 3 x = 0 \\ x (4 x + 3) = 0 \end{aligned}$
${\begin{matrix} x = 0 \\ 4 x + 3 = 0 \end{matrix} {\begin{matrix} x = 0 \\ 4 x = - 3 \end{matrix} {\begin{matrix} x = 0 \\ x = - \frac{3}{4} \end{matrix}$
d): $\begin{aligned} (x - \frac{1}{3}) (x + \frac{1}{3}) = \frac{8}{9} \\ x^{2} - {(\frac{1}{3})}^{2} = \frac{8}{9} \\ x^{2} - \frac{1}{9} = \frac{8}{9} \\ x^{2} = \frac{8}{9} + \frac{1}{9} \\ x^{2} = 1 \\ x = \pm 1 . \end{aligned}$

gimli

2022-11-01 12:36

Задание 10, стр 49

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