391 Вправа 391Алгебра 8 клас Бевз 2025 · § 8 · стор. 79

Умова

Знайди корені рівняння.

391. Знайди корені рівняння.

а) $\frac{1 - 2 x}{1 + 2 x} =$ $\frac{1 + 2 x}{1 - 2 x} + \frac{16}{4 x^{2} - 1}$ ;

$\frac{1 - 2 x}{1 + 2 x} - \frac{1 + 2 x}{1 - 2 x} - \frac{16}{4 x^{2} - 1} =$ $0$ ;

$\frac{{1 - 2 x}^{∖ 2 x - 1}}{2 x + 1} + \frac{{1 + 2 x}^{∖ 2 x + 1}}{2 x - 1} - \frac{16}{(2 x - 1) (2 x + 1)}$ =

= $\frac{(1 - 2 x) (2 x - 1) + (1 + 2 x) (2 x + 1) - 16}{(2 x - 1) (2 x + 1)}$ =

= $\frac{4 x - 4 x^{2} - 1 + 4 x^{2} + 4 x + 1 - 16}{(2 x - 1) (2 x + 1)} =$ $\frac{8 x - 16}{(2 x - 1) (2 x + 1)}$ =0;

${\begin{aligned} 8 x - 16 = 0 \\ (2 x - 1) (2 x + 1) \neq 0 \end{aligned}$ ;

${\begin{aligned} x = 2 \\ x \neq - \frac{1}{2}, x \neq \frac{1}{2} \end{aligned}$ ;

$x =$ $2$ ;

б) $\frac{2 z}{2 z + 1} - \frac{2 z + 1}{1 - 2 z} =$ $\frac{1}{4 z^{2} - 1}$ ;

$\frac{2 z}{2 z + 1} - \frac{2 z + 1}{1 - 2 z} - \frac{1}{4 z^{2} - 1} =$ $0$ ;

$\frac{{2 z}^{∖ 2 z - 1}}{2 z + 1} + \frac{{2 z + 1}^{∖ 2 z + 1}}{2 z - 1} - \frac{1}{(2 z - 1) (2 z + 1)}$ =

= $\frac{2 z (2 z - 1) + (2 z + 1) (2 z + 1) - 1}{(2 z - 1) (2 z + 1)}$ =

= $\frac{4 z^{2} - 2 z + 4 z^{2} + 4 z + 1 - 1}{(2 z - 1) (2 z + 1)} =$ $\frac{8 z^{2} + 2 z}{(2 z - 1) (2 z + 1)}$ =0;

${\begin{aligned} 2 z (4 z + 1) = 0 \\ (2 z - 1) (2 z + 1) \neq 0 \end{aligned}$ ;

${\begin{aligned} z = - \frac{1}{4} або z = 0 \\ z \neq - \frac{1}{2}, z \neq \frac{1}{2} \end{aligned}$ ;

$z =$ $- 0, 25 або z =$ $0$ ;

в) $\frac{1}{3 - 2 x} - \frac{4}{3 + 2 x} =$ $\frac{9 - x^{2}}{4 x^{2} - 9}$ ;

$\frac{1}{3 - 2 x} - \frac{4}{3 + 2 x} - \frac{9 - x^{2}}{4 x^{2} - 9} =$ $0$ ;

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

= $\frac{- (2 x + 3) - 4 (2 x - 3) - (9 - x^{2})}{(2 x - 3) (2 x + 3)}$ =

= $\frac{- 2 x - 3 - 8 x + 12 - 9 + x^{2}}{(2 x - 3) (2 x + 3)} =$ $\frac{x^{2} - 10 x}{(2 x - 3) (2 x + 3)}$ =0;

${\begin{aligned} x (x - 10) = 0 \\ (2 x - 3) (2 x + 3) \neq 0 \end{aligned}$ ;

${\begin{aligned} x = 0 або x = 10 \\ x \neq - \frac{3}{2}, x \neq \frac{3}{2} \end{aligned}$ ;

$x =$ $0 або x =$ $10$ ;

г) $\frac{1}{{(x + 5)}^{2}} - \frac{3}{{(x - 5)}^{2}} =$ $\frac{2}{25 - x^{2}}$ ;

$\frac{1}{{(x + 5)}^{2}} - \frac{3}{{(x - 5)}^{2}} - \frac{2}{25 - x^{2}} =$ $0$ ;

$\frac{1^{∖ {(x - 5)}^{2}}}{{(x + 5)}^{2}} - \frac{3^{∖ {(x + 5)}^{2}}}{{(x - 5)}^{2}} + \frac{2^{∖ (x - 5) (x + 5)}}{(x - 5) (x + 5)}$ =

= $\frac{{(x - 5)}^{2} - 3 {(x + 5)}^{2} + 2 (x - 5) (x + 5)}{{(x - 5)}^{2} {(x + 5)}^{2}}$ =

= $\frac{x^{2} - 10 x + 25 - 3 x^{2} - 30 x - 75 + 2 x^{2} - 50}{{(x - 5)}^{2} {(x + 5)}^{2}}$ =

= $\frac{- 40 x - 100}{{(x - 5)}^{2} {(x + 5)}^{2}}$ =0;

${\begin{aligned} - 40 x - 100 = 0 \\ {(x - 5)}^{2} {(x + 5)}^{2} \neq 0 \end{aligned}$ ;

${\begin{aligned} x = - \frac{5}{2} \\ x \neq - 5, x \neq 5 \end{aligned}$ ;

$x =$ $- 2, 5$ .