402 Вправа 402Алгебра 8 клас Бевз 2025 · § 8 · стор. 80

402. Розв’яжи систему рівнянь.

а) ${\displaystyle \{\begin{array}{rl}& \frac{1}{x-y}+\frac{1}{x+y}=\frac{5}{8},\\ & \frac{1}{x-y}-\frac{1}{x+y}=\frac{3}{8};\end{array}}$

Нехай ${\displaystyle u=}$${\displaystyle \frac{1}{x-y},v=}$${\displaystyle \frac{1}{x+y}}$:

${\displaystyle \{\begin{array}{rl}& u+v=\frac{5}{8},\\ & u-v=\frac{3}{8};\end{array}}$ додамо:

${\displaystyle 2u=}$${\displaystyle 1}$;

${\displaystyle u=}$${\displaystyle \frac{1}{2}}$;

${\displaystyle v=}$${\displaystyle \frac{5}{8}-\frac{1}{2}=}$${\displaystyle \frac{1}{8}}$;

тоді ${\displaystyle x-y=}$${\displaystyle 2}$;

${\displaystyle x+y=}$${\displaystyle 8}$:

${\displaystyle \{\begin{array}{rl}& x-y=2,\\ & x+y=8;\end{array}}$ додамо:

${\displaystyle 2x=}$${\displaystyle 10}$;

${\displaystyle x=}$${\displaystyle 5}$;

${\displaystyle y=}$${\displaystyle 8-5=}$${\displaystyle 3}$;

${\displaystyle x=}$${\displaystyle 5,y=}$${\displaystyle 3}$; відповідь ${\displaystyle \left(5;3\right)}$;

б) ${\displaystyle \{\begin{array}{rl}& \frac{10}{z-5}+\frac{1}{x+2}=1,\\ & \frac{25}{z-5}+\frac{3}{x+2}=2;\end{array}}$

Нехай ${\displaystyle u=}$${\displaystyle \frac{1}{z-5},v=}$${\displaystyle \frac{1}{x+2}}$:

${\displaystyle \{\begin{array}{rl}& 10u+v=1,\\ & 25u+3v=2;\end{array}\{\begin{array}{rl}& v=1-10u,\\ & 25u+3v=2;\end{array}}$;

${\displaystyle 25u+3(1-10u)=}$${\displaystyle 2}$;

${\displaystyle 25u+3-30u=}$${\displaystyle 2}$;

${\displaystyle -5u=}$${\displaystyle 2-3}$;

${\displaystyle -5u=}$${\displaystyle -1}$;

${\displaystyle u=}$${\displaystyle \frac{1}{5}}$;

${\displaystyle v=}$${\displaystyle 1-10\cdot \frac{1}{5}=}$${\displaystyle -1}$;

тоді ${\displaystyle z-5=}$${\displaystyle 5}$;

${\displaystyle x+2=}$${\displaystyle -1}$:

${\displaystyle z=}$${\displaystyle 10}$;

${\displaystyle x=}$${\displaystyle -3}$;

${\displaystyle z=}$${\displaystyle 10,x=}$${\displaystyle -3}$; відповідь ${\displaystyle z=}$${\displaystyle 10,x=}$${\displaystyle -3}$.