400 Вправа 400Алгебра 8 клас Бевз 2025 · § 8 · стор. 80

400. Розв’яжи систему рівнянь.

а) ${\displaystyle \{\begin{array}{rl}& \frac{1}{x}+\frac{1}{y}=\frac{3}{4},\\ & \frac{1}{x}-\frac{1}{y}=\frac{1}{4};\end{array}}$

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

${\displaystyle \{\begin{array}{rl}& u+v=\frac{3}{4},\\ & u-v=\frac{1}{4};\end{array}}$ додамо рівняння:

${\displaystyle 2u=}$${\displaystyle \frac{3}{4}+\frac{1}{4}}$;

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

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

${\displaystyle v=}$${\displaystyle \frac{3}{4}-u=}$${\displaystyle \frac{3}{4}-\frac{1}{2}=}$${\displaystyle \frac{1}{4}}$;

${\displaystyle x=}$${\displaystyle \frac{1}{u}=}$${\displaystyle 2,y=}$${\displaystyle \frac{1}{v}=}$${\displaystyle 4}$; відповідь ${\displaystyle \left(2;4\right)}$;

б) ${\displaystyle \{\begin{array}{rl}& \frac{5}{x}+\frac{2}{y}=2,\\ & \frac{10}{x}-\frac{6}{y}=-1;\end{array}}$

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

${\displaystyle \{\begin{array}{rl}& 5u+2v=2,\\ & 10u-6v=-1;\end{array}}$ перше рівняння помножимо на 3:

${\displaystyle \{\begin{array}{rl}& 15u+6v=6,\\ & 10u-6v=-1;\end{array}}$ додамо:

${\displaystyle 25u=}$${\displaystyle 5}$;

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

${\displaystyle 5\cdot \frac{1}{5}+2v=}$${\displaystyle 2}$;

${\displaystyle 1+2v=}$${\displaystyle 2}$;

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

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

${\displaystyle x=}$${\displaystyle \frac{1}{u}=}$${\displaystyle 5,y=}$${\displaystyle \frac{1}{v}=}$${\displaystyle 2}$; відповідь ${\displaystyle \left(5;2\right)}$.