401 Вправа 401Алгебра 8 клас Бевз 2025 · § 8 · стор. 80

401. Розв’яжи систему рівнянь.

а) ${\displaystyle \{\begin{array}{rl}& \frac{5}{x}+\frac{4}{y}=51,\\ & \frac{1}{x}-\frac{8}{y}=8;\end{array}}$

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

${\displaystyle \{\begin{array}{rl}& 5u+4v=51,\\ & u-8v=8;\end{array}\{\begin{array}{rl}& 5u+4v=51,\\ & u=8+8v;\end{array}}$;

${\displaystyle 5(8+8v)+4v=}$${\displaystyle 51}$;

${\displaystyle 40+40v+4v=}$${\displaystyle 51}$;

${\displaystyle 44v=}$${\displaystyle 51-40}$;

${\displaystyle 44v=}$${\displaystyle 11}$;

${\displaystyle v=}$${\displaystyle \frac{1}{4}}$;

${\displaystyle u=}$${\displaystyle 8+8\cdot \frac{1}{4}=}$${\displaystyle 10}$;

${\displaystyle x=}$${\displaystyle \frac{1}{10}=}$${\displaystyle 0,1,y=}$${\displaystyle 4}$; відповідь ${\displaystyle \left(0,1;4\right)}$;

б) ${\displaystyle \{\begin{array}{rl}& \frac{4}{x}+\frac{2}{y}=4,\\ & \frac{6}{x}-\frac{4}{y}=-1;\end{array}}$

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

${\displaystyle \{\begin{array}{rl}& 4u+2v=4,\\ & 6u-4v=-1;\end{array}}$ перше рівняння помножимо на 2:

${\displaystyle \{\begin{array}{rl}& 8u+4v=8,\\ & 6u-4v=-1;\end{array}}$ додамо:

${\displaystyle 14u=}$${\displaystyle 7}$;

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

${\displaystyle 4\cdot \frac{1}{2}+2v=}$${\displaystyle 4}$;

${\displaystyle 2+2v=}$${\displaystyle 4}$;

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

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

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