Question

Решите уравнения, сократите дроби и найдите значения

Answer 1

Задание 1.

1. $\frac{42}{x^{2}+5x}-\frac{3}{x^{2}-5x} =\frac{7}{x}$

$\frac{42}{x(x+5)}-\frac{3}{x(x-5)}=\frac{7}{x} | * x$

$\frac{42}{x+5}-\frac{3}{x-5}=7$

$\frac{42(x-5)-3(x+5)}{x^{2}-25}=7$

$\frac{42x-210-3x-15}{x^{2}-25} =7$

$\frac{39x-195}{x^{2}-25} =7 | * (x^{2}-25)$

$39x-195=7(x^{2} -25)$

$39-195=7x^{2} -175$

$7x^{2} -39x+20=0$

$D=b^{2}-4ac=(-39)^{2}-4*7*20=1521-560=961$

$x_{1}=\frac{39-\sqrt{961} }{2*7}=\frac{39-31}{14} =\frac{8}{14}=\frac{4}{7} =0.57$

$x_{2}=\frac{39+\sqrt{961} }{2*7}=\frac{39+31}{14}=\frac{70}{14}=\frac{10}{2}=5$

2. $\frac{x}{x+8}+\frac{x+8}{x-8}=\frac{x^{2}+x+27}{x^{2} -64}$

$\frac{x(x-8)+(x+8)(x-8)}{x^{2}-64}=\frac{x^{2}+x+72}{x^{2}-64}$ |* ${x^{2}-64}$

$\frac{x^{2}-8x+x^{2}+2x+64}{x^{2}-64}=\frac{x^{2}+x+72}{x^{2}-64}$

$2x^{2}-6x+64=x^{2}+x+72$

$x^{2} -7x-8=0$

$D=b^{2}-4ac=(-7^{2}-4*1*(-8)=49+32=81$

$x_{1}=\frac{7-\sqrt{81}}{2*1}=\frac{7-9}{2}=\frac{-2}{2}=-1$

$x_{2}=\frac{7+\sqrt{81}}{2*1}=\frac{7+9}{2}=\frac{16}{2} =8$