Question

Два уравнения, помогите пожалуйста)

Answer 1

а)

$\displaystyle \tt \frac{3x+4}{x^2-16}=\frac{x^2}{x^2-16}\:\:\:\:\:|\:x\ne-4, \: x\ne4\\\\ \displaystyle \tt 3x+4=x^2\\\displaystyle \tt -x^2+3x+4=0\\\displaystyle \tt x^2-3x-4=0\\\displaystyle \tt D=(-3)^2-4\cdot1\cdot(-4)=9+16=25\\\displaystyle \tt \sqrt{D}=\sqrt{25}=5\\\\ \displaystyle \tt \bold{x_1}=\frac{3+5}{2}=\frac{8}{2}=\bold{4}\\\\ \displaystyle \tt \bold{x_2}=\frac{3-5}{2}=\frac{-2}{2}=\bold{-1}$

$\displaystyle \tt x=4$ - не подходит по ОДЗ

Ответ: $\displaystyle \tt x=-1$

б)

$\displaystyle \tt \frac{3}{x-5}+\frac{8}{x}=2\:\:\:\:\:|\:x\ne5, \: x\ne0\\\\ \displaystyle \tt \frac{3}{x-5}+\frac{8}{x}-2=0\\\\ \displaystyle \tt \frac{3x+8(x-5)-2x(x-5)}{x(x-5)}=0\\\\\displaystyle \tt \frac{3x+8x-40-2x^2+10x}{x(x-5)}=0\\\\\displaystyle \tt \frac{21x-40-2x^2}{x(x-5)}=0\\\\ \displaystyle \tt 21x-40-2x^2=0\\\displaystyle \tt -2x^2+21x-40=0\\\displaystyle \tt 2x^2-21x+40=0\\\displaystyle \tt D=(-21)^2-4\cdot2\cdot40=441-320=121$

$\displaystyle \tt \sqrt{D}=\sqrt{121}=11\\\\ \displaystyle \tt \bold{x_1}=\frac{21+11}{2\cdot2}=\frac{32}{4}=\bold{8}\\\\ \displaystyle \tt \bold{x_2}=\frac{21-11}{2\cdot2}=\frac{10}{5}=\frac{5}{2}=\bold{2,5}$

Ответ: $\displaystyle \tt x_1=8; \: x_2=2,5$