Question

[Nguyễn Dương Ánh Hiền]

\(\frac{2}{x+2}-\frac{2x^2+16}{x^3+8}=\frac{5}{x^2-2x+4}\)

\(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)

\(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)

Answer 1

[Edogawa Conan]

a) Đk: x \(\ne\)-2

Ta có: \(\frac{2}{x+2}-\frac{2x^2+16}{x^2+8}=\frac{5}{x^2-2x+4}\)

<=> \(\frac{2\left(x^2-2x+4\right)-\left(2x^2+16\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{5\left(x+2\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)

<=> 2x² - 4x + 8 - 2x² - 16 = 5x + 10

<=> -4x - 8 = 5x + 10

<=> -4x - 5x = 10 + 8

<=> -9x = 18

<=> x = -2 (ktm)

=> pt vô nghiệm

b) Đk: x \(\ne\)2; x \(\ne\)-3

Ta có: \(\frac{1}{x-2}-\frac{6}{x+3}=\frac{5}{6-x^2-x}\)

<=> \(\frac{x+3}{\left(x-2\right)\left(x+3\right)}-\frac{6\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{5}{\left(x-2\right)\left(x+3\right)}\)

<=> x + 3 - 6x + 12 = -5

<=> -5x = -5 - 15

<=> -5x = -20

<=> x = 4 

vậy S = {4}

c) Đk: x \(\ne\)8; x \(\ne\)9; x \(\ne\)10; x \(\ne\)11

Ta có: \(\frac{8}{x-8}+\frac{11}{x-11}=\frac{9}{x-9}+\frac{10}{x-10}\)

<=> \(\left(\frac{8}{x-8}+1\right)+\left(\frac{11}{x-11}+1\right)=\left(\frac{9}{x-9}+1\right)+\left(\frac{10}{x-10}+1\right)\)

<=> \(\frac{x}{x-8}+\frac{x}{x-11}-\frac{x}{x-9}-\frac{x}{x-10}=0\)

<=> \(x\left(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\right)=0\)

<=> x = 0 (vì \(\frac{1}{x-8}+\frac{1}{x-11}-\frac{1}{x-9}-\frac{1}{x-10}\ne0\)

Vậy S = {0}