`(x+2)/(x-2)-(x-1)/x=8/(x^2-2x)`
`đk:x ne 0,x ne 2`
Nhân 2 vế cho `x(x-2)` ta có:
`x(x+2)-(x-1)(x-2)=8`
`<=>x^2+2x-(x^2-3x+2)=8`
`<=>x^2+2x-x^2+3x-2=8`
`<=>5x-2=8`
`<=>5x=10`
`<=>x=2(l` do `x ne 2)`
Vậy pt vô nghiệm
\(\dfrac{x+2}{x-2}\) - \(\dfrac{x-1}{x}\) = \(\dfrac{8}{x^2-2x}\)
\(\dfrac{x+2}{x-2}\) -\(\dfrac{x-1}{x}\) = \(\dfrac{8}{x\left(x-2\right)}\)
Đkxđ : x ≠ 0
x - 2 ≠ 0
\(\Rightarrow\) x ≠ 2
MTC : x (x - 2)
Qiuy đồng mẫu thức hai vế của phương trình :
\(\Rightarrow\) \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}\) - \(\dfrac{\left(x-1\right)\left(x-2\right)}{x\left(x-2\right)}\) = \(\dfrac{8}{x\left(x-2\right)}\)
Suy ra : x (x + 2) - (x - 1) (x - 2) = 8
\(\Leftrightarrow\) x2 + 2x - (x2 - 3x + 2) = 8
\(\Leftrightarrow\) x2 + 2x - x2 + 3x - 2 - 8 = 0
\(\Leftrightarrow\) 5x - 10 = 0
\(\Leftrightarrow\) 5x = 10
\(\Leftrightarrow\) x = \(\dfrac{10}{5}=2\)
Vậy S = \(\left\{2\right\}\)
Chúc bạn học tốt