$\frac{12}{x^2-4}$ – $\frac{x+1}{x-2}$ + $\frac{x+7}{x+2}$ = 0
giúp mk vshhuhu
$\frac{12}{x^2-4}$ – $\frac{x+1}{x-2}$ + $\frac{x+7}{x+2}$ = 0 giúp mk vshhuhu
By Quinn
By Quinn
$\frac{12}{x^2-4}$ – $\frac{x+1}{x-2}$ + $\frac{x+7}{x+2}$ = 0
giúp mk vshhuhu
$\text{Đáp án + Giải thích các bước giải:}$
`(12)/(x^{2}-4)-(x+1)/(x-2)+(x+7)/(x+2)=0` `(ĐKXĐ:x\ne±2)`
`<=>(12)/((x-2)(x+2))-((x+1)(x+2))/((x-2)(x+2))+((x+7)(x-2))/((x+2)(x-2))=(0)/((x-2)(x+2)`
`=>12-(x+1)(x+2)+(x+7)(x-2)=0`
`<=>12-(x^{2}+x+2x+2)+(x^{2}+7x-2x-14)=0`
`<=>12-(x^{2}+3x+2)+x^{2}+5x-14=0`
`<=>12-x^{2}-3x-2+x^{2}+5x-14=0`
`<=>2x-4=0`
`<=>2x=4`
`<=>x=2(KTM)`
`\text{Vậy phương trình vô nghiệm}`
ĐKXĐ: \(x^2-4\ne 0↔x^2\ne 4↔x\ne \pm 2\)
\(\dfrac{12}{x^2-4}-\dfrac{x+1}{x-2}+\dfrac{x+7}{x+2}=0\\↔\dfrac{12}{(x+2)(x-2)}-\dfrac{(x+1)(x+2)}{(x-2)(x+2)}+\dfrac{(x+7)(x-2)}{(x-2)(x+2)}=0\\→12-(x^2+3x+2)+(x^2+5x-14)=0\\↔12-x^2-3x-2+x^2+5x-14=0\\↔-4+(x^2-x^2)+(-3x+5x)=0\\↔2x=4\\↔x=2(KTM)\)
Vậy pt vô nghiệm \(S=\varnothing\)