x/(2x ^ 2 - 5x + 3) - (2x + 2)/(2x ^ 2 - 9x + 7) = 5/(2x ^ 2 - 5x + 3) - (x + 7)/(2x ^ 2 - 9x + 7)

tính giúp t với

Đáp án+Giải thích các bước giải:

 $\color{#a8b0b4}{⸻}\color{#91999d}{⸻}\color{#7a8286}{⸸}\color{#636b6f}{⛧}
\color{#4c5356}{𝕯}\color{#434a4d}{ψ}\color{#3b4244}{ภ}\color{#32393b}{ค}
\color{#2a3133}{๓}\color{#21282a}{๏}\color{#1a2123}{ภ}\color{#141a1c}{ร}
\color{#0f1516}{ฬ}\color{#0a1011}{๏}\color{#05090a}{ɭ}\color{#020406}{๔}
\color{#05090a}{⛧}\color{#0a1011}{⸸}\color{#141a1c}{⸻}\color{#1a2123}{⸻}$

`\frac{x}{2x^{2}-5x+3}-\frac{2x+2}{2x^{2}-9x+7}=\frac{5}{2x^{2}-5x+3}-\frac{x+7}{2x^{2}-9x+7}`

Điều kiện: `x\ne1;x\ne 3/2;x\ne7/2`

`\frac{x}{2x^{2}-5x+3}-\frac{5}{2x^{2}-5x+3}=\frac{2x+2}{2x^{2}-9x+7}-\frac{x+7}{2x^{2}-9x+7}`

`\frac{x-5}{2x^{2}-5x+3}=\frac{2x+2-x-7}{2x^{2}-9x+7}`

`\frac{x-5}{2x^{2}-5x+3}-\frac{x-5}{2x^{2}-9x+7}=0`

`(x-5).(\frac{1}{2x^{2}-5x+3}-\frac{1}{2x^{2}-9x+7})=0`

`+)TH1: x-5=0`

`x=5(tmđk)`

`+)TH2:\frac{1}{2x^{2}-5x+3}-\frac{1}{2x^{2}-9x+7}=0`

`\frac{1}{2x^{2}-5x+3}=\frac{1}{2x^{2}-9x+7}`

`2x^{2}-9x+7=2x^{2}-5x+3`

`2x^{2}-2x^{2}-9x+5x=3-7`

`-4x=-4`

`x=1(ktmđk)`

Vậy `S={5}`