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Bài 3. (1,5 điểm) Cho \( P = \frac{5x - 2}{x^2 - 4} - \frac{3}{x + 2} + \frac{x}{x - 2} \)

a) Rút gọn biểu thức \( P \)

b) Tìm \( x \) biết \( P = \frac{4}{3} \)

Answer 1

\(P=\dfrac{5x-2}{x^2-4}-\dfrac{3}{x+2}+\dfrac{x}{x-2}\\ P=\dfrac{5x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{3}{x+2}+\dfrac{x}{x-2}\\ P=\dfrac{5x-2-3\cdot\left(x-2\right)+x\cdot\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\\ P=\dfrac{5x-2-3x+6+x^2+2x}{\left(x+2\right)\left(x-2\right)}\\ P=\dfrac{x^2+4x+4}{\left(x-2\right)\left(x+2\right)}\\ P=\dfrac{\left(x+2\right)^2}{\left(x+2\right)\left(x-2\right)}\\ P=\dfrac{x+2}{x-2}\)

vì P = \(\dfrac{4}{3}\) nên:

\(\dfrac{x+2}{x-2}=\dfrac{4}{3}\\ \Rightarrow3\cdot\left(x+2\right)=4\cdot\left(x-2\right)\\ \Rightarrow3x+6=4x-8\\ \Rightarrow8+6=4x-3x\\ \Rightarrow x=14\)

Answer 2

`a) P = ( 5x-2)/(x^2 -4) - 3/(x+2) + x/(x-2)` ( ĐKXĐ\(x\ne\pm2\))

`=> P = (5x-2)/[(x-2)(x+2)] - [3(x-2)]/[(x-2)(x+2)] + [x*(x+2)]/[(x-2)(x+2)]`

`=> P = (5x-2 - 3x + 6 + x^2 + 2x)/[(x-2)(x+2)]`

`=> P = (x^2 +4x+4)/[(x-2)(x+2)]`

`=> P = [(x+2)^2]/[(x-2)(x+2)]`

`=> P = (x+2)/(x-2)`

Vậy ` P = (x+2)/(x-2)`

`b)` Có : `P =4/3`

`=> (x+2)/(x -2) =4/3`

`=> 3(x+2) = 4(x-2)`

`=> 3x + 6 = 4x - 8`

`=> 3x - 4x = -6 -8`

`=> -x = -14`

`=> x =14` (thỏa mãn ĐKXĐ)`

Vậy `x=14`

Answer 3

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