Giải các phương trình sau a, 3x – 1 – 5 ( x + 2 ) = x – 4 b, ( 5x – 2 ) ( 3 – 2x ) = 4×2 — 9 c, 1/x+2 – 3/x-2 = 4x + 10/x2 — 4

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1. Đáp án: a.$ x=-\dfrac73$

                b.$x\in\{-\dfrac17,\dfrac32\}$

                c.$x=-3$

   Giải thích các bước giải:

   a.$3x-1-5(x+2)=x-4$

   $\to -2x-11=x-4$

   $\to -3x=7$

   $\to x=-\dfrac73$

   b.$(5x-2)(3-2x)=4x^2-9$

   $\to 5\cdot \:3x-5\cdot \:2xx-2\cdot \:3+2\cdot \:2x=4x^2-9$

   $\to -10x^2+19x-6=4x^2-9$

   $\to -14x^2+19x+3=0$

   $\to -14x^2-2x+21x+3=0$

   $\to -2x(7x+1)+3(7x+1)=0$

   $\to (-2x+3)(7x+1)=0$

   $\to x\in\{-\dfrac17,\dfrac32\}$

   c.ĐKXĐ: $x\ne \pm2$

   Ta có:
   $\dfrac1{x+2}-\dfrac{3}{x-2}=\dfrac{4x+10}{x^2-4}$

   $\to\dfrac1{x+2}-\dfrac{3}{x-2}=\dfrac{4x+10}{(x-2)(x+2)}$

   $\to \dfrac{1}{x+2}\left(x+2\right)\left(x-2\right)-\dfrac{3}{x-2}\left(x+2\right)\left(x-2\right)=\dfrac{4x+10}{(x-2)(x+2)}\left(x+2\right)\left(x-2\right)$

   $\to x-2-3\left(x+2\right)=2\left(2x+5\right)$

   $\to -2x-8=4x+10$

   $\to -6x=18$

   $\to x=-3$

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