giải phương trình chứa ẩn ở mẫu a) 3x-2/x+7=6x+1/2x-3 h)x+1/x-2-x-1/x+2=2.(x ²+2)/x ²-4 i) 2x+1/x-1=5.(x-1)/x+1 j) x-1/x+2-x/x-2=5x-2/4-x ² giúp em v

0 bình luận về “giải phương trình chứa ẩn ở mẫu a) 3x-2/x+7=6x+1/2x-3 h)x+1/x-2-x-1/x+2=2.(x ²+2)/x ²-4 i) 2x+1/x-1=5.(x-1)/x+1 j) x-1/x+2-x/x-2=5x-2/4-x ² giúp em v”

1. Đáp án:

   $\begin{array}{l}
   a)\dfrac{{3x – 2}}{{x + 7}} = \dfrac{{6x + 1}}{{2x – 3}}\\
   Dk:x \ne  – 7;x \ne \dfrac{3}{2}\\
    \Rightarrow \left( {3x – 2} \right).\left( {2x – 3} \right) = \left( {x + 7} \right)\left( {6x + 1} \right)\\
    \Rightarrow 6{x^2} – 9x – 4x + 6 = 6{x^2} + x + 42x + 7\\
    \Rightarrow 43x + 13x = 6 – 7\\
    \Rightarrow 56x =  – 1\\
    \Rightarrow x =  – \dfrac{1}{{56}}\left( {tmdk} \right)\\
   \text{Vậy}\,x = \dfrac{{ – 1}}{{56}}\\
   h)\dfrac{{x + 1}}{{x – 2}} – \dfrac{{x – 1}}{{x + 2}} = \dfrac{{2\left( {{x^2} + 2} \right)}}{{{x^2} – 4}}\\
   Dkxd:x \ne 2;x \ne  – 2\\
    \Rightarrow \dfrac{{\left( {x + 1} \right)\left( {x + 2} \right) – \left( {x – 1} \right)\left( {x – 2} \right)}}{{\left( {x + 2} \right)\left( {x – 2} \right)}}\\
    = \dfrac{{2{x^2} + 4}}{{\left( {x + 2} \right)\left( {x – 2} \right)}}\\
    \Rightarrow {x^2} + 3x + 2 – \left( {{x^2} – 3x + 2} \right) = 2{x^2} + 4\\
    \Rightarrow 6x = 2{x^2} + 4\\
    \Rightarrow {x^2} – 3x + 2 = 0\\
    \Rightarrow \left( {x – 1} \right)\left( {x – 2} \right) = 0\\
    \Rightarrow \left[ \begin{array}{l}
   x = 1\left( {tm} \right)\\
   x = 2\left( {ktm} \right)
   \end{array} \right.\\
   \text{Vậy}\,x = 1\\
   j)\dfrac{{2x + 1}}{{x – 1}} = \dfrac{{5\left( {x – 1} \right)}}{{x + 1}}\\
   Dkxd:x \ne 1;x \ne  – 1\\
    \Rightarrow \left( {2x + 1} \right)\left( {x + 1} \right) = 5\left( {x – 1} \right)\left( {x – 1} \right)\\
    \Rightarrow 2{x^2} + 3x + 1 = 5\left( {{x^2} – 2x + 1} \right)\\
    \Rightarrow 3{x^2} – 13x + 4 = 0\\
    \Rightarrow \left( {3x – 1} \right)\left( {x – 4} \right) = 0\\
    \Rightarrow \left[ \begin{array}{l}
   x = \dfrac{1}{3}\left( {tm} \right)\\
   x = 4\left( {tm} \right)
   \end{array} \right.\\
   \text{Vậy}\,x = \dfrac{1}{3};x = 4\\
   j)\dfrac{{x – 1}}{{x + 2}} – \dfrac{x}{{x – 2}} = \dfrac{{5x – 2}}{{4 – {x^2}}}\\
   DKxd:x \ne 2;x \ne  – 2\\
    \Rightarrow \dfrac{{\left( {x – 1} \right)\left( {x – 2} \right) – x\left( {x + 2} \right)}}{{\left( {x – 2} \right)\left( {x + 2} \right)}} = \dfrac{{2 – 5x}}{{\left( {x – 2} \right)\left( {x + 2} \right)}}\\
    \Rightarrow {x^2} – 3x + 2 – {x^2} – 2x = 2 – 5x\\
    \Rightarrow 2 – 5x = 2 – 5x\\
   \text{Vậy phương trình đúng với mọi x#2;x#-2}
   \end{array}$

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