Tim x biêt a,3x+2 (5-x)=0 b,x (2x-1)(x+5)-(2x^+1)x+4,5)=3,5 c,(x-2)^-(x-3)(x+3)=6 d,4 (x-3)^-(2x-1)(2x+1)=10

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1. Giải thích các bước giải:

   Ta có:

   \(\begin{array}{l}
   a,\\
   3x + 2\left( {5 – x} \right) = 0\\
    \Leftrightarrow 3x + 10 – 2x = 0\\
    \Leftrightarrow x + 10 = 0\\
    \Leftrightarrow x =  – 10\\
   b,\\
   x\left( {2x – 1} \right)\left( {x + 5} \right) – \left( {2{x^2} + 1} \right)x + 4,5 = 3,5\\
    \Leftrightarrow x\left( {2{x^2} + 9x – 5} \right) – \left( {2{x^2} + 1} \right)x – 1 = 0\\
    \Leftrightarrow x\left[ {2{x^2} + 9x – 5 – 2{x^2} – 1} \right] – 1 = 0\\
    \Leftrightarrow x\left( {9x – 6} \right) – 1 = 0\\
    \Leftrightarrow 9{x^2} – 6x – 1 = 0\\
    \Leftrightarrow x = \frac{{1 \pm \sqrt 2 }}{3}\\
   c,\\
   {\left( {x – 2} \right)^2} – \left( {x – 3} \right)\left( {x + 3} \right) = 6\\
    \Leftrightarrow \left( {{x^2} – 4x + 4} \right) – \left( {{x^2} – 9} \right) = 6\\
    \Leftrightarrow  – 4x + 13 = 6\\
    \Leftrightarrow 4x = 7\\
    \Leftrightarrow x = \frac{7}{4}\\
   d,\\
   4{\left( {x – 3} \right)^2} – \left( {2x – 1} \right)\left( {2x + 1} \right) = 10\\
    \Leftrightarrow 4\left( {{x^2} – 6x + 9} \right) – \left( {4{x^2} – 1} \right) = 10\\
    \Leftrightarrow  – 24x + 27 = 0\\
    \Leftrightarrow x = \frac{{27}}{{24}} = \frac{9}{8}
   \end{array}\)

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