Tìm x biết a) (x-3).(x+2)+(x-1).(x+5)=11 b)(3x-2).(x+1)-(x-2)-(3x+5)=32

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1. `a)`

   `(x-3).(x+2)+(x-1).(x+5)=11`

   `⇔x^2-x-6+x^2+4x-5=11`

   `⇔2x^2+3x-22=0`

   `⇔(x-(\sqrt(185)-3)/4)(x-(-\sqrt(185)-3)/4)=0`

   `⇔`\(\left[ \begin{array}{l}x=(\sqrt(185)-3)/4\\x=(-\sqrt(185)-3)/4)\end{array} \right.\) 

   `b)`

   `(3x-2).(x+1)-(x-2)-(3x+5)=32`

   `⇔3x^2+x-2-x+2-3x-5-32=0`

   `⇔3x^2-3x-37=0`

   `⇔(x-(\sqrt(453)+3)/6)(x-(\sqrt(453)-3)/6)=0`

   `⇔`\(\left[ \begin{array}{l}x=(\sqrt(453)+3)/6\\x=(\sqrt(453)-3)/6)\end{array} \right.\) 

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2. Đáp án:

   a) \(\left[ \begin{array}{l}
   x = \dfrac{{ – 3 + \sqrt {185} }}{4}\\
   x = \dfrac{{ – 3 – \sqrt {185} }}{4}
   \end{array} \right.\)

   b) x=12

   Giải thích các bước giải:

   \(\begin{array}{l}
   a)\left( {x – 3} \right).\left( {x + 2} \right) + \left( {x – 1} \right).\left( {x + 5} \right) = 11\\
    \to {x^2} – x – 6 + {x^2} + 4x – 5 = 11\\
    \to 2{x^2} + 3x – 22 = 0\\
    \to 2{x^2} + 2.x\sqrt 2 .\dfrac{3}{{2\sqrt 2 }} + \dfrac{9}{8} – \dfrac{{185}}{8} = 0\\
    \to {\left( {x\sqrt 2  + \dfrac{3}{{2\sqrt 2 }}} \right)^2} = \dfrac{{185}}{8}\\
    \to \left[ \begin{array}{l}
   x\sqrt 2  + \dfrac{3}{{2\sqrt 2 }} = \dfrac{{\sqrt {370} }}{4}\\
   x\sqrt 2  + \dfrac{3}{{2\sqrt 2 }} =  – \dfrac{{\sqrt {370} }}{4}
   \end{array} \right.\\
    \to \left[ \begin{array}{l}
   x = \dfrac{{ – 3 + \sqrt {185} }}{4}\\
   x = \dfrac{{ – 3 – \sqrt {185} }}{4}
   \end{array} \right.\\
   b)\left( {3x – 2} \right).\left( {x + 1} \right) – \left( {x – 2} \right).\left( {3x + 5} \right) = 32\\
    \to 3{x^2} + x – 2 – 3{x^2} + x + 10 = 32\\
    \to 2x – 24 = 0\\
    \to x = 12
   \end{array}\)

   ( câu b t sửa dấu “-” thành dấu nhân nhé thì hợp lý hơn )

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