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ღ kekio ღ

05/08/2017 at 20:41
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1,  \(x^3-25x=0\)
2,  \(2x\left(x-1\right)-3x+3=0\)
3,  \(x^2-36+\left(x^2-12x+36\right)=0\)




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    CTC 05/08/2017 at 21:05

    1) \(x^3-25x=0\)

    \(\Leftrightarrow x\cdot\left(x^2-25\right)=0\)

    \(\Rightarrow x\cdot\left(x-5\right)\cdot\left(x+5\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

    2) \(2x\left(x-1\right)-3x+3=0\)

    \(\Leftrightarrow2x\left(x-1\right)-3\left(x-1\right)=0\)

    \(\Leftrightarrow\left(2x-3\right)\left(x-1\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}2x-3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=1\end{matrix}\right.\)

    3) \(x^2-36+\left(x^2-12x+36\right)=0\)

    \(\Leftrightarrow x^2-36+\left(x-6\right)^2=0\)

    \(\Leftrightarrow\left(x-6\right)\cdot\left(x+6\right)+\left(x-6\right)^2=0\)

    \(\Leftrightarrow\left(x-6\right)\cdot\left[\left(x+6\right)+\left(x-6\right)\right]=0\)

    \(\Leftrightarrow\left(x-6\right)\cdot2x=0\)

    \(\Rightarrow\left[{}\begin{matrix}x-6=0\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=0\end{matrix}\right.\)

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    nguyễn thị thanh mai 06/08/2017 at 16:21
    1,  x3−25x=0

     x^3-25x=0

    X^3-X=0:25

    X^3-X=0

    X.X.X-X=0

    3X-X=0

    X-X=0:3

    X-X=0

    =>X=0


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