Làm hộ mik bài 4, 5 zớii :<

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Làm hộ mik bài 4, 5 zớii :<
lam-ho-mik-bai-4-5-zoii

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Khoii Minh 1 year 2020-10-14T04:52:33+00:00 1 Answers 82 views 0

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    2020-10-14T04:53:41+00:00

    $Bài4_{}$

    $a)x^2-4=0_{}$

    $⇔(x-2)(x+2)=0_{}$

    $⇔x=±2_{}$

    $b)_{}$ $x^2+10x+16=0_{}$

    $⇔x^2+8x+2x+16=0_{}$

    $⇔x.(x+8)+2(x+8)_{}$

    $⇔(x+8).(x+2)=0_{}$

    $⇔_{}$ \(\left[ \begin{array}{l}x+8=0\\x+2=0\end{array} \right.\)$⇔_{}$  \(\left[ \begin{array}{l}x=-8\\x=-2\end{array} \right.\) 

    $c)_{}$ $(x^2+2x)^2-x^2-2x-2=0_{}$

    $⇔x^4+4x^3+4x^2-x^2-2x-2=0_{}$

    $⇔x^4+x^3+3x^3+3x^2-2x-2=0_{}$

    $⇔x^3(x+1)+3x^2(x+1)-2(x=1)=0_{}$

    $⇔(x+1)(x^3+3x^2-2)=0_{}$

    $⇔(x+1)(x^3+x^2+2x^2-2+2x-2x)=0_{}$

    $⇔(x+1)[ x^2(x+1)+2x(x+1)-2(x+1)]=0 _{}$

    $⇔(x+1)(x+1)(x^2+2x-2)=0_{}$

    $⇔(x+1)^2.(x^2+2x-2)=0_{}$

    $⇔_{}$ \(\left[ \begin{array}{l}(x+1)^2=0\\x^2+2x-2=0\end{array} \right.\) $⇔_{}$  \(\left[ \begin{array}{l}x=-1\\x=-1±\sqrt{3}\end{array} \right.\) 

    #Xin hay nhất ạ

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