Câu hỏi của Ly Le

Question

tìm x biết :

a.$5\left(x+3\right)-6x-2x^2=0$

b.$\left(x-2004\right)=8016x-4x^2$

c.$\left(x+1\right)^2=x+1$

Bùi Mạnh Khôi · Accepted Answer

a ) $5\left(x+3\right)-6x-2x^2=0$

$\Leftrightarrow5\left(x+3\right)-2x\left(3+x\right)=0$

$\Leftrightarrow\left(5-2x\right)\left(x+3\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}5-2x=0\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\x=-3\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\x=-3\end{matrix}\right.$

Vậy ...

b ) $\left(x-2004\right)=8016x-4x^2$

$\Leftrightarrow x-2004=-4x\left(x-2004\right)$

$\Leftrightarrow x-2004+4x\left(x-2004\right)=0$

$\Leftrightarrow\left(x-2004\right)\left(4x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-2004=0\4x+1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2004\4x=-1\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2004\x=-\dfrac{1}{4}\end{matrix}\right.$

Vậy ...

c ) $\left(x+1\right)^2=x+1$

$\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0$

$\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)-1\right]=0$

$\Leftrightarrow\left[{}\begin{matrix}x+1=0\x+1-1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-1\x=0\end{matrix}\right.$

Vậy ...Câu trả lời: a ) $5\left(x+3\right)-6x-2x^2=0$

$\Leftrightarrow5\left(x+3\right)-2x\left(3+x\right)=0$

$\Leftrightarrow\left(5-2x\right)\left(x+3\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}5-2x=0\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\x=-3\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\x=-3\end{matrix}\right.$

Vậy ...

b ) $\left(x-2004\right)=8016x-4x^2$

$\Leftrightarrow x-2004=-4x\left(x-2004\right)$

$\Leftrightarrow x-2004+4x\left(x-2004\right)=0$

$\Leftrightarrow\left(x-2004\right)\left(4x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-2004=0\4x+1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2004\4x=-1\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=2004\x=-\dfrac{1}{4}\end{matrix}\right.$

Vậy ...

c ) $\left(x+1\right)^2=x+1$

$\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)=0$

$\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)-1\right]=0$

$\Leftrightarrow\left[{}\begin{matrix}x+1=0\x+1-1=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=-1\x=0\end{matrix}\right.$

Vậy ...

Yukru · Answer

a) $5\left(x+3\right)-6x-2x^2=0$

$\Rightarrow5\left(x+3\right)-2x\left(3+x\right)=0$

$\Rightarrow\left(x+3\right)\left(5-2x\right)=0$

$\Rightarrow\left[{}\begin{matrix}x+3=0\5-2x=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=-3\x=\dfrac{5}{2}\end{matrix}\right.$

b) $\left(x-2004\right)=8016x-4x^2$

$\Rightarrow\left(x-2004\right)=4x\left(2004-x\right)$

$\Rightarrow\left(x-2004\right)-4x\left(2004-x\right)=0$

$\Rightarrow\left(x-2004\right)+4x\left(x-2004\right)=0$

$\Rightarrow\left(x-2004\right)\left(1+4x\right)=0$

$\Rightarrow\left[{}\begin{matrix}x-2004=0\1+4x=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=2004\x=-\dfrac{1}{4}\end{matrix}\right.$

c) $\left(x+1\right)^2=x+1$

$\Rightarrow\left(x+1\right)^2-\left(x+1\right)=0$

$\Rightarrow\left(x+1\right)\left(x+1-1\right)=0$

$\Rightarrow\left(x+1\right)x=0$

$\Rightarrow\left[{}\begin{matrix}x+1=0\x=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=-1\x=0\end{matrix}\right.$Câu trả lời: a) $5\left(x+3\right)-6x-2x^2=0$

$\Rightarrow5\left(x+3\right)-2x\left(3+x\right)=0$

$\Rightarrow\left(x+3\right)\left(5-2x\right)=0$

$\Rightarrow\left[{}\begin{matrix}x+3=0\5-2x=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=-3\x=\dfrac{5}{2}\end{matrix}\right.$

b) $\left(x-2004\right)=8016x-4x^2$

$\Rightarrow\left(x-2004\right)=4x\left(2004-x\right)$

$\Rightarrow\left(x-2004\right)-4x\left(2004-x\right)=0$

$\Rightarrow\left(x-2004\right)+4x\left(x-2004\right)=0$

$\Rightarrow\left(x-2004\right)\left(1+4x\right)=0$

$\Rightarrow\left[{}\begin{matrix}x-2004=0\1+4x=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=2004\x=-\dfrac{1}{4}\end{matrix}\right.$

c) $\left(x+1\right)^2=x+1$

$\Rightarrow\left(x+1\right)^2-\left(x+1\right)=0$

$\Rightarrow\left(x+1\right)\left(x+1-1\right)=0$

$\Rightarrow\left(x+1\right)x=0$

$\Rightarrow\left[{}\begin{matrix}x+1=0\x=0\end{matrix}\right.$

$\Rightarrow\left[{}\begin{matrix}x=-1\x=0\end{matrix}\right.$

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