Lần sau ghi tách ra tí bạn ơi ;v
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1. a) \(5x-20y=5\left(x-4y\right)\)
b) \(5\left(x-1\right)-3x\left(x-1\right)=\left(x-1\right)\left(5-3x\right)\)
c) \(x\left(x+1\right)-5x-5=x\left(x+1\right)-5\left(x+1\right)\)
\(=\left(x+1\right)\left(x-5\right)\)
d) \(\left(x+y\right)^2-\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)\left(x+y-x+y\right)\)
\(=4xy\)
e) \(\left(3x+1\right)^2-\left(x+1\right)^2\)
\(=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\)
\(=2x\left(4x+2\right)\)
2. a) \(x+5x^2=0\)
\(\Leftrightarrow x\left(1+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\1+5x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-1}{5}\end{matrix}\right.\)
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b) \(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x+1-x^2-2x-1=0\)
\(\Leftrightarrow-x^2-x=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-x=0\\x+1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
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c) \(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
Vì \(x^2+1>0\Rightarrow x=0\)
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d) \(x^3-0,25x=0\)
\(\Leftrightarrow x\left(x^2-0,25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-0,25=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)
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e) \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
Vậy...
