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a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
2x3-7x2+5x=0
<=>x*(2x2-7x+5)=0
<=>x*(2x2-2x-5x+5)=0
<=>x*[(2x2-2x)-(5x-5)]=0
<=>x*[2x*(x-1)-5*(x-1)]=0
<=>x*(x-1)*(2x-5)=0
Ta có: \(7x^2-28=0\)
\(\Leftrightarrow7\left(x^2-4\right)=0\)
\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-2\right\}\)
\(\Leftrightarrow x\left(x^2-7x-8\right)=0\\ \Leftrightarrow x\left(x^2-8x+x-8\right)=0\\ \Leftrightarrow x\left(x-8\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\\x=-1\end{matrix}\right.\)
1. (x + 5)2 - (x + 5)(x - 2) = 0
<=> (x + 5 - x + 2)(x + 5) = 0
<=> 7(x + 5) = 0
<=> x + 5 = 0
<=> x = -5
2. x3 + 7x2 + 6x = 0
<=> x3 + x2 + 6x2 + 6x = 0
<=> x2(x + 1) + 6x(x + 1) = 0
<=> (x2 + 6x)(x + 1) = 0
<=> x(x + 6)(x + 1) = 0
<=> \(\left[{}\begin{matrix}x=0\\x+6=0\\x+1=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=0\\x=-6\\x=-1\end{matrix}\right.\)
3. (x + 1)2 - (2x + 3)2 = 0
<=> (x + 1 + 2x + 3)(x + 1 - 2x - 3) = 0
<=> (3x + 4)(-2 - x) = 0
<=> \(\left[{}\begin{matrix}3x+4=0\\-2-x=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\dfrac{-4}{3}\\x=-2\end{matrix}\right.\)
\(7x^2-14x=0\)
\(\Rightarrow7x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7x=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)