\(b,x^3-x^2-x+1=0\)
\(\Rightarrow\left(x^3-x^2\right)-\left(x-1\right)=0\)
\(\Rightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x-1\right)^2\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(c,2x^2-5x-7=0\)
\(\Rightarrow2x^2+2x-7x-7=0\)
\(\Rightarrow\left(2x^2+2x\right)-\left(7x+7\right)=0\)
\(\Rightarrow2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(2x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{2}\end{matrix}\right.\)
a)\(3\left(x-1\right)^2-3x\left(x-5\right)-2=0\)
\(3\left(x^2-2x+1\right)-\left(3x^2-15x\right)-2=0\)
\(3x^2+6x+3-3x^2+15x-2=0\)
\(9x+1=0\)
=>\(9x=1\)=>\(x=\dfrac{-1}{9}\)
Vậy...
b)\(x^3-x^2-x+1=0\)
\(\left(x^3-x^2\right)-\left(x-1\right)=0\)
\(x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right).\left(x^2-1\right)=0\)
\(\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\left(x-1\right)^2\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy...
c)\(2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(\left(2x^2+2x\right)-\left(7x+7\right)=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(2x-7\right)\left(x+1\right)=0\)
=)\(\left[{}\begin{matrix}2x-7=0\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-1\end{matrix}\right.\)
Vậy...
a. 3(x-1)2-3x(x-5)-2=0
=>3(x-1)2=3x(x-5)+2
<=>3x2-6x+3=3x2-15x+2
<=>-6x+3=-15x+2
=>9x=-1
=>x=\(\dfrac{-1}{9}\)
