Tìm x, biết:
\(x ^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy x=2 ; x=-3
\(y^3\left(x-1\right)-7y^3+7xy^3\)
\(=y^3\left(x-1\right)-\left(7y^3-7xy^3\right)\)
\(=y^3\left(x-1\right)-7y^3\left(1-x\right)\)
\(=y^3\left(x-1\right)+7y^3\left(x-1\right)\)
\(=\left(y^3+7y^3\right)\left(x-1\right)\)
\(=y^3\left(1+7\right)\left(x-1\right)\)
x2+x-6=0
=>x2+3x-2x-6=0
=>x(x+3)-2(x+3)=0
=> (x-2)(x+3)=0
=>\(\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
vậy ....
y3(x-1)-7y3+7xy3
=y3(x-1)+(7xy3-7y3)
= y3(x-1)+7y3(x-1)
=(x-1)(y3+7y3)
= (x-1)8y3
