Bài 1:
a: \(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
b: \(3x^2+9x-30\)
\(=3\left(x^2+3x-10\right)\)
\(=3\left(x^2+5x-2x-10\right)\)
\(=3\left[x\left(x+5\right)-2\left(x+5\right)\right]\)
\(=3\left(x+5\right)\left(x-2\right)\)
c: \(2x^2+5x+2\)
\(=2x^2+4x+x+2\)
\(=2x\left(x+2\right)+\left(x+2\right)\)
\(=\left(x+2\right)\left(2x+1\right)\)
Bài 2:
a: \(x^2\left(x-5\right)+5-x=0\)
=>\(x^2\left(x-5\right)-\left(x-5\right)=0\)
=>\(\left(x-5\right)\left(x^2-1\right)=0\)
=>\(\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}x-5=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\\x=-1\end{matrix}\right.\)
b: \(\left(x+3\right)\left(x^2-3x+5\right)=x^2+3x\)
=>\(\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-3x+5-x\right)=0\)
=>\(\left(x+3\right)\left(x^2-4x+5\right)=0\)
=>x+3=0(Do \(x^2-4x+5=\left(x-2\right)^2+1>0\forall x\))
=>x=-3
Bạn phải làm 1 2 bài j đấy đi chứ giúp hết thì bạn học làm cái j?
