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\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
Lời giải:
$x^4-x^3-2x-4=(x^4+x^3)-(2x^3+2x^2)+(2x^2+2x)-(4x+4)$
$=x^3(x+1)-2x^2(x+1)+2x(x+1)-4(x+1)$
$=(x+1)(x^3-2x^2+2x-4)$
$=(x+1)[x^2(x-2)+2(x-2)]=(x+1)(x-2)(x^2+2)$
\(=x^3+x^2+x^2+x+x+1=x^2\left(x+1\right)+x\left(x+1\right)+x+1\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
x3 + 2x2 + 2x + 1
= (x3 + 1) + (2x2 + 2x)
= (x + 1)(x2 + x + 1) + 2x(x + 1)
= (x + 1)(x2 + x + 1 + 2x)
= (x + 1)(x2 + 3x + 1)
Chúc bạn học tốt
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
\(x^3+2-2x^2-x=\left(x^3-2x^2\right)-\left(x-2\right)=x^2\left(x-2\right)-\left(x-2\right)=\left(x^2-1\right)\left(x-2\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
`2x^2+x-3`
`=2x^2-2x+3x-3`
`=2x(x-1)+3(x-1)`
`=(x-1)(2x+3)`
\(2x^2+x-3\)
\(=2x^2+3x-2x-3\)
\(=\left(2x+3\right)\left(x-1\right)\)