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\(x^3-3x^2+6x-4\)
\(=x^3-2x^2+4x-x^2+2x-4\)
\(=\left(x^3-2x^2+4x\right)-\left(x^2-2x+4\right)\)
\(=x\left(x^2-2x+4\right)-\left(x^2-2x+4\right)\)
\(=\left(x-1\right)\left(x^2-2x+4\right)\)
x^3 - 3x^2 + 6x - 4
<=> x^3-3x^2+3x-1+3x-3
<=>(x-1)^3+3(x-1)
<=>(x-1)+((x-1)^2+3)
<=>(x-1)+(x^2-2x+4)
\(=x^5-2x^4+x^3-x^4+2x^3-x^2\)
\(=x^3\left(x^2-2x+1\right)-x^2\left(x^2-2x+1\right)\)
\(=\left(x^2-2x+1\right)\left(x^3-x^2\right)\)
\(=\left(x-1\right)^2x^2\left(x-1\right)=\left(x-1\right)^3x^2\)
\(=x^2\left(x^3-1\right)-3x^3\left(x-1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1-3x\right)\)
\(=x^2\left(x-1\right)\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)\left(x-1\right)^2\)
\(=x^2\left(x-1\right)^3\)
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(7xy^5\left(x-1\right)-3x^2y^4\left(1-x\right)+5xy^3\left(x-1\right)\)
\(=7xy^5\left(x-1\right)+3x^2y^4\left(x-1\right)+6xy^3\left(x-1\right)\)
\(=\left(x-1\right)\left(7xy^5+3x^2y^4-6xy^3\right)=xy\left(x-1\right)\left(7y^4+3xy^3-6y^2\right)\)
`a)3x-3a+yx-ya`
`=3(x-a)+y(x-a)`
`=(x-a)(y+3)`
`b)x^2-9-4(x+3)`
`=(x-3)(x+3)-4(x+3)`
`=(x+3)(x-3-4)`
`=(x+3)(x-7)`
\(C=x^3+3x^2-x-3\)
\(C=x^2\left(x+3\right)-\left(x+3\right)\)
\(C=\left(x^2-1\right)\left(x+3\right)\)
\(C=\left(x+1\right)\left(x-1\right)\left(x+3\right)\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
\(x^4-3x^3-x+3\)
\(=x^4-x^3-2x^3+2x-3x+3\)
\(=\)\(x^3\left(x-1\right)-2x\left(x^2-1\right)-3\left(x-1\right)\)
\(=x^3\left(x-1\right)-2x\left(x-1\right)\left(x+1\right)-3\left(x-1\right)\)
\(=\left[x^3-2x\left(x+1\right)-3\right]\left(x-1\right)\)
\(=\left[x^3-2x^2-2x-3\right]\left(x-1\right)\)
\(=\)\(\left[x^3-3x^2+x^2-3x+x-3\right]\left(x-1\right)\)
\(=\left[x^2\left(x-3\right)+x\left(x-3\right)+\left(x-3\right)\right]\left(x-1\right)\)
\(=\left[\left(x-3\right)\left(x^2+x+1\right)\right]\left(x-1\right)\)
\(x^4-3x^3-x+3\)
= \(x\left(x-3\right)-\left(x-3\right)\)
= \(\left(x-1\right)\left(x-3\right)\)