phân tích đa thức thành nhân tử x^3-3x^2+4x-2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
x3-3x2-4x+12=(x3-3x2)-(4x-12)=x2(x-3)-4(x-3)=(x-3)(x2-4)=(x-3)(x-2)(x+2)
\(x^3-3x^2-4x+12\)
\(=\left(x^3-3x^2\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(x^4-4x^3-2x^2-3x+2\)
\(\Leftrightarrow x^4+x^3-5x^3+x^2-5x^2+2x^2-5x+2x+2\)
\(\Leftrightarrow x^4+x^3+x^2-5x^3-5x^2-5x+2x^2+2x+2\)
\(\Leftrightarrow x^2\left(x^2+x+1\right)-5x\left(x^2+x+1\right)+2\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x^2-5x+2\right)\left(x^2+x+1\right)\)
Xin tick ạ !!!
\(x^3-3x^2+4x-2=x^3-x^2-2x^2+2x-2=x^2\left(x-1\right)-2x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+2\right)\)
x3 - 3x2 + 4x - 2
= x3 - x2 - 2x2 + 2x + 2x - 2
= x2( x -1 ) - 2x( x - 1 ) + 2( x - 1 )
= ( x - 1 )( x2 - 2x + 2 )
a: \(2y\left(x+2\right)-3x-6\)
\(=2y\left(x+2\right)-3\left(x+2\right)\)
\(=\left(x+2\right)\left(2y-3\right)\)
b: \(3\left(x+4\right)-x^2-4x\)
\(=3\left(x+4\right)-\left(x^2+4x\right)\)
\(=3\left(x+4\right)-x\left(x+4\right)\)
\(=\left(x+4\right)\left(3-x\right)\)
c: \(2\left(x+5\right)-x^2-4x\)
\(=2x+10-x^2-4x\)
\(=-x^2-2x+10\)
\(=-x^2-2x-1+11\)
\(=11-\left(x^2+2x+1\right)\)
\(=11-\left(x+1\right)^2\)
\(=\left(\sqrt{11}-x-1\right)\left(\sqrt{11}+x+1\right)\)
d: \(x^2+6x-3x-18\)
\(=\left(x^2+6x\right)-\left(3x+18\right)\)
\(=x\left(x+6\right)-3\left(x+6\right)\)
\(=\left(x+6\right)\left(x-3\right)\)
Làm theo kiểu PP số 7 nhé bạn
\(x^3-3x^2+4x-2\)
\(=x^3-x^2-2x^2+2x+2x-2\)
\(=x^2-\left(x-1\right)-2x\left(x-1\right)+2\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+2\right)\)
cái này hay nha
nhưng mik chỉ bik đáp án
\(=\left(x-1\right)\left(x^2-2x+2\right)\)
e) \(=x^2\left(x+1\right)-2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x^2-2x+3\right)\)
g) \(=x^2\left(3x-1\right)-x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(x^2-x+4\right)\)
h) \(=3x^2\left(2x+1\right)-x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(3x^2-x+1\right)\)
i) \(=2x^2\left(2x+1\right)+2x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(2x^2+2x+1\right)\)
\(x^3-3x^2+4x-2\)
\(=x^3-2x^2+2x-1x^2+2x-2\)
\(=x\left(x^2-2x+2\right)-1\left(x^2-2x+2\right)\)
\(=\left(x-1\right)\left(x^2-2x+2\right)\)