Phân tích đa thức thành nhân tử:
a) 12x2-3xy-8xz+2yz
b) (x2+x+1) (x2+x+2)-12
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Phân tích đa thức thành nhân tử:
a) (x-1)(x-2)(x-3)(x-4)+1
b) (x2+3x+2)(x2+7x+12)+1
c) 12x2-3xy-8xz+2yz
a) \(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+1\)
\(A=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]+1\)
\(A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)
Đặt \(a=x^2-5x+5\)
\(\Leftrightarrow A=\left(a-1\right)\left(a+1\right)+1\)
\(\Leftrightarrow A=a^2-1^2+1\)
\(\Leftrightarrow A=a^2\)
Thay \(a=x^2-5x+5\)vào A ta có :
\(A=\left(x^2-5x+5\right)^2\)
b) \(B=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1\)
\(B=\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)+1\)
\(B=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]+1\)
\(B=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
Làm tương tự câu a)
c) \(12x^2-3xy-8xz+2yz\)
\(=3x\left(4x-y\right)-2z\left(4x-y\right)\)
\(=\left(4x-y\right)\left(3x-2z\right)\)
\(1,\\ 12x^6y^3:4x^3y=3x^3y^2\\ \left(x+1\right)\left(x^2-x+1\right)=x^3+1\\ 2x^2y\left(x^2+3xy\right)=3x^4y+6x^3y^2\\ 2,\\ a,=2xy\left(2x+3y-4\right)\\ b,=\left(x-3\right)\left(x+y\right)\\ c,=\left(x-2\right)\left(x+2\right)+y\left(x-2\right)=\left(x+y+2\right)\left(x-2\right)\\ d,=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\\ 3,\\ a,\Leftrightarrow x^2-x^2+2x=2\\ \Leftrightarrow2x=2\Leftrightarrow x=1\\ b,\Leftrightarrow\left(x-2\right)\left(x-2+1\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
\(=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
\(=\left(x-1-y+z\right)\left(x-1+y-z\right)\)
\(x^2-2x+1-y^2+2yz-z^2\)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
\(=\left(x-1-y+z\right)\left(x-1+y-z\right)\)
a, \(12x^2-3xy-8xz+2yz=3x\left(4x-y\right)-2z\left(4x-y\right)=\left(4x-y\right)\left(3x-2z\right)\)
b: =(x^2+x)^2+3(x^2+x)+2-12
=(x^2+x)^2+3(x^2+x)-10
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)