phân tích các đa thức sau thành nhân tử
A = \(9a^2+b^2-6ab-1\)
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`a, x^3 + 4x = x(x^2+4)`
`b, 6ab - 9ab^2 = 3ab(2-b)`
`c, 2a(x-1) + 3b(1-x)`
`= (2a-3b)(x-1)`
`d, (x-y)^2 - x(y-x)`
`= (x-y+x)(x-y)`
`= (2x-y)(x-y)`
a: \(3abc^3-6a^2b^3c+12a^3bc\)
\(=3abc\cdot c^2-3abc\cdot2ab^2+3abc\cdot4a^2\)
\(=3abc\left(c^2-2ab^2+4a^2\right)\)
b: \(27-8y^3\)
\(=3^3-\left(2y\right)^3\)
\(=\left(3-2y\right)\left(9+6y+4y^2\right)\)
c: Sửa đề: \(4x^2+4x-y^2+1\)
\(=\left(4x^2+4x+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+1+y\right)\left(2x+1-y\right)\)
d: \(3a^2\cdot\left(x-2\right)-6ab\cdot\left(2-x\right)\)
\(=3a^2\cdot\left(x-2\right)+6ab\cdot\left(x-2\right)\)
\(=\left(x-2\right)\left(3a^2+6ab\right)\)
\(=3a\left(a+2b\right)\left(x-2\right)\)
a) \(36a^4-y^2=\left(6a^2-y\right)\left(6a^2+y\right)\)
b) \(6x^2+x-2=2x\left(3x+2\right)-1\left(3x+2\right)=\left(3x+2\right)\left(2x-1\right)\)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a
\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)
b
\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)
a: 8x^3-1/125y^3
=(2x)^3-(1/5y)^3
=(2x-1/5y)(4x^2+2/5xy+1/25y^2)
b: =(2y-x)^3
a) \(=\left(2x-1\right)^2\)
b) \(=x\left(y^2-x^2+2x-1\right)=x\left[y^2-\left(x-1\right)^2\right]=x\left(y-x+1\right)\left(y+x-1\right)\)
bằng (3a-b)^2-1
bằng (3a-b+1)(3a-b+1)
\(A=9a^2-6ab+b^2-1\)
\(A=\left(3a-b\right)^2-1\)
\(A=\left(3a-b-1\right)\left(3a-b+1\right)\)
P/s haphuong