a, Tính (4x 3 y 2 – 3xy 3 + 6x 2 y 2 ) : 2xy 2
b, Tìm m để 2x 3 + x 2 – x + m – 2 chia hết cho x + 2
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Bài 3:
A(x)⋮B(x)
=>\(3x^2+5x+m\) ⋮x-2
=>\(3x^2-6x+11x-22+m+22\) ⋮x-2
=>m+22=0
=>m=-22
Bài 2:
a: \(2x^3-8x^2+8x\)
\(=2x\left(x^2-4x+4\right)\)
\(=2x\left(x-2\right)^2\)
b: 2xy+2x+yz+z
=2x(y+1)+z(y+1)
=(y+1)(2x+z)
c: \(x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
=(x+1-y)(x+1+y)
Câu 1:
a:\(\left(4x-1\right)\left(2x^2-x-1\right)\)
\(=8x^3-4x^2-4x-2x^2+x+1\)
\(=8x^3-6x^2-3x+1\)
b: \(\left(4x^3+8x^2-2x\right):2x\)
\(=\frac{4x^3}{2x}+\frac{8x^2}{2x}-\frac{2x}{2x}\)
\(=2x^2+4x-1\)
c: \(\left(6x^3-7x^2-16x+12\right):\left(2x+3\right)\)
\(=\left(6x^3+9x^2-16x^2-24x+8x+12\right):\left(2x+3\right)\)
\(=\left\lbrack3x^2\left(2x+3\right)-8x\left(2x+3\right)+4\left(2x+3\right)\right\rbrack:\left(2x+3\right)\)
\(=3x^2-8x+4\)
c) Ta có: \(P=x^3+y^3+6xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+6xy\)
\(=\left(x+y\right)^3-3xy\left(x+y-2\right)\)
\(=2^3=8\)
a) \(3xy-6xy^2=3xy\left(1-2y\right)\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(x^3-x^2+2\)
d) \(x^2+4x+4-y^2=\left(x^2+4x+4\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
e) \(x^3+4x^2+4x=x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)
f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)
g) \(6x^2-12x=6x\left(x-2\right)\)
h) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
\(a,=2x^2-\dfrac{3}{2}y+3x\)
\(b,\)bt để chia hết cho x+2 là:\(2x^3+x^2-x+10\)
\(\Rightarrow m=12\)