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Giải
1) 3xy2 : 5x = \(\frac{3}{5}\)y2
2) 15x4yz3 : 4xyz = \(\frac{15}{4}\)x3z2
3) (4x2y2 - 12xy3 - 7x) : 3x = \(\frac{4}{3}\)xy2 - 4y3 - \(\frac{7}{3}\)
4) (14x4y2 - 12xy3 - x) : 4x = \(\frac{7}{2}\)x3y2 - 3y3 - \(\frac{1}{4}\)
5) (6x2 + 13x - 5) : (2x + 5) = (3x - 1)(2x + 5) : (2x + 5) = 3x - 1
6) (2x4 + x3 - 5x2 - 3x - 3) : (x2 - 3)
= 2x4 + x2 - 6x2 + x3 - 3 - 3x : x2 - 3
= x2(2x2 + x + 1) - 3(2x2 + x + 1) : x2 - 3
= (2x2 + x + 1)(x2 - 3) : x2 - 3
= 2x2 + x + 1
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
1.
= 4x\(^{^{ }2}\)-4x-9x+9
=4x(x-1)-9(x-1)
=(4x-9)(x-1)
Bài 1:
a)
$A=(3x-5)(2x+11)-(2x+3)(3x+7)$
$=6x^2+33x-10x-55-(6x^2+14x+9x+21)$
$=-76$
b)
$B=4x(3x-2)-3x(4x+1)=12x^2-8x-(12x^2-3x)=-5x$
c)
$C=(x+3)(x-3)-(x-1)^2=(x^2-9)-(x^2-2x+1)=2x-10$
Bài 2:
a)
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b)
$x^3-5x^2+x-5=x^2(x-5)+(x-5)=(x^2+1)(x-5)$
c)
$x^2-2xy+y^2-9=(x-y)^2-3^2=(x-y-3)(x-y+3)$
2x^4+x^3-5x^2-3x-3 2x^2+x+1 x^2-3 2x^4 -6x^2 - x^3+x^2-3x-3 x^3 -3x - x^2-3 x^2-3 x^2-3 - 0
a, \(\left(x^3-3x^2+x-3\right):\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2+1\right):\left(x-3\right)\)
\(=x^2+1\)
b, \(\left(x^3-2x^2+5x-10\right):\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+5\right):\left(x-2\right)\)
\(=x^2+5\)
c. \(\left(15x^5y^2+25x^4y^3-30x^3y^2\right):5x^3y^2\)
\(=3x^2+5xy-6\)
d, \(\left(2x^3+5x^2-2x+3\right):\left(2x^2-x+1\right)\)
\(=\left[\left(2x^3-x^2+x\right)+\left(6x^2-3x+3\right)\right]:\left(2x^2-x+1\right)\)
\(=\left(2x^2-x+1\right)\left(x+3\right):\left(2x^2-x+1\right)=x+3\)
a: [x3 - 3x2 +x-3] : [x-3]
Ta có:\(x^3-3x^2+x-3=x^2\left(x-3\right)+\left(x-3\right)=\left(x-3\right)\left(x^2+1\right)\)
\(\Rightarrow\left(x^3-3x^2+x-3\right):\left(x-3\right)=x^2+1\)
b: [x3 - 2x2 + 5x - 10 ] : [x-2]
Tương tự ta có:
\(x^3-2x^2+5x-10=\left(x-2\right)\left(x^2+5\right)\)
\(\Rightarrow\left(x^3-2x^2+5x-10\right):\left(x-2\right)=x^2+5\)