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a) (x + 3y) (2x2y - 6xy2)
= (x + 3y) + 2xy (x - 3y)
= 2xy [(x + 3y) (x - 3y)]
= 2xy (x2 - 3y2)
b) (6x5y2 - 9x4y3 + 15x3y4) : 3x3y2
= (6x5y2 : 3x3y2) + (-9x4y3 : 3x3y2) + (15x3y4 : 3x3y2)
= [(6 : 3) (x5 : x3) (y2 : y2)] + [(-9 : 3) (x4 : x3) (y3 : y2)] + [(15 : 3) (x3 : x3) (y4 : y2)]
= 2x2 + (-3xy) + 5y2
= 2x2 - 3xy + 5y2
#Học tốt!!!
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
f: \(=\dfrac{2x^3-10x^2-11x^2+55x+12x-60}{x-5}=2x^2-11x+12\)
a) Ta có: \(4x^2-6x\)
\(=2x\left(2x-3\right)\)
b) Ta có: \(9x^4y^3+3x^2y^4\)
\(=3x^2y^3\left(3x^2+y\right)\)
c) Ta có: 3(x-y)-5x(y-x)
=3(x-y)+5x(x-y)
=(x-y)(3+5x)
d) Ta có: \(x^3-2x^2+5x\)
\(=x\left(x^2-2x+5\right)\)
e) Ta có: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=\left(x+3y\right)\left(5-15x\right)\)
\(=5\left(x+3y\right)\cdot\left(1-3x\right)\)
f) Ta có: \(2x^2\left(x+1\right)+4\left(x+1\right)\)
\(=\left(x+1\right)\left(2x^2+4\right)\)
\(=2\left(x+1\right)\left(x^2+2\right)\)
a) Ta có: \(\dfrac{1-x}{x^2-2x+1}+\dfrac{x+1}{x-1}\)
\(=\dfrac{1-x}{\left(x-1\right)^2}-\dfrac{x+1}{1-x}\)
\(=\dfrac{1-x}{\left(1-x\right)^2}-\dfrac{x+1}{1-x}\)
\(=\dfrac{1-x-1}{1-x}=\dfrac{-x}{1-x}=\dfrac{x}{x-1}\)
b) Ta có: \(\dfrac{2x}{3y^4z}\cdot\left(-\dfrac{4y^2z}{5x}\right)\cdot\left(-\dfrac{15y^3}{8xz}\right)\)
\(=\dfrac{2x\cdot4y^2z\cdot15y^3}{3y^4z\cdot5x\cdot8xz}\)
\(=\dfrac{120xy^5z}{120x^2y^4z^2}=\dfrac{y}{xz}\)
a) (4x-1)(2-x)-(2x-1)2
= 8x-4x2-2+x-(4x2-4x+1) = -8x2+13x-3
b) (15x4y5-30x3y4+35x3y4):(5x3y3)
= 3xy2-6y+7y = 3xy2+y
a: \(=8x-4x^2-2+2x-4x^2+4x-1\)
\(=-8x^2+14x-3\)