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![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3x(x + 1) - 5y(x + 1)
= (x + 1)(3x - 5y)
b) 3x(x - 6) - 2(x - 6)
= (x - 6)(3x - 2)
c) 4y(x - 1) - (1 - x)
= 4y(x - 1) + (x - 1)
= (x - 1)(4y + 1)
d) (x - 3)³ + 3 - x
= (x - 3)³ - (x - 3)
= (x - 3)[(x - 3)² - 1]
= (x - 3)(x - 3 - 1)(x - 3 + 1)
= (x - 3)(x - 4)(x - 2)
e) 7x(x - y) - (y - x)
= 7x(x - y) + (x - y)
= (x - y)(7x + 1)
h) 3x³(2y - 3z) - 15x(2y - 3z)²
= (2y - 3z)[3x³ - 15x(2y - 3x)]
= 3x(2y - 3x)[x² - 5(2y - 3x)]
= 3x(2y - 3x)(x² - 10y + 3x)
= 3x(2y - 3x)(x² + 3x - 10y)
k) 3x(x + 2) + 5(-x - 2)
= 3x(x + 2) - 5(x + 2)
= (x + 2)(3x - 5)
l) 18x²(3 + x) + 3(x + 3)
= (x + 3)(18x² + 3)
= 3(x + 3)(6x² + 1)
m) 7x(x - y) - (y - x)
= 7x(x - y) + (x - y)
= (x - y)(7x + 1)
n) 10x(x - y) - 8y(y - x)
= 10x(x - y) + 8y(x - y)
= (x - y)(10x + 8y)
= 2(x - y)(5x + 4y)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,4y\left(x-1\right)-\left(1-x\right)\)
\(=4y\left(x-1\right)+\left(x-1\right)\)
\(=\left(4y+1\right)\left(x-1\right)\)
\(b,18x^2\left(3+x\right)+3\left(x+3\right)\)
\(=\left(18x^2+3\right)\left(3+x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(N=\dfrac{3x^5-4x^4+6x^3}{-2x^2}=-\dfrac{3}{2}x^3+2x^2-3x\)
b: \(N=\dfrac{\left(6x^4y^5-3x^3y^4+\dfrac{1}{2}x^4y^3z\right)}{-\dfrac{1}{3}x^2y^3}=-18x^2y^2+9xy-\dfrac{3}{2}x^2z\)
c: \(\Leftrightarrow N\cdot\left(y-x\right)=\left(x-y\right)^3\)
\(\Leftrightarrow N=\dfrac{\left(x-y\right)^3}{y-x}=-\left(y-x\right)^2\)
d: \(\Leftrightarrow N\cdot\left(y^2-x^2\right)=\left(y^2-x^2\right)^2\)
hay \(N=y^2-x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)
\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)
\(=\dfrac{2y^2+8y+12}{y-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
h) \(=3x\left(2y-3z\right)\left[x^2-5\left(2y-3z\right)\right]=3x\left(2y-3z\right)\left(x^2-10y+15z\right)\)
k) \(=\left(x+2\right)\left(3x-5\right)\)
l) \(=\left(18^2+3\right)\left(x+3\right)=327\left(x+3\right)\)
m) \(=7xy\left(2x-3y+4xy\right)\)
n) \(=2\left(x-y\right)\left(5x-4y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (x - 3)3 + 3 - x
= (x - 3)3 - (x - 3)
= (x - 3)[(x - 3)2 - 1]
= (x - 3)(x - 3 - 1)(x - 3 + 1)
= (x - 3)(x - 4)(x - 2).
b) 33(2y - 3z) - 15x(2y - 3z)2
= 27(2y - 3z) - 15x(2y - 3z)2
= 3(2y - 3z)[9 - 5x(2y - 3z)]
= 3(2y - 3z)(9 - 10xy + 15xz)
d) 18x2(3 + x) + 3(x + 3)
= (x + 3)(18x2 + 3)
= 3(x + 3)(6x2 + 1).
a, (x-3)\(^3\) + 3-x
= (x-3)\(^3\) - (x-3)
= (x-3)*[(x-3)\(^2\) -1]
= (x-3)* ( x\(^2\) - 6x + 9 -1)
= (x-3)*(x\(^2\) - 6x +8)
= (x-3)*[(x\(^2\) -2x) - (4x - 8)]
= (x-3)*[x(x-2) - 4(x-2)]
= (x-3)*(x-2)*(x-4)
b, 3\(^3\)(2y-3z) - 15x(2y-3z)\(^2\)
= (2y-3z) *[3\(^3\) -15x(2y-3z)]
= (2y - 3z)*(27 -30xy +45xz)
= (2y -3z)*3*(9 -10xy +15xz)
= 3*(2y-3z)*(9-10xy+15xz)
c, đề sai
d, 18x\(^2\)(3+x) + 3(x+3)
= 18x\(^2\)(x+3) +3(x+3)
= (x+3)(18x\(^2\) +3)
= (x+3)*3*[(3x)\(^2\) +1)]
= 3(x+3)*[(3x)\(^2\) +1)]