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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)
a) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\) (sửa \(\dfrac{x}{2}\rightarrow x^2\))
\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)
\(=x^6-\dfrac{1}{27}\)
b) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)
\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\dfrac{1}{27}x^3+8y^3\)
Lưu ý : Áp dụng hằng đẳng thức đáng nhớ \(a^3\pm b^3=...\)
\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
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 \(N=y^2-x^2\)
c) \(x^6-3x^4y+3x^2y^2-y^3\)
\(=\left(x^2\right)^3-3\cdot\left(x^2\right)^2\cdot y+3\cdot x^2\cdot y^2-y^3\)
\(=\left(x^2-y\right)^3\)
d) \(\left(x-y\right)^3+\left(x-y\right)^2+\dfrac{1}{2}\left(x-y\right)+\dfrac{1}{27}\)
\(=\left(x-y\right)^3+3\cdot\dfrac{1}{3}\cdot\left(x-y\right)^2+3\cdot\left(\dfrac{1}{3}\right)^2\cdot\left(x-y\right)+\left(\dfrac{1}{3}\right)^3\)
\(=\left(x-y+\dfrac{1}{3}\right)^3\)
1.(3x+1)2=9x2+6x+1
2.(2y+1)2=4y2+4y+1
3.(x+1)2=x2+2x+1
4.(4y+1)2=16y2+8y+1