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P = x(x - y) - x + y2(x - y) - y2 + 5
P = x - x + y2 - y2 + 5
P = 5
Q = x2(x - y) - x2 + y2(x - y) - y2 + 5(x - y) - 2015
Q = 5 - 2015
Q = -2010
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
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\(a)\left(x+3y\right)\left(x-2y\right)\\ =x^3-2xy+3xy-6y^2\\ =x^2+xy-6y^2\\ b)\left(2x-y\right)\left(y-5x\right)\\ = 2xy-10x^2-y^2+5xy\\ =7xy-10x^2-y^2\\ c)\left(2x-5y\right)\left(y^2-2xy\right)\\ =2xy^2-4x^2y-5y^3+10xy^2\\ =12xy^2-4x^2y-5y^2\\ d)\left(x-y\right)\left(x^2-xy-y^2\right)\\ =x^3-x^2y-xy^2-x^2y+xy^2+y^3\\ =x^3-2x^2y+y^3\)
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\(\frac{5x+y^2}{x^2y}-\frac{5y-x^2}{xy^2}\)
\(=\frac{y\left(5x+y^2\right)-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\frac{5xy+y^3-5xy+x^3}{x^2y^2}\)
\(=\frac{x^3+y^3}{x^2y^2}\)
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\(=\dfrac{y\left(5x+y^2\right)-x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)
\(=\dfrac{y^3+x^3}{x^2y^2}\)
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a) \(-xy\cdot2x^3y^4\cdot-\dfrac{5}{4}x^2y^3\)
\(=\left(-1\cdot2\cdot-\dfrac{5}{4}\right)\cdot\left(x\cdot x^3\cdot x^2\right)\cdot\left(y\cdot y^4\cdot y^3\right)\)
\(=\dfrac{5}{2}x^6y^8\)
Bậc là: \(6+8=14\)
Hệ số: \(\dfrac{5}{2}\)
Biến: \(x^6y^8\)
b) \(5xyz\cdot4x^3y^2\cdot-2x^5y\)
\(=\left(5\cdot4\cdot-2\right)\cdot\left(x\cdot x^3\cdot x^5\right)\cdot\left(y\cdot y^2\cdot y\right)\cdot z\)
\(=-40x^9y^4z\)
Bậc là: \(9+4=13\)
Hệ số: \(-40\)
Biến: \(x^9y^4z\)
c) \(-2xy^5\cdot-x^2y^2\cdot7x^2y\)
\(=\left(-2\cdot-1\cdot7\right)\cdot\left(x\cdot x^2\cdot x^2\right)\cdot\left(y^5\cdot y^2\cdot y\right)\)
\(=14x^6y^8\)
Bậc là: \(6+8=14\)
Hệ số: \(14\)
Biến: \(x^6y^8\)
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a) \(x^2+4x+4-y^2\)
\(=\left(x^2+2.x.2+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(a,=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\\ b=\left(x-2y\right)^2-16=\left(x-2y-4\right)\left(x-2y+4\right)\\ c,=x\left(x^2+2xy+y^2\right)=x\left(x+y\right)^2\\ d,=5\left(x+y\right)-\left(x+y\right)^2=\left(5-x-y\right)\left(x+y\right)\\ e,=x^4\left(x-1\right)+x^2\left(x-1\right)\\ =x^2\left(x^2+1\right)\left(x-1\right)\)
\(\dfrac{5x+y^2}{x^2y}-\dfrac{5y+x^2}{xy^2}\)
\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)
\(=\dfrac{5xy+y^3}{x^2y^2}-\dfrac{5xy+x^3}{x^2y^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)
\(=\dfrac{\left(5xy-5xy\right)+x^3+y^3}{x^2y^2}\)
\(=\dfrac{x^3+y^3}{x^2y^2}\)
\(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}=\dfrac{y\left(5x+y^2\right)}{y\cdot x^2y}-\dfrac{x\left(5y-x^2\right)}{x\cdot xy^2}\)
\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}=\dfrac{x^3+y^3}{x^2y^2}\) \(\left(x,y\ne0\right)\)