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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A = {x^2} + 2y - 2x - xy \\= \left( {{x^2} - 2x} \right) + \left( {2y - xy} \right) \\= x\left( {x - 2} \right) + y\left( {2 - x} \right) \\= x\left( {x - 2} \right) - y\left( {x - 2} \right) \\= (x-y)(x-2) \)
Thay \(x = 2022,y = 2020\) vào A ta được:
\(A = (2022 - 2020)(2022-2) = 2.2020 = 4040\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đây nhé ta thêm bớt:
\(x^2+xy+y^2=x^2+y^2+2xy-xy=\left(x+y\right)^2-xy=\left(-2\right)^2-xy=4-xy\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](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)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)
\(=\dfrac{x\left(x-2\right)-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-2x-x^2-2x}{\left(x-2\right)\left(x+2\right)}=-\dfrac{4x}{x^2-4}\)
Bài 2:
1: \(x^2y^2-8-1\)
\(=x^2y^2-9\)
\(=\left(xy-3\right)\left(xy+3\right)\)
2: \(x^3y-2x^2y+xy-xy^3\)
\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)
\(=xy\left(x^2-2x+1-y^2\right)\)
\(=xy\left[\left(x-1\right)^2-y^2\right]\)
\(=xy\left(x-1-y\right)\left(x-1+y\right)\)
3: \(x^3-2x^2y+xy^2\)
\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)
\(=x\left(x^2-2xy+y^2\right)=x\left(x-y\right)^2\)
4: \(x^2+2x-y^2+1\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1+y\right)\left(x+1-y\right)\)
5: \(x^2+2x-4y^2+1\)
\(=\left(x^2+2x+1\right)-4y^2\)
\(=\left(x+1\right)^2-4y^2\)
\(=\left(x+1-2y\right)\left(x+1+2y\right)\)
6: \(x^2-6x-y^2+9\)
\(=\left(x^2-6x+9\right)-y^2\)
\(=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(2\left(x-1\right)^3-\left(x-1\right)=\left(x-1\right)\left(2\left(x-1\right)^2-1\right)\)
2) \(y\left(x-2y\right)^2+xy^2\left(2y-x\right)=\left(2y-x\right)\left(2\left(2y-x\right)+1\right)=\left(2y-x\right)\left(4y-2x+1\right)\)
3) \(xy\left(x+y\right)-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\) (xem lại đề sửa -2x thành -x mới đúng)
4) \(xy\left(x-3y\right)-2x+6y=xy\left(x-3y\right)-2\left(x-3y\right)=\left(x-3y\right)\left(xy-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`x^2 + xy - 2x - 2y`
`= (x^2 - 2x) + (xy - 2y)`
`= x(x - 2) + y(x - 2)`
`= (x + y)(x - 2)`
____
`x^2 - xy - 6x + 6y`
`= (x^2 - 6x) - (xy - 6y)`
`= x(x - 6) - y(x - 6)`
`= (x - y)(x - 6)`
____
`5xy^2 - 5x + y^2 - 1`
`= (5xy^2 + y^2) - (5x + 1)`
`= y^2(5x + 1) - (5x + 1)`
`= (y^2 - 1)(5x + 1)`
`= (y - 1)(y + 1)(5x + 1)`
a: =(x^2+xy)-(2x+2y)
=x(x+y)-2(x+y)
=(x+y)(x-2)
b: =(x^2-xy)-(6x-6y)
=x(x-y)-6(x-y)
=(x-y)(x-6)
c: =5xy^2+y^2-5x-1
=y^2(5x+1)-(5x+1)
=(5x+1)(y^2-1)
=(5x+1)(y+1)(y-1)
giúp em vớiiiiiii
vì (x+y)^2-xy=x^2+y^2+2xy-xy=x^2+y^2+xy