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2x2 + 2y2 = 5xy
=> 2x2 + 2y2 - 5xy = 0
=> (x - 2y)(2x - y) = 0
x = 2y (loại)
y = 2x
E = \(\dfrac{x+2x}{x-2x}\)=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
(2\(x^3\) - 2\(xy\)) - (\(x^2\) +5\(xy\) - \(x^2\) - \(x^3\))
= 2\(x^3\) - 2\(xy\) - (5\(xy\) - \(x^3\))
= 2\(x^3\) - 2\(xy\) - 5\(xy\) + \(x^3\)
= 3\(x^3\) - 7\(xy\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,C=A+B\\ =4x^2+3y^2-5xy+3x^2+2y^2+2x^2y^2\\ =\left(4x^2+3x^2\right)+\left(3y^2+2y^2\right)-5xy+2x^2y^2\\ =7x^2+5y^{^2}-5xy+2x^2y^2\\ b,C+A=B\\ =>C=B-A\\ =\left(3x^2+2y^2+2x^2y^2\right)-\left(4x^2+3y^2-5xy\right)\\ =3x^2+2y^2+2x^2y^2-4x^2-3y^2+5xy\\ =\left(3x^2-4x^2\right)+\left(2y^2-3y^2\right)+2x^2y^2+5xy\\ =-x^2-y^2+2x^2y^2+5xy\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`C = A + B`
`C = 4x^2 + 3y^2 - 5xy + 3x^2 + 2y^2 + 2x^2y^2`
`= (4x^2 + 3x^2) + (3y^2 + 2y^2) - 5xy + 2x^2y^2`
`= 7x^2 + 5y^2 - 5xy + 2x^2y^2`
`b)`
`C + A = B`
`=> C = B - A`
`C = (3x^2 + 2y^2 + 2x^2y^2)-(4x^2 + 3y^2 - 5xy)`
`= 3x^2 + 2y^2 + 2x^2y^2 - 4x^2 - 3y^2 + 5xy`
`= (3x^2 - 4x^2) + (2y^2 - 3y^2) + 2x^2y^2 + 5xy`
`= -x^2 - y^2 + 2x^2y^2 + 5xy`
![](https://rs.olm.vn/images/avt/0.png?1311)
`#3107.101107`
`5x^2y(3x^3 - 4y + 5xy) - 15x^5y + 20x^2y^2`
`= 15x^5y - 20x^2y^2 + 25x^3y^2 - 15x^5y + 20x^2y^2`
`= (15x^5y - 15x^5y) + (-20x^2y^2 + 20x^2y^2) + 25x^3y^2`
`= 25x^3y^2`
_______
`(8x^5y^2 + 4x^3y^3 - 2x^6y^2) \div 2x^3y`
`= 4x^2y + 2y^2 - x^3y`