Cho x+y=1 tính P sao cho 2x^2×y+2×x×y^2-2xy+5
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a) Ta có: A = (x + y)3 + 2x2 + 4xy + 2y2
A = 73 + 2(x2 + 2xy + y2)
A = 343 + 2(x + y)2
A = 343 + 2. 72
A = 343 + 98 = 441
b) B = (x - y)3 - x2 + 2xy - y2
=> B = (-5)3 - (x2 - 2xy + y2)
=> B = -125 - (x - y)2
=> B = -125 - (-5)2
=> B = -125 - 25 = -150
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
![](https://rs.olm.vn/images/avt/0.png?1311)
vì x=5+y => x-y=5
đặt \(A=x^2+y\left(y-2x\right)+75\)
\(=x^2+y^2-2xy+75\)
\(=\left(x-y\right)^2+75\)
\(=5^2+75\)
=100
b) đặt \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+65\)
\(=x^2+2x+y^2-2y-2xy+65\)
\(=\left(x^2+y^2-2xy\right)+\left(2x-2y\right)+65\)
\(=\left(x-y\right)^2+2\left(x-y\right)+65\)
\(=5^2+2.5+65\)
=100
![](https://rs.olm.vn/images/avt/0.png?1311)
a) P = \(x^2+3x+y^2-3y-2xy+90\)
= \(\left(x-y\right)^2+3\left(x-y\right)+90\)
= \(5^2+3.5+90=130\)
b) P = \(4x^2+9y^2-12xy-12x+24xy-18y+118\)
= \(4x^2+9y^2+12xy-12x-18y+118\)
= \(\left(2x+3y\right)^2-6\left(2x+3y\right)+118\)
= \(\left(-7\right)^2-6.\left(-7\right)+118=209\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=x^2+2xy+y^2-4x-4y+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
b) \(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2.7+37=100\)
c) \(C=x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10=25\)
a) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(M=\left(x+y\right)^3+2x^2+4xy+2y^2\)
\(=7^3+2\left(x^2+2xy+y^2\right)\)
\(=343+2\left(x+y\right)^2\)
\(=343+2.7^2\)
\(=343+98=441\)
b) \(N=\left(x-y\right)^3-x^2+2xy-y^2\)
\(=\left(-5\right)^3-\left(x-y\right)^2\)
\(=-125-\left(-5\right)^2\)
\(=-125-25=-150\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P=2x^2y+2xy^2-2xy+5\)
\(P=2xy\left(x+y-1\right)+5\)
Thay x + y = 1 ta có :
\(P=2xy\left(1-1\right)+5\)
\(P=2xy\cdot0+5\)
\(P=5\)
Vậy....