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a, Với x-y=7 thì
\(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37=7^2+2.7+37\)
\(=49+14+37=100\)
Vậy A=100
b, Với x+2y=5 thì
\(B=x^2+4y^2-2x+10+4xy-4y\)
\(=x^2+4y^2-2x+2x+4y+4xy-4y=x^2+4y^2+4xy\)
\(=x^2+2.x.2y+\left(2y\right)^2=\left(x+2y\right)^2=5^2=25\)
Vậy B=25
a) Ta có:
\(A=x^2+2xy+y^2-4x-4y+1\)
\(A=\left(x+y\right)^2-4\left(x+y\right)+1\)
Thay x + y = 3 vào A
\(A=3^2-4.3+1\)
\(A=9-12+1\)
\(A=-2\)
b) Sửa đề:
\(B=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(B=x^2+2x+y^2-2y-2xy+37\)
\(B=\left(x^2+y^2+1+2x-2y-2xy\right)+36\)
\(B=\left(x-y+1\right)^2+36\)
Thay x - y = 7 vào B
\(B=\left(7+1\right)^2+36\)
\(B=100\)
c) Ta có:
\(C=x^2+4y^2-2x+10+4xy-4y\)
\(C=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)
\(C=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
Thay x + 2y = 5 vào C
\(C=5^2-2.5+10\)
\(C=25-10+10\)
\(C=25\)
Ta có: x - y = 7 ⇔ x = 7 + y
⇒ A = x ( x+2) + y ( y-2) - 2xy +37
⇔ A = (7 + y)( y+9) + y ( y-2) - 2(7+ y)y +37
⇔ A = 7y + 63 + y2 + 9y + y2 - 2y - 14y -2y2 +37
⇔ A = 63 + 37 = 100
Ta có: x+ 2y = 5 ⇔ x = 5 - 2y
⇒ B = x2 +4y2 - 2x +10 + 4xy - 4y
⇔ B = x2 + 4xy + 4y2 - 2x +10 - 4y
⇔ B = (x + 2y)2 - 2(x -5 + 2y)
⇔ B = (5 - 2y + 2y)2 - 2(5 - 2y -5 + 2y)
⇔ B = 52 = 25
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) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
\(a.\)
\(x\left(x+z\right)+y\left(y-z\right)-2xy+37\)
\(=x^2+xz+y^2-yz-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+z\left(x-y\right)+37\)
\(=\left(x-y\right)^2+z\left(x-y\right)+37\)
\(=7^2+x.7^2+37\)
\(=86+49x\)
\(b.\)
\(x^2+4y^2-2x+10+4xy-4y\)
\(=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
\(=25\)
a: \(M=\left(x+y\right)^3+2\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3+2\left(x+y\right)^2\)
\(=7^3+2\cdot49=441\)
b: \(A=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2\cdot7+37\)
\(=49+14+37=100\)
a, \(A=x^2+2xy+y^2-4x-4y+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
Thay x + y = 3
\(\Leftrightarrow A=9-12+1=-2\)
Vậy A = -2 khi x + y = 3
b, \(B=x^2+4y^2-2x+10+4xy-4y\)
\(=x^2+4xy+4y^2-2x-4y+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
Thay x + 2y = 5 có:
\(B=25-10+10=25\)
Vậy B = 25 khi x + 2y = 5
\(B=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
\(=25\)
a, Ta có
A= x(x+2)+y(y-2)-2xy +37
=x2+2x+y2-2y-2xy+37
=x2-2xy+y2+2(x-y)+37
=(x-y)2+2(x-y)+37
Vì x-y=7
=>(x-y)2+2(x-y)+37=72+14+37=100
KL
b, Ta có B=x2+4y2-2x+10+4xy-4y
=x2+4xy+4y2-2x-4y+10
=(x+2y)2-2(x+2y)+10
Vì x+2y=5
=>(x+2y)2-2(x+2y)+10=52-10+10=25
KL