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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
\(C=5^2-2.5+10\)
\(C=25-10+10\)
\(C=25\)
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
Câu 2:
\(B=x^2+2x+y^2-2x-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+2\left(x-y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=7^2+2\cdot7+37=49+37+14=100\)
Câu 3:
\(C=\left(x^2+4xy+4y^2\right)-2\left(x+2y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2\cdot5+10=25\)
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, 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