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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) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
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, 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, \(A=x^2+2xy+y^2-4x-4y+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
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
x2 + 4y2 - 2x + 10 + 4xy - 4y
= (x2 + 4xy + 4y2) - 2x + 10 - 4y)
= (x + 2y)2 - (2x + 4y) + 10
= 52 - 2(x + 2y) + 10
= 25 - 10 + 10
= 25
\(x^2+4y^2-2x+10+4xy-4y\) =\(x^2+4xy+4y^2-2\left(x+2y\right)+10\)
=\(\left(x+2y\right)^2-2\left(x+2y\right)+10\)
=\(5^2-2\cdot5+10=25\)
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
\(C=5^2-2.5+10\)
\(C=25-10+10\)
\(C=25\)
\(B=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
\(=25\)
A = x2 - x + 1
A = x2 - 2.x.\(\frac{1}{2}\)+\(\frac{1}{4}\) +\(\frac{3}{4}\)
A = \(\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
B = (x - 2)(x - 4) + 3
B = x2 - 4x - 2x + 8 + 3
B = x2 - 6x + 11
B = x2 - 2.3.x + 9 + 3
B = \(\left(x-3\right)^2+3>0\)
C = 2x2 - 4xy + 4y2 + 2x + 5
C = (x2 - 4xy + 4y2) + x2 + 2x + 5
C = (x - 2y)2 + (x2 + 2x + 1) + 4
C = (x - 2y)2 + (x + 1)2 + 4
Xét biểu thức C thấy :
Có 2 hạng tử không âm (vì là bình phương)
Vậy C > 0
B=\(x^2+4y^2-2x+10+4xy-4y\)
B=\(x^2+4xy+4y^2-2\left(x+2y\right)+10\)
B=\(\left(x+2y\right)^2-2\left(5\right)+10\)
B=\(5^2-10+10\)
B=25
\(B=x^2+4y^2-2x+10+4xy-4y\)
\(B=x^2+4y^2-2x+2x+4y+4xy\)
\(B=x^2+4y^2+4xy\)
\(B=x^2+2.x.2y+\left(2y\right)^2=\left(x+2y\right)^2\)
\(B=5^2\)
\(B=25\)