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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) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
A=(x^2+4xy+4y^2)-(2x+4y)+10
A=(x+2y)^2-2(x+2y)+1+9
A=(x+2y-1)^2+9
A=(5-1)^2+9=16+9=25
a)
\(A=4x-x^2+3=-\left(x^2-4x-3\right)=-\left(x^2-4x+4\right)+7=-\left(x-2\right)^2+7\le7\)
Daaus = xayr ra khi: x = 2
b) \(B=4x^2-12x+15=4\left(x^2-3x+9\right)-21=4\left(x-3\right)^2-21\ge-21\)
Dấu = xảy ra khi x = 3
c) \(C=4x^2+2y^2-4xy-4y+1=\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3=\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)
Dấu = xảy ra khi
2x = y và y = 2
=> x = 1 và y = 2
a) A = \(-x^2+4x+3=-\left(x-2\right)^2+7\le7\)
Dấu "=" <=> x = 2
b) \(4x^2-12x+15=\left(2x-3\right)^2+6\ge6\)
Dấu "=" xảy ra <=> \(x=\dfrac{3}{2}\)
c) \(4x^2+2y^2-4xy-4y+1\)
= \(\left(4x^2-4xy+y^2\right)+\left(y^2-4y+4\right)-3\)
= \(\left(2x-y\right)^2+\left(y-2\right)^2-3\ge-3\)
Dấu "=" <=> \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)
Theo đề ta có : x + 2y = 5
và A = \(x^2-4y^2-2x+10+4xy-4y\)
\(=\left(x^2-4y^2+4xy\right)\) - \(\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
=25
k nha mn!
TA có A=\(A=x^2-4y^2-2x+10+4xy-4xy\)
\(=\left(x^2-4y^2=4xy\right)-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
\(=25\)
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
\(x^2+4y^2-2x+10+4xy-4y\) =\(x^2+4xy+4y^2-2\left(x+2y\right)+10\)
=\(\left(x+2y\right)^2-2\left(x+2y\right)+10\)
=\(5^2-2\cdot5+10=25\)
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
\(B=x^2+4y^2-2x+10+4xy-4y\)
\(B=x^2+4y^2-2x+2x+4y+4xy\)
\(B=x^2+4y^2+4xy\)
\(B=x^2+2.x.2y+\left(2y\right)^2=\left(x+2y\right)^2\)
\(B=5^2\)
\(B=25\)
Ta có
\(C=\left(x^2+2.x.2y+\left(2y\right)^2\right)-\left(2x+4y\right)+10\)
\(\Rightarrow C=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(\Rightarrow C=5^2-2.5+10\)
\(\Rightarrow C=25-10+10=25\)
\(C=x^2+4y^2-2x+10+4xy-4y\)
\(=\left[x^2+2.x.2y+\left(2y\right)^2\right]-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2\right)+10\)
\(=5^2-2.5+10\)
\(=5^2-10+10\)
\(=25-10+10\)
\(=25\)
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