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![](https://rs.olm.vn/images/avt/0.png?1311)
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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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
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=\(A=x^2-4y^2-2x+10+4xy-4xy\)
\(=\left(x^2-4y^2=4xy\right)-\left(2x+4y\right)+10\)
\(=\left(x+2y\right)^2-2\left(x+2y\right)+10\)
\(=5^2-2.5+10\)
\(=25\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+4y^2-5x+10y-4xy+20\)
\(=x^2-4xy+4y^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}-\frac{25}{4}+20\)
\(=\left(x-2y\right)^2-2.\frac{5}{2}\left(x-2y\right)+\frac{25}{4}+\frac{55}{4}\)
\(=\left(x-2y-\frac{5}{2}\right)^2+\frac{55}{4}\)Thay x - 2y = 5 ta được :
\(=\left(5-\frac{5}{2}\right)^2+\frac{55}{4}=20\)
\(B=x^2-2xy-2x+2y+y^2\)
\(=x^2-2xy+y^2-2\left(x-y\right)\)
\(=\left(x-y\right)^2-2\left(x-1\right)\)Thay x = y + 1 => x - y = 1 ta được :
\(=1-2=-1\)
![](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) \(A=x^2+2xy+y^2-4x-4v+1\)
\(=\left(x+y\right)^2-4\left(x+y\right)+1\)
\(=3^2-4.3+1=-2\)
\(A=x^2+4y^2-2x+10+4xy-4y=\left(x^2+4xy+4y^2\right)-\left(2x+4y\right)+10=\left(x+2y\right)^2-2\left(x+2y\right)+10=5^2-2\cdot5+10=25-10+10=25\)