chuyển về dạng tổng hai bình phương: x^2-2x+2+4y^2+4y
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a, \(25x^2+5xy+\frac{1}{4}y^2=\left(5x\right)^2+2.5x.\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\)
\(=\left(5x+\frac{1}{2}y\right)^2\)
b, \(9x^2+12x+4=\left(3x\right)^2+2.3x.2+2^2=\left(3x+2\right)^2\)
c, \(x^2-6x+5-y^2-4y=\left(x^2-6x+9\right)-\left(y^2+4y+4\right)\)
\(=\left(x-3\right)^2-\left(y+2\right)^2=\left(x-y-5\right)\left(x+y-1\right)\)
d, \(\left(2x-y\right)^2+4\left(x+y\right)^2-4\left(2x-y\right)\left(x+y\right)\)
\(=\left(2x-y\right)^2-2\left(2x-y\right)\left(2x+2y\right)+\left(2x+2y\right)^2\)
\(=\left(2x-y+2x+2y\right)^2=\left(4x+y\right)^2\)
\(\left(2x-4y\right)^2+2\left(2x-4y\right)+1=\left(2x-4y+1\right)^2\)
x^2-4y^2-6x+4y+10
= x^2- 2.x.3 + 9 + 4y^2-2.2y+1
= ( x - 3)^2+ (2y-1) ^2
a/ \(=\left(9x^2+30x+25\right)+\left(x^2+10x+25\right)=\)
\(=\left(3x+5\right)^2+\left(x+5\right)^2\)
b/ \(=\left(16x^2+8x+1\right)+\left(y^2-4y+4\right)=\left(4x+1\right)^2+\left(y-2\right)^2\)
c/
a) \(2x^2+2b^2=x^2+b^2+x^2+b^2=x^2+2xb+b^2+x^2-2xb+b^2=\left(x+b\right)^2+\left(x-b\right)^2\)
a, Đề sai bạn ơi phải là cộng 16 chứ không phải cộng 4
b,B= (x-2y+1)^2
1. \(x^2-2x+2+4y^2+4y\)
\(=\left(x^2-2x+1\right)+\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2+\left(2y+1\right)^2\)
2. \(4x^2-4x+y^2+2y+2\)
\(=\left(4x^2-4x+1\right)+\left(y^2+2y+1\right)\)
\(=\left(2x-1\right)^2+\left(y+1\right)^2\)
3. \(4x^2+4x+4y^2+4y+2\)
\(=\left(4x^2+4x+1\right)+\left(4y^2+4y+1\right)\)
\(=\left(2x+1\right)^2+\left(2y+1\right)^2\)
4. \(4x^2+y^2+12x+4y+13\)
\(=\left(4x^2+12x+9\right)+\left(y^2+4y+4\right)\)
\(=\left(2x+3\right)^2+\left(y+2\right)^2\)
\(x^2-2x+2+4y^2+4y\)
\(=\left(x^2-2x+1\right)+\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2+\left(2y+1\right)^2\)
\(4x^2-4x+y^2+2y+2\)
\(=\left(2x-1\right)^2+\left(y+1\right)^2\)
`x^2-2x+4y^2+4y+2`
`=x^2-2x+1+4y^2+4y+1`
`=(x-1)^2+(2y+1)^2`
Ta có: \(x^2-2x+2+4y^2+4y\)
\(=x^2-2x+1+4y^2+4y+1\)
\(=\left(x-1\right)^2+\left(2y+1\right)^2\)