a)  \(x^2-4x+5+y^2+2y=\left(x^2-4x+4\right)+\left(y^2+2y+1\right)\)

\(=\left(x-2\right)^2+\left(y+1\right)^2\)

b)  \(2x^2+y^2-2xy+10x+25=\left(x^2+10x+25\right)+\left(x^2-2xy+y^2\right)\)

\(=\left(x+5\right)^2+\left(x-y\right)^2\)

c)  \(2x^2+2y^2=\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)=\left(x-y\right)^2+\left(x+y\right)^2\)

bạn ơi , bạn lấy bài này ở đâu vậy bạn

tớ ko biết

\(a.x^2-4x+4+y^2+2y+1\)

\(=\left(x-2\right)^2+\left(y+1\right)^2\)

\(b.x^2+10x+25+x^2-2xy+y^2\)

\(=\left(x+5\right)^2+\left(x-y\right)^2\)

Bạn Đỗ Thanh Tùng ơi, bạn sai đề rồi nha

Cái này là viết dưới dạng tổng hai bình phương ạ

a)x²-4x+5+y²+2y=x²-4x+4+y²+2y+1=(x-2)²+(y+1)²

b)2x²+y²-2xy+10x+25=x²-2xy+y²+x²+10x+25=(X+Y)²+(X+5)²

c)a²+2ab+5b²+4b+1=a²+2ab+b²+4b²+4b+1=(a+b)²+(2b+1)²

d)2x²+2b²+4x+4b+4=2x²+4x+2+2b²+4b+2=(\(\sqrt{2}x+\sqrt{2}\))²+(\(\sqrt{2}b+\sqrt{2}\))²

e)X⁴+13-6x²+4y+y²=x⁴-6x²+9+y²+4y+4=(x²-3)²+(y+2)²

f)-6x+9x²-8y+4y+y2+5= 9x²-6x+1+4y²-8y+4= (3x-1)²+(2y-2)²

tôi chỉ rút gọn đc như thế là cùng rồi

ko còn cách nào khác đâu

chắc bạn đang học những hằng đẳng thức đáng nhớ

a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)

\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)

\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)

\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)

\(=\left(x^2+9x+19\right)^2\)

b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)

\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)

\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)

c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)

\(=x^2-2x\left(y+2\right)+\left(y+2\right)^2\)

\(=\left(x-y-2\right)^2\)

d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)

\(=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)

\(=\left(x+y+1\right)^2\)

\(A=x^2+4x+5=\left(x+2\right)^2+1\ge1\)

Dấu \("="\Leftrightarrow x=-2\)

\(B=x^2+10x-1=\left(x+5\right)^2-26\ge-26\)

Dấu \("="\Leftrightarrow x=-5\)

\(C=5-4x+4x^2=\left(2x-1\right)^2+4\ge4\)

Dấu \("="\Leftrightarrow x=\dfrac{1}{2}\)

\(D=x^2+y^2-2x+6y-3=\left(x-1\right)^2+\left(y+3\right)^2-13\ge-13\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

\(E=2x^2+y^2+2xy+2x+3=\left(x+y\right)^2+\left(x+1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow x=-y=-1\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)

\(A=x^2+4x+5\)

\(=x^2+4x+4+1\)

\(=\left(x+2\right)^2+1\ge1\forall x\)

Dấu '=' xảy ra khi x=-2

\(C=4x^2-4x+5\)

\(=4x^2-4x+1+4\)

\(=\left(2x-1\right)^2+4\ge4\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)

......................?

mik ko biết

mong bn thông cảm !$$%

......................?

mik ko biết

mong bn thông cảm 

nha ................

kho lam a?

a: \(=x^2-4x+4+y^2+2y+1\)

\(=\left(x-2\right)^2+\left(y+1\right)^2\)

b: \(=x^2+10x+25+x^2-2xy+y^2\)

\(=\left(x+5\right)^2+\left(x-y\right)^2\)

c: \(=a^2+2ab+b^2+4b^2+4b+1\)

\(=\left(a+b\right)^2+\left(2b+1\right)^2\)

d: \(=2\left(x^2+b^2\right)\)

a) Ta có: \(x^2+10x+26+y^2+2y=\left(x^2+10x+25\right)+\left(y^2+2y+1\right)\)

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

b) Ta có: \(a^2+5b^2+2ab+4b+1=\left(a^2+2ab+b^2\right)+\left(4b^2+4b+1\right)\)

\(=\left(a+b\right)^2+\left(2b+1\right)^2\)

c) Ta có: \(4x^2+4x+10+6y+y^2=\left(4x^2+4x+1\right)+\left(y^2+6y+9\right)\)

\(=\left(2x+1\right)^2+\left(y+3\right)^2\)

a) \(x^2+10x+26+y^2+2y=x^2+2.5.x+5^2+y^2+2.y.1+1^2\) = \(\left(x+5\right)^2+\left(y+1\right)^2\)

b) \(a^2+5b^2+2ab+4b+1=a^2+2ab+b^2+4b^2+4b+1\)

= \(\left(a+b\right)^2+\left(2b+1\right)^2\)

c) \(4x^2+4x+10+6y+y^2=4x^2+4x+1+y^2+6y+9\)

= \(\left(2x+1\right)^2+\left(y+3\right)^2\)

a, \(x^2+10x+26+y^2+2y\)

= \(x^2+10x+25+1+y^2+2y\)

= \((x^2+10x+25)+(y^2+2y+1)\)

= \(\left(x+5\right)^2+(y+1)^2\)

b, \(a^2+5b^2+2ab+4b+1\)

= \(a^2+4b^2+b^2+2ab+4b+1\)

= \((a^2+2ab+b^2)+(4b^2+4b+1)\)

= \((a+b)^2+(2b+1)^2\)

c,\(4x^2+4x+10+6y+y^2\)

= \(4x^2+4x+9+1+6y+y^2\)

= \((4x^2+4x+1)+(y^2+6y+9)\)

= \((2x+1)^2+(y+3)^2\)