=(xy²)²+2xy².1+1²=(xy²+1)²

Lời giải:

a. $x^2+y^2+4y+13-6x$

$=(x^2-6x+9)+(y^2+4y+4)$

$=(x-3)^2+(y+2)^2$

b.

$4x^2-4xy+1+2y^2-2y$

$=(4x^2-4xy+y^2)+(y^2-2y+1)$

$=(2x-y)^2+(y-1)^2$

c.

$x^2-2xy+2y^2+2y+1$

$=(x^2-2xy+y^2)+(y^2+2y+1)$

$=(x-y)^2+(y+1)^2$

a. \(x^2+y^2+4y+12-6x=\left(x^2-6x+9\right)+\left(y^2+4y+4\right)=\left(x-3\right)^2+\left(y+2\right)^2\)b. \(4x^2-4xy+1+2y^2-2y=\left(4x^2-4xy+y^2\right)+\left(y^2-2y+1\right)=\left(2x-y\right)^2+\left(y-1\right)^2\)c. \(x^2-2xy+2y^2+2y+1=\left(x^2-2xy+y^2\right)+\left(y^2+2y+1\right)=\left(x-y\right)^2+\left(y+1\right)^2\)

a: \(x^2-6x+y^2+4y+13\)

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

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

b: \(4x^2-4xy+1+2y^2-2y\)

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

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

c: \(x^2-2xy+2y^2+2y+1\)

\(=x^2-2xy+y^2+y^2+2y+1\)

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

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) 2x² + y² - 2xy + 10x + 25

= (x² + y² - 2xy) + (x² + 10x + 25)

= (x - y)² + (x + 5)²

các bn xem đúng ko nhé mk làm bừa nên lên olm hỏi lại mọi người giúp giùm câu b) nha!!

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\(x^2+2y^2+2xy-2y+2\)

\(=\left(\frac{x^2}{2}+2xy+2y^2\right)+\left(\frac{x^2}{2}-2x+2\right)\)

\(=\left(\frac{x}{\sqrt{2}}+\sqrt{2}y\right)^2+\left(\frac{x}{\sqrt{2}}-\sqrt{2}\right)^2\)

\(5+2\sqrt{6}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)

\(6+2\sqrt{5}=\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(5+2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^2\)

\(6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)