a) \(6x^2y+9+x^4y^2=\left(x^2y+3\right)^2\)

b) \(-4xy+4x^2+y^2=\left(2x-y\right)^2\)

c) \(25y^4-10y^2+1=\left(5y^2-1\right)^2\)

\(a,=\left(x^2y+3\right)^2\\ b,=\left(2x+y\right)^2\\ c,=\left(5y^2-1\right)^2\)

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

\(minA=4\Leftrightarrow x=2\)

\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)

\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)

\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)

\(minC=-8\Leftrightarrow x=-1\)

\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)

\(maxD=-4\Leftrightarrow x=1\)

\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)

\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)

\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)

\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)

\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)

\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)

\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

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

b, \(x^2+4xy+4y^2=\left(x+2y\right)^2\)

c, \(4x^2+4xy+y^2=\left(2x+y\right)^2\)

d, \(x^2+12xy+36y^2=\left(x+6y\right)^2\)

e, \(x^2-12xy+36y^2=\left(x-6y\right)^2\)

a, \(4x^2-4x+1\)

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

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

b, \(x^2+4xy+4y^2\)

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

\(=x.\left(x+2y\right)+2y.\left(x+2y\right)\)

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

Chúc bạn học tốt!!! (bạn nhờ mình giải chi tiết bài này á)

c, \(4x^2+4xy+y^2\)

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

\(=2x.\left(2x+y\right)+y.\left(2x+y\right)\)

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

d, \(x^2+12xy+36y^2\)

\(=x^2+6xy+6xy+36y^2\)

\(=x.\left(x+6y\right)+6y.\left(x+6y\right)\)

\(=\left(x+6y\right)^2\)

e, \(x^2-12xy+36y^2\)

\(=x^2-6xy-6xy+36y^2\)

\(=x.\left(x-6y\right)-6y.\left(x-6y\right)\)

\(=\left(x-6y\right)^2\)

a)  \(x^2+10x+25+y^2+1+2y=\left(x+5\right)^2+\left(y+1\right)^2\)

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

c)  \(x^2-6x+13+y^2+4y=\left(x-3\right)^2+\left(y+2\right)^2\)

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

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

Với x = -5; y = 2 thì: \(A=5\cdot2^2-\left(-5\right)\cdot2=20+10=30\)

b) \(B=-3x\left(x^2+y^2\right)+2y\left(x^2-y\right)=-3x^3-3xy^2+2yx^2-2y^2=-3x^3+2x^2y-3xy^2-2y^2\)

Với x = 1; y = 2 thì: \(B=-3\cdot1^3+2\cdot1^2\cdot2-3\cdot1\cdot2^2-2\cdot2^2=-3+4-12-8=-19\)

Viết các đơn thức sau dưới dạng thu gọn 

a)3x.5y².x²