Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\left(x^2y+3\right)^2\\ b,=\left(2x+y\right)^2\\ c,=\left(5y^2-1\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)
\(minA=4\Leftrightarrow x=2\)
\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)
\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)
\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)
\(minC=-8\Leftrightarrow x=-1\)
\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)
\(maxD=-4\Leftrightarrow x=1\)
\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)
\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)
\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)
\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)
\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)
\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)
\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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/
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
b)\(4x^2+4x+5+y^2-4y\)
\(=\left[\left(2x\right)^2+4x+1\right]+\left(y^2-4y+4\right)\)
\(=\left(2x+1\right)^2+\left(y-2\right)^2\)
c) \(4x^2+5y^2+4xy-12y+9\)
\(=\left(4x^2+4xy+y^2\right)+\left(4y^2-12y+9\right)\)
\(=\left(2x+y\right)^2+\left(2y-3\right)^2\)