\(\left(\dfrac{2x}{2x+y}-\dfrac{4x^2}{4x^2+4xy+y^2}\right):\left(\dfrac{2x}{4x^2-y^2}+\dfrac{1}{y-2x}\right)\)

=\(\left[\dfrac{2x}{2x+y}-\dfrac{4x^2}{\left(2x+y\right)^2}\right]:\left[\dfrac{2x}{\left(2x-y\right)\left(2x+y\right)}-\dfrac{1}{2x-y}\right]\)

\(=\left[\dfrac{2x\left(2x+y\right)}{\left(2x+y\right)^2}-\dfrac{4x^2}{\left(2x+y\right)^2}\right]:\left[\dfrac{2x}{\left(2x-y\right)\left(2x+y\right)}-\dfrac{y+2x}{\left(2x-y\right)\left(y+2x\right)}\right]\)

\(=\left[\dfrac{4x^2+2xy-4x^2}{\left(2x+y\right)^2}\right]:[\dfrac{2x-y-2x}{\left(2x-y\right)\left(2x+y\right)}]\)

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

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

\(=\dfrac{2x\left(2x-y\right)}{-\left(2x+y\right)}\)

\(\dfrac{4x^2-2xy}{-2x-y}\)

a/ \(4x^2+2y^2-4xy+4x-2y+5=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+2\left(2x-y\right)+1+4=0\)

\(\Leftrightarrow\left(2x-y\right)^2+2\left(2x-y\right)+1+4=0\)

\(\Leftrightarrow\left(2x-y+1\right)^2+4=0\)

Với mọi x, y ta có :

\(\left(2x-y+1\right)^2\ge0\Leftrightarrow\left(2x-y+1\right)^2+4>0\)

\(\Leftrightarrow pt\) vô nghiệm

a: \(=\dfrac{5\left(x^2+2xy+y^2\right)}{3\left(x^3+y^3\right)}\)

\(=\dfrac{5\left(x+y\right)^2}{3\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{5\left(x+y\right)}{3\left(x^2-xy+y^2\right)}\)

b: \(=\dfrac{x^2-4xy+4y^2-4}{2x\left(x-2y+2\right)}=\dfrac{\left(x-2y-2\right)\left(x-2y+2\right)}{2x\left(x-2y+2\right)}\)

\(=\dfrac{x-2y-2}{2x}\)

c: \(=\dfrac{2\left(x^2+5x+1\right)}{x\left(x-2\right)\left(x+2\right)}\)

Giải sơ qua:

1)\(B=4x^2-4xy+2y^2+1=\left(2x-y\right)^2+y^2+1\ge1\)

2) có vẻ sai đề

Đúng đề hết nhé

\(4x^2+2y^2+2y-4xy+1=0\)

\(\Leftrightarrow\left(4x^2-4xy+y^2\right)+\left(y^2+2y+1\right)=0\)

\(\Leftrightarrow\left(2x-y\right)^2+\left(y+1\right)^2=0\)

Vì \(\hept{\begin{cases}\left(2x-y\right)^2\ge0;\forall;x,y\\\left(y+1\right)^2\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(2x-y\right)^2+\left(y+1\right)^2\ge0;\forall x,y\)

Do đó \(\left(2x-y\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(2x-y\right)^2=0\\\left(y+1\right)^2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2x+1=0\\y=-1\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=-1\end{cases}}\)

Vậy ...

a)\(2x^2+y^2+4x-2y-2xy+10=2x^2+y^2+4x-2y\left(x+1\right)+10\)

\(=y^2-2y\left(x+1\right)+2\left(x^2+2x+1\right)+8\)

\(=y^2-2y\left(x+1\right)+2\left(x+1\right)^2+8\)

\(=\left(y+x+1\right)^2+\left(x+1\right)^2+8\ge8\)

Dấu "=" xảy ra khi x=-1 và y=0

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

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

Dấu "=" xảy ra khi x=1/2 và y=-1/2

a, \(=12x^5+9x^3y^2-6x^2y^3-20x^4y-15x^2y^3-10xy^4-24x^3y^2-18xy^4+12y^5\)

(tự rút gọn cái :P)

b, \(8x^3+4x^2y-2xy^2-y^3\)

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

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

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

Mấy cái còn lại nhân tung ra là được mà :))))

làm luôn đi cậu

\(\Leftrightarrow4x^2-4xy+y^2+y^2+2y+1=0\)

\(\Leftrightarrow\left(2x-y\right)^2+\left(y+1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-y=0\\y+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-\frac{1}{2}\\y=-1\end{matrix}\right.\)

\(4x^2+2y^2+2y-4xy+1=0\\ \Rightarrow\left(4x^2+y^2-4xy\right)+\left(y^2+2y+1\right)=0\\ \Rightarrow\left(2x-y\right)^2+\left(y+1\right)^2=0\\ \Rightarrow\left\{{}\begin{matrix}2x-y=0\\y+1=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}2x=y\\y=-1\end{matrix}\right.\\ \Rightarrow x=\frac{-1}{2};y=-1\)

Lời giải:

a)

\(A=4x^2-8x+17=4(x^2-2x+1)+13\)

\(=4(x-1)^2+13\)

Vì \((x-1)^2\geq 0, \forall x\Rightarrow A\geq 4.0+13=13\)

Vậy GTNN của $A$ là $13$ tại \((x-1)^2=0\Leftrightarrow x=1\)

b)

\(B=3x^2-5x-1=3(x^2-\frac{5}{3}x+\frac{5^2}{6^2})-\frac{37}{12}\)

\(=3(x-\frac{5}{6})^2-\frac{37}{12}\)

Vì \((x-\frac{5}{6})^2\ge 0, \forall x\Rightarrow B\geq 3.0-\frac{37}{12}=-\frac{37}{12}\)

Vậy GTNN của $B$ là \(\frac{-37}{12}\) khi \(x=\frac{5}{6}\)

c)

\(C=5x^2-4xy-y^2-4x+21\)

\(=(4x^2-4xy+y^2)+(x^2-4x+4)+17\)

\(=(2x-y)^2+(x-2)^2+17\)

Vì \((2x-y)^2\geq 0, (x-2)^2\geq 0, \forall x,y\)

\(\Rightarrow C\geq 0+0+17=17\)

Vậy GTNN của $C$ là $17$ tại \(\left\{\begin{matrix} (2x-y)^2=0\\ (x-2)^2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2\\ y=4\end{matrix}\right.\)