K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
20 tháng 11 2018
\(\frac{2x+y}{2x^2-xy}+\frac{8y}{y^2-4x^2}+\frac{2x-y}{2x^2+xy}\)
\(=\frac{2x+y}{x\left(2x-y\right)}-\frac{8y}{\left(2x-y\right)\left(2x+y\right)}+\frac{2x-y}{x\left(2x+y\right)}\)
\(=\frac{\left(2x+y\right)^2-8xy+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{4x^2+4xy+y^2-8xy+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\frac{8x^2-8xy+2y^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{2\left(4x^2-4xy+y^2\right)}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\frac{2\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}=\frac{2\left(2x-y\right)}{x\left(2x+y\right)}\)
NN
0
LT
1
10 tháng 8 2020
\(\left(\sqrt{2x}-y\right)^2=\left(\sqrt{2x}\right)^2-2\cdot\sqrt{2x}\cdot y+y^2=2x-2\sqrt{2x}\cdot y+y^2\)
\(\left(\sqrt{2x}+\sqrt{8y}\right)^2=\left(\sqrt{2x}\right)^2+2\left(\sqrt{2x}\right)\left(\sqrt{8y}\right)+\left(\sqrt{8y}\right)^2=2x+2\sqrt{16xy}+8y\)
Không chắc nha :)
L
1
DT
3 tháng 1 2022
\(2x+\dfrac{y}{2x^2}-xy+\dfrac{8y}{y^2}-4x^2+2x-\dfrac{y}{2x^2}+xy\\ =4x+\dfrac{8y}{y^2}-4x^2=4x+\dfrac{8}{y}-4x^2\)
\(=4x-4x^2+\dfrac{8}{y}=4x\left(1-x\right)+\dfrac{8}{y}\)
PT
23 tháng 11 2017
a) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{8y}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-8y}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{\left(2x+y\right)\left(2x+y\right)-8yx+\left(2x-y\right)\left(2x-y\right)}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{8x^2-8xy+2y^2}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{2\left(4x^2-4xy+y^2\right)}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{2\left(2x-y\right)^2}{x\left(2x+y\right)\left(2x-y\right)}\)
\(=\dfrac{2\left(2x-y\right)}{x\left(2x+y\right)}\)
b) \(\dfrac{1}{x^2+3x+2}+\dfrac{2x}{x^2+4x+3}+\dfrac{1}{x^2+5x+6}\)
\(=\dfrac{1}{x^2+x+2x+2}+\dfrac{2x}{x^2+x+3x+3}+\dfrac{1}{x^2+2x+3x+6}\)
\(=\dfrac{1}{x\left(x+1\right)\left(x+2\right)}+\dfrac{2x}{x\left(x+1\right)+3\left(x+1\right)}+\dfrac{1}{x\left(x+2\right)+2\left(x+2\right)}\)
\(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{2x}{\left(x+1\right)\left(x+3\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{x+3+2x\left(x+2\right)+x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{x+3+2x^2+4x+x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2x^2+6x+4}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2\left(x^2+3x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{2}{x+3}\)
19 tháng 10 2018
Tìm GTNN
Câu 1 :
\(C=2x^2-5x+1\)
\(C=2\left(x^2-\frac{5}{2}x+\frac{1}{2}\right)\)
\(C=2\left(x^2-2\cdot x\cdot\frac{5}{4}+\frac{25}{16}-\frac{17}{16}\right)\)
\(C=2\left[\left(x-\frac{5}{4}\right)^2-\frac{17}{16}\right]\)
\(C=2\left(x-\frac{5}{4}\right)^2-\frac{17}{8}\ge\frac{-17}{8}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-\frac{5}{4}=0\Leftrightarrow x=\frac{5}{4}\)
Câu 2 :
\(D=x^2+2x+y^2-8y-4\)
\(D=x^2+2\cdot x\cdot1+1^2+y^2-2\cdot y\cdot4+4^2-21\)
\(D=\left(x+1\right)^2+\left(y-2\right)^2-21\ge-21\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x+1=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)
Tìm GTLN :
Câu 1 :
\(C=-2x^2+2x-1\)
\(C=-2\left(x^2-x+\frac{1}{2}\right)\)
\(C=-2\left(x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)\)
\(C=-2\left[\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\right]\)
\(C=-2\left(x-\frac{1}{2}\right)^2-\frac{1}{2}\)
\(C=-\frac{1}{2}-2\left(x-\frac{1}{2}\right)^2\le-\frac{1}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
19 tháng 10 2018
Câu 2 :
\(D=-x^2-y^2-x+y-4\)
\(D=-\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)-\left(y^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)-\frac{7}{2}\)
\(D=-\left(x+\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2-\frac{7}{2}\)
\(D=\frac{-7}{2}-\left[\left(x+\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2\right]\le\frac{-7}{2}\forall x;y\)
Dấu "=' xảy ra \(\Leftrightarrow\hept{\begin{cases}x+\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=\frac{1}{2}\end{cases}}}\)
\(\left(\sqrt{2x}-y\right)^2=2x-2\sqrt{2x}.y+y^2\)
\(\left(\sqrt{2x}+\sqrt{8y}\right)^2=2x+8y+2\sqrt{16xy}\)