Có \(2x^2+2y^2=5xy\)

\(\Leftrightarrow2x^2-2y^2-5xy=0\)

\(\Leftrightarrow2x^2-4xy-xy+2y^2=0\)

\(\Leftrightarrow2x\left(x-2y\right)-y\left(x-2y\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-2y=0\\2x-y=0\end{matrix}\right.\)

TH1: Với \(x-2y=0\) hay \(x=2y\) thì:

\(E=\dfrac{2y+y}{2y-y}=\dfrac{3y}{y}=3\) ( loại do \(0< x< y\) nên \(E=\dfrac{x+y}{x-y}< 0\) )

TH2: Với \(2x-y=0\)  hay \(2x=y\) thì:

\(E=\dfrac{x+2x}{x-2x}=\dfrac{3x}{-x}=-3\left(tm\right)\)

Vậy \(E=-3\)

a: =-3x^2y*x^2y+3x^2y*2xy

=-3x^4y^2+6x^3y^2

b: =x^3-x^2y+x^2y+y^2=x^3+y^2

c: =x*4x^3-x*5xy+2x*x

=4x^4-5x^2y+2x^2

d: =x^3+x^2y+2x^3+2xy

=3x^3+x^2y+2xy

b: \(B=\dfrac{3y+5}{y-1}-\dfrac{-y^2-4y}{y-1}+\dfrac{y^2+y+7}{y-1}\)

\(=\dfrac{3y+5+y^2+4y+y^2+y+7}{y-1}\)

\(=\dfrac{2y^2+8y+12}{y-1}\)

Lời giải:

Đặt $x=ty$ ($0< t< 2$)

\(2x^2+y^2=5xy\)

\(\Leftrightarrow 2t^2y^2+y^2-5ty^2=0\)

\(\Leftrightarrow y^2(2t^2-5t+1)=0\Rightarrow 2t^2-5t+1=0\) (Do $y\neq 0$)

\(\Leftrightarrow 2(t-\frac{5}{4})^2=\frac{17}{8}\Rightarrow t-\frac{5}{4}=\pm \frac{\sqrt{17}}{4}\)

\(\Rightarrow t=\frac{5\pm \sqrt{17}}{4}\). Mà $0< t< 2$ nên $t=\frac{5-\sqrt{17}}{4}$

Do đó:

\(D=\frac{x+y}{x-y}=\frac{ty+y}{ty-y}=\frac{y(t+1)}{y(t-1)}=\frac{t+1}{t-1}=\frac{\frac{5-\sqrt{17}}{4}+1}{\frac{5-\sqrt{17}}{4}-1}=\frac{1-\sqrt{17}}{2}\)

Lời giải:

Đặt $x=ty$ ($0< t< 2$)

\(2x^2+y^2=5xy\)

\(\Leftrightarrow 2t^2y^2+y^2-5ty^2=0\)

\(\Leftrightarrow y^2(2t^2-5t+1)=0\Rightarrow 2t^2-5t+1=0\) (Do $y\neq 0$)

\(\Leftrightarrow 2(t-\frac{5}{4})^2=\frac{17}{8}\Rightarrow t-\frac{5}{4}=\pm \frac{\sqrt{17}}{4}\)

\(\Rightarrow t=\frac{5\pm \sqrt{17}}{4}\). Mà $0< t< 2$ nên $t=\frac{5-\sqrt{17}}{4}$

Do đó:

\(D=\frac{x+y}{x-y}=\frac{ty+y}{ty-y}=\frac{y(t+1)}{y(t-1)}=\frac{t+1}{t-1}=\frac{\frac{5-\sqrt{17}}{4}+1}{\frac{5-\sqrt{17}}{4}-1}=\frac{1-\sqrt{17}}{2}\)

bn có viết nhầm 5xy thành 4xy ko

Ta có :

\(2x^2+2y^2=5xy\)

\(\Rightarrow2x^2+2y^2-5xy=0\)

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

\(\Rightarrow2x\left(x-2y\right)+y\left(2y-x\right)=0\)

\(\Rightarrow\left(x-2y\right)\left(2x-y\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2y=0\\2x-y=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2y\\2x=y\end{matrix}\right.\)

*) Với \(x=2y\) ta có:

\(M=\dfrac{2y+y}{2y-y}=\dfrac{3y}{y}=3\)

*) Với \(2x=y\) ta có:

\(M=\dfrac{x+2x}{x-2x}=\dfrac{3x}{-x}=-3\)

Vậy \(M=3\) hoặc \(M=-3\)

\(=\dfrac{25y^2+6xy+10x^2}{10x^2y^3}\)

=25y2+6xy+10x²/10x²y³

2y² + 2x² = 5xy 
<=> 2y² - 5xy + 2x² = 0 
<=> (x - 2y)(2x - y) = 0 
=> x = 2y hoặc y = 2x 
Thay vào biểu thức ta có: 
+) Nếu x = 2y => (x - y)/(x + y) = (2y - y)/(2y + y) = y/3y = 1/3 
+) Nếu y = 2x => (x - y)/(x + y) = (x - 2x)/(x + 2x) = -x/3x = -1/3