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a) ĐKXĐ : \(x+y\ne0\)
\(x^2-2y^2=xy\)
\(x^2-y^2-y^2-xy=0\)
\(\left(x-y\right)\left(x+y\right)-y\left(y+x\right)=0\)
\(\left(x+y\right)\left(x-2y\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+y=0\left(Loai\right)\\x-2y=0\left(Chon\right)\end{matrix}\right.\)
Với x - 2y = 0 ta có x = 2y
Thay x = 2y vào A ta có :
\(A=\dfrac{2y-y}{2y+y}=\dfrac{y}{3y}=\dfrac{1}{3}\)
ta có:
\(\left(3x-2y\right)^2=9x^2-12xy+4y^2=20xy-12xy=8xy\)
\(\Rightarrow3x-2y=\sqrt{8xy}\)(1)
\(\left(3x+2y\right)^2=9x^2+12xy+4y^2=20xy+12xy=32xy\)
\(\Rightarrow3x+2y=\sqrt{32xy}\)(2)
từ (1) và (2)
\(\Rightarrow\frac{3x-2y}{3x+2y}=\frac{\sqrt{8xy}}{\sqrt{32xy}}=0,5\)
) \(\dfrac{x^3+8y^3}{2y+x}\)
\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)
\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)
\(=x^2+2xy+4y^2\)
b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)
\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)
\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)
\(=\dfrac{3a-1}{2\left(a-4\right)}\)
c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)
\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2}\)
d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)
\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)
\(=x^2-10x+25+7x+14-x^2-2x\)
\(=39-5x\)
e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)
\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)
\(=\dfrac{3x+2x+1}{x-2}\)
\(=\dfrac{5x+1}{x-2}\)
h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
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:
\(D=\frac{x+y}{x-y}=\frac{ty+y}{ty-y}=\frac{y(t+1)}{y(t-1)}=\frac{t+1}{t-1}=\frac{\frac{5-\sqrt{17}}{4}+1}{\frac{5-\sqrt{17}}{4}-1}=\frac{1-\sqrt{17}}{2}\)
có 3x^2+2y^2=7xy
=>3x^2+2y^2-7xy=0
=>(3x^2-6xy)+(2y^2-xy)=0
=>3x(x-2y)-y(x-2y)=0
=>(x-2y)(3x-y)=0
=>x-2y=0 hoặc 3x-y=0
=>x=2y hoặc y=3x
Xét TH x=2y vài A ta được 3x+y/7y-x+6x-9y/2x+y
=6y+y/7y-2y+12y-9y/4y+y
=7y/5y+3y/5y
=7/5+3/5
=10/5
=1/2
Xét TH y=3x có
3x+y/7y-x+6x-9y/2x+y
=3x+3x/(21x-x)+(6x-27x)/2x+3x
=6x/20x-21x/5x
=3/10-21/5
=3/10-42/10
=-39/10