\(x^2\)+\(\dfrac{1}{y^2}\)=x+\(\dfrac{1}{y}\)=3-\(\dfrac{x}{y}\)
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23 tháng 7 2021
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
MH
18 tháng 2 2022
\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)
\(=\dfrac{x^2+\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}:\left(\dfrac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\right)\)
\(=\dfrac{x^2+x^2-1}{x\left(x+1\right)}.\dfrac{x\left(x+1\right)}{x^2-x^2+1}\)
\(=\dfrac{\left(2x^2-1\right)x\left(x+1\right)}{x\left(x+1\right)}=2x^2-1\)
16 tháng 11 2021
a: \(A=\dfrac{x^2+2xy+y^2-x^2+xy+2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{3y^2+3xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{3y}{x-y}\)
3 tháng 1 2021
a) Ta có: \(\left(x-\dfrac{1}{1-x}\right):\dfrac{x^2-x+1}{x^2-2x+1}\)
\(=\left(x+\dfrac{1}{x-1}\right):\dfrac{x^2-x+1}{\left(x-1\right)^2}\)
\(=\dfrac{x^2-x+1}{x-1}\cdot\dfrac{\left(x-1\right)^2}{x^2-x+1}\)
\(=x-1\)
b) Ta có: \(\left(1+\dfrac{x}{y}+\dfrac{x^2}{y^2}\right)\left(1-\dfrac{x}{y}\right)\cdot\dfrac{y^2}{x^3-y^3}\)
\(=\left(\dfrac{y^2}{y^2}+\dfrac{xy}{y^2}+\dfrac{x^2}{y^2}\right)\cdot\left(\dfrac{y-x}{y}\right)\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{x^2+xy+y^2}{y^2}\cdot\dfrac{-\left(x-y\right)}{y}\cdot\dfrac{y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{-1}{y}\)
LM
14 tháng 6 2017
\(\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\right]:\dfrac{x^3+y^3}{x^2y^2}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{x+y}.\dfrac{x+y}{xy}\right].\dfrac{x^2y^2}{x^3+y^3}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\left[\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{2}{xy}\right].\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\dfrac{y^2+x^2+2xy}{x^2y^2}.\dfrac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
\(=\dfrac{\left(x+y\right)^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}-\dfrac{x+y}{x^2-xy+y^2}\)
=\(=\dfrac{x+y}{x^2-xy+y^2}-\dfrac{x+y}{x^2-xy+y^2}=0\)
Lời giải:
PT $\Leftrightarrow (x+\frac{1}{y})^2-\frac{2x}{y}=x+\frac{1}{y}=3-\frac{x}{y}$
Đặt $x=a; \frac{1}{y}=b$ thì:
$(a+b)^2-2ab=a+b=3-ab$
$\Rightarrow 2(3-ab)-(a+b)=(a+b)^2-2ab$
$\Leftrightarrow 6-(a+b)=(a+b)^2$
$\Leftrightarrow (a+b)^2+(a+b)-6=0$
$\Leftrightarrow (a+b+3)(a+b-2)=0$
$\Rightarrow a+b=-3$ hoặc $a+b=2$
Nếu $a+b=-3$ thì:
$9-3ab=-3=3-ab\Rightarrow ab=4=6$ (vô lý)
Nếu $a+b=2$ thì:
$4-2ab=2=3-ab\Rightarrow ab=1$
Thay $a=2-b$ vào thì: $(2-b)b=1$
$\Leftrightarrow b^2-2b+1=0\Leftrightarrow (b-1)^2=0$
$\Leftrightarrow b=1$
$\Rightarrow a=2-b=1$
Vậy $(a,b)=(1,1)\Leftrightarrow (x,y)=(1,1)$
tại sao lại có được như hàng thứ 5 vậy