Question

Answer 1

ĐK: x>=1

đặt\(\dfrac{1}{x+y}=a\)

     \(\dfrac{1}{\sqrt{x-1}}=b\)

(*) <=> a+b=2

           2a-b=1

=> a=1 , b=1

=>\(\dfrac{1}{x+y}=1\)  

    \(\dfrac{1}{\sqrt{x-1}}=1\)

<=>x + y=1

    \(\sqrt{x-1}=1\)

<=>x= 1-y

      x-1=1 ( bình phương 2 vế)

<=>x= 1-y

      1-y-1=1

<=> x=1-y

      y=-1

<=>x=2

      y=-1

Answer 2

đk : x khác -y ; x > 1 

Đặt \(\dfrac{1}{x+y}=u;\dfrac{1}{\sqrt{x-1}}=v\)

\(\Leftrightarrow\left\{{}\begin{matrix}u+v=2\\2u-v=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}u=1\\y=1\end{matrix}\right.\)

Theo cách đặt 

x + y = 1 ; \(\sqrt{x-1}=1\Leftrightarrow x-1=1\Leftrightarrow x=2\)(tm)

=> y = -1