@Arakawa White
@DƯƠNG PHAN KHÁNH DƯƠNG
@Nguyễn Việt Lâm
@Nguyễn Huy Tú
giúp với ạ !
a) \(K=\dfrac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}+\dfrac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{x+y}{\sqrt{xy}}\)
\(K=\dfrac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-\left(x+y\right)\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(K=\dfrac{x^2-x\sqrt{xy}-y^2-y\sqrt{xy}-x^2+y^2}{\sqrt{xy}\left(x-y\right)}\)
\(K=\dfrac{-\sqrt{xy}\left(x-y\right)}{\sqrt{xy}\left(x-y\right)}=-1\)
Có gì đó hơi sai sai
câu a kết quả là \(\dfrac{x+y}{x-y}\)
mình cần câu b, c
a: \(K=\dfrac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}+\dfrac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{x+y}{\sqrt{xy}}\)
\(=\dfrac{y\sqrt{y}\left(\sqrt{y}+\sqrt{x}\right)+x\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)-\left(x+y\right)\left(y-x\right)}{\sqrt{xy}\left(y-x\right)}\)
\(=\dfrac{y^2+y\sqrt{xy}+x\sqrt{xy}-x^2+x^2-y^2}{\sqrt{xy}\left(y-x\right)}\)
\(=\dfrac{\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(y-x\right)}=\dfrac{-x-y}{x-y}\)
b: 2x^2+2y^2=5xy
=>2x^2-5xy+2y^2=0
=>2x^2-4xy-xy+2y^2=0
=>(x-2y)(2x-y)=0
=>x=2y hoặc y=2x(loại)
KHi x=2y thì \(A=\dfrac{-2y-y}{2y-y}=-3\)