a) \(B=\left(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}+\dfrac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x,y\ge0;x\ne y\right)\)

\(B=\left[\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{x-y}\right]:\dfrac{x-2\sqrt{xy}+y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)

\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(B=\left[\left(\sqrt{x}+\sqrt{y}\right)-\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\right]:\dfrac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}}\)

\(B=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)

\(B=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)

\(B=\dfrac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x+\sqrt{xy}+y}\)

\(B=\dfrac{\sqrt{xy}}{x+\sqrt{xy}+y}\)

b) Xét tử: 

\(\sqrt{xy}\ge0\forall x,y\) (xác định) (1) 

Xét mẫu: 

\(x+\sqrt{xy}+y\)

\(=\left(\sqrt{x}\right)^2+2\cdot\dfrac{1}{2}\sqrt{y}\cdot\sqrt{x}+\left(\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)

\(=\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2+\dfrac{3}{4}y\)

Mà: \(\left(\sqrt{x}+\dfrac{1}{2}\sqrt{y}\right)^2\ge0\forall x,y\) (xác định), còn: \(\dfrac{3}{4}y\ge0\) vì theo đkxđ thì \(y\ge0\) (2) 

Từ (1) và (2) ⇒ B luôn không âm với mọi x,y (\(B\ge0\)) (đpcm) 

a: \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5}+1-\sqrt{5}+1\)

=2

c: \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)

d: \(\dfrac{y-2\sqrt{y}+1}{\sqrt{y}-1}=\sqrt{y}-1\)

a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)

b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

1: \(A=\dfrac{x-2\sqrt{xy}+y}{x-y}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)

2: Thay \(x=3+2\sqrt{2}\) và \(y=3-2\sqrt{2}\) vào A, ta được:

\(A=\dfrac{\sqrt{2}+1-\sqrt{2}+1}{\sqrt{2}+1+\sqrt{2}-1}=\dfrac{2}{2\sqrt{2}}=\dfrac{\sqrt{2}}{2}\)

\(A=\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+y}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)}{\left(x-y\right)\cdot\left(\sqrt{x}-\sqrt{y}\right)}\)

\(=\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{x+\sqrt{xy}+y}{x+y}\cdot\dfrac{x+\sqrt{xy}-\sqrt{xy}+y}{x-y}\)

\(=\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}-\dfrac{x+\sqrt{xy}+y}{x-y}\)

\(=\dfrac{\sqrt{xy}+y-x-\sqrt{xy}-y}{x-y}=\dfrac{-x}{x-y}\)

a \(A=\dfrac{\left(x-y\right)\left(\sqrt{x}+\sqrt{y}\right)-x\sqrt{x}+y\sqrt{y}}{x-y}\cdot\dfrac{\left(\sqrt{x}+\sqrt{y}\right)}{x-\sqrt{xy}+y}\)

\(=\dfrac{x\sqrt{x}+x\sqrt{y}-y\sqrt{x}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\cdot\dfrac{1}{x-\sqrt{xy}+y}\)

\(=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

b: căn xy>=0

x-căn xy+y

=x-2*căn x*1/2*căn y+1/4*y+3/4y

=(căn x-1/2*căn y)^2+3/4y>0

=>A>=0

\(P=\dfrac{x\left(\sqrt{y}-\sqrt{z}\right)-y\left(\sqrt{x}-\sqrt{z}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}+\dfrac{z}{\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{x\sqrt{y}-x\sqrt{z}-y\sqrt{x}+y\sqrt{z}+z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)-\sqrt{z}\left(x-y\right)+z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{xy}-\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)+z\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{\left(\sqrt{xy}-\sqrt{zx}-\sqrt{zy}+z\right)}{\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{y}-\sqrt{z}\right)-\sqrt{z}\left(\sqrt{y}-\sqrt{z}\right)}{\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

\(=\dfrac{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{z}\right)}{\left(\sqrt{x}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)

=1

Câu a có sai đề nên mk có sửa lại nha ​

Nếu có thêm điều kiện \(y>1\) thì kết quả là \(\dfrac{1}{x-1}\)