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2 tháng 8 2018

a) ta có : \(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)

b) ta có : \(\dfrac{x-\sqrt{3x}+3}{x\sqrt{x}+3\sqrt{3}}=\dfrac{x-\sqrt{3x}+3}{\left(\sqrt{x}+\sqrt{3}\right)\left(x-\sqrt{3x}+3\right)}=\dfrac{1}{\sqrt{x}+\sqrt{3}}\)

c) ta có : \(\dfrac{2}{\sqrt{5}-\sqrt{3}}+\dfrac{3}{\sqrt{6}+\sqrt{3}}\)

\(=\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}+\dfrac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)

\(=\dfrac{2\left(\sqrt{5}+\sqrt{3}\right)}{2}+\dfrac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\) \(=\sqrt{5}+\sqrt{3}+\sqrt{6}-\sqrt{3}=\sqrt{5}+\sqrt{6}\)

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: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)

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

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

c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)

\(=\sqrt{x}+2-\sqrt{x}-2=0\)

7 tháng 2 2022

a) Rút gọn được \(\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

c) \(H=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\Rightarrow H^2=\dfrac{xy}{\left(x-\sqrt{xy}+y\right)^2}\)

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

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

Do \(\left\{{}\begin{matrix}\sqrt{xy}\ge0\\\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\\\left(x-\sqrt{xy}+y\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow H^2-H=-\dfrac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)^2}{\left(x-\sqrt{xy}+y\right)^2}\le0\Rightarrow H^2\le H\)

Mà \(H\ge0\left(cmt\right)\Rightarrow H\le\sqrt{H}\)

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

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

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

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

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

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

15 tháng 10 2021

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) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)

\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)

\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)

\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)

\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)

=-1

 

Bài 1: 

a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:

\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)

b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x+\sqrt{x}-6-x+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)

\(=\dfrac{3}{\sqrt{x}+3}\)

c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)

\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)

\(\Leftrightarrow3-\sqrt{x}>0\)

\(\Leftrightarrow\sqrt{x}< 3\)

hay x<9

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)

6 tháng 12 2023

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)