a)  B= \(\frac{1}{\sqrt{a}}\)(ĐKXĐ: a,b>0)   B) Khi a= \(6+2\sqrt{5}\)thì B=\(\frac{1}{\sqrt{\left(\sqrt{5}+1\right)^2}}\)=\(\frac{1}{\sqrt{5}+1}\)     C) Do \(\sqrt{a}>0\)\(\Rightarrow\frac{1}{\sqrt{a}}>0\)\(\Rightarrow\frac{1}{\sqrt{a}}>-1\)

giải rõ hộ với bạn

Ta có: \(M=\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}-\dfrac{a}{\sqrt{a}+\sqrt{b}}+\dfrac{b}{\sqrt{a}-\sqrt{b}}\)

\(=\dfrac{a\sqrt{a}-b\sqrt{b}-a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}}{\left(\sqrt{a}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)

\(M=\dfrac{a\sqrt{a}-b\sqrt{b}-a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\\ M=\dfrac{a\sqrt{b}+b\sqrt{a}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ M=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)

\(\left(1-a\right)\left(1-b\right)+2\sqrt{ab}=1\\ \Leftrightarrow1-a-b+ab+2\sqrt{ab}=1\\ \Leftrightarrow a+b-ab-2\sqrt{ab}=0\\ \Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=ab\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{a}-\sqrt{b}=\sqrt{ab}\\\sqrt{a}-\sqrt{b}=-\sqrt{ab}\end{matrix}\right.\)

Với \(\sqrt{a}-\sqrt{b}=\sqrt{ab}\Leftrightarrow M=\dfrac{\sqrt{ab}}{\sqrt{ab}}=1\)

Với \(\sqrt{a}-\sqrt{b}=-\sqrt{ab}\Leftrightarrow M=\dfrac{\sqrt{ab}}{-\sqrt{ab}}=-1\)

\(M=\dfrac{a\sqrt{a}-b\sqrt{b}-a\left(\sqrt{a}-\sqrt{b}\right)+b\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

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

\(\left(1-a\right)\left(1-b\right)+2\sqrt{ab}=1\)

\(\Leftrightarrow a+b-ab-2\sqrt{ab}=0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2=ab\Leftrightarrow\sqrt{a}-\sqrt{b}=\sqrt{ab}\)

\(M=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}}{\sqrt{ab}}=1\)

a: \(P=-5\sqrt{\dfrac{160}{90}}=-5\cdot\dfrac{4}{3}=-\dfrac{20}{3}\)

b: \(Q=\sqrt{a}-\sqrt{b}+2\sqrt{b}=\sqrt{a}+\sqrt{b}\)

 Trước hết ta rút gọn D :

 \(D=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)(ĐKXĐ : \(a\ne0,b\ne0,ab\ne1\))

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}-\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{\left(1-\sqrt{ab}\right)\left(1+\sqrt{ab}\right)}:\frac{1+a+b+ab}{1-ab}\)

\(=\frac{2\sqrt{a}\left(b+1\right)}{1-ab}.\frac{1-ab}{\left(a+1\right)\left(b+1\right)}=\frac{2\sqrt{a}}{a+1}\)

a) Với \(a=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow D=\frac{2\sqrt{\left(\sqrt{3}-1\right)^2}}{4-2\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)}{5-2\sqrt{3}}\)

b) Ta có : \(\left(\sqrt{a}-1\right)^2\ge0\Leftrightarrow a+1\ge2\sqrt{a}\Leftrightarrow\frac{2\sqrt{a}}{a+1}\le1\)

Suy ra Max D = 1 <=> a = 1

Em gửi ảnh ạ !

Em gửi ảnh trên ạ !!!!!

a, Ta có \(x=49\Rightarrow\sqrt{x}=7\)

Thay vào biểu thức A ta được : 

\(A=\frac{7.4}{7-1}=\frac{28}{6}=\frac{14}{3}\)

b, Với \(x\ge0;x\ne1\)

\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)( đpcm )