a: ĐKXĐ: a>=0; b>=0; ab<>1

Ta có: \(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\)

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

\(=\frac{\sqrt{a}+a\cdot\sqrt{b}+\sqrt{b}+b\cdot\sqrt{a}+\sqrt{a}-a\cdot\sqrt{b}-\sqrt{b}+b\cdot\sqrt{a}}{1-ab}=\frac{2\cdot\sqrt{a}+2b\cdot\sqrt{a}}{1-ab}\)

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

Ta có: \(D=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)

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

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

b: \(a=\frac{2}{2+\sqrt3}=\frac{2\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}\)

\(=\frac{4-2\sqrt3}{4-3}=4-2\sqrt3=\left(\sqrt3-1\right)^2\)

Thay \(a=\left(\sqrt3-1\right)^2\) vào D, ta được:

\(D=\frac{2\cdot\sqrt{\left(\sqrt3-1\right)^2}}{\left(\sqrt3-1\right)^2+1}\)

\(=\frac{2\left(\sqrt3-1\right)}{4-2\sqrt3+1}=\frac{2\sqrt3-2}{5-2\sqrt3}=\frac{\left(2\sqrt3-2\right)\left(5+2\sqrt3\right)}{\left(5-2\sqrt3\right)\left(5+2\sqrt3\right)}\)

\(=\frac{10\sqrt3+12-10-4\sqrt3}{25-12}=\frac{6\sqrt3+2}{13}\)

c: \(\frac{1}{D}=\frac{a+1}{2\sqrt{a}}\)

=>\(\frac{1}{D}-1=\frac{a+1-2\sqrt{a}}{2\sqrt{a}}=\frac{\left(\sqrt{a}-1\right)^2}{2\sqrt{a}}\ge0\forall a\) thỏa mãn ĐKXĐ

=>\(\frac{1}{D}\ge1\forall a\) thỏa mãn ĐKXĐ

=>D<=1∀a thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{a}-1=0\)

=>a=1(nhận)

Ta có: \(M=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2\)=\(2a+2b\le2\)

\(Max\)\(M=2\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{a}+\sqrt{b}\\a+b=1\end{matrix}\right.\)\(\Leftrightarrow a=b=\dfrac{1}{2}\)

\(M=\left(\sqrt[]{a}+\sqrt[]{b}\right)^2;a+b\le1\left(a;b>0\right)\)

Áp dụng Bất đẳng thức Bunhiacopxki cho 2 cặp số \(\left(1;\sqrt[]{a}\right);\left(1;\sqrt[]{b}\right)\)

\(M=\left(1.\sqrt[]{a}+1.\sqrt[]{b}\right)^2\le\left(1^2+1^2\right)\left(a+b\right)\le2\)  \(\left(a+b\le1\right)\)

\(\Rightarrow M=\left(\sqrt[]{a}+\sqrt[]{b}\right)^2\le2\)

Dấu "=" xảy ra khi và chỉ khi

\(\dfrac{1}{\sqrt[]{a}}=\dfrac{1}{\sqrt[]{b}}\Leftrightarrow a=b=1\)

\(\Rightarrow GTLN\left(M\right)=2\left(khi.a=b=1\right)\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)

\(P=\sqrt{\dfrac{yz}{x^2+1}}+\sqrt{\dfrac{zx}{y^2+1}}+\sqrt{\dfrac{xy}{z^2+1}}\)

\(P=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}+\sqrt{\dfrac{zx}{y^2+xy+yz+zx}}+\sqrt{\dfrac{xy}{z^2+xy+yz+zx}}\)

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

\(P\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right)+\dfrac{1}{2}\left(\dfrac{z}{y+z}+\dfrac{x}{x+y}\right)+\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{3}{2}\)

\(P_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(a=b=c=\sqrt{3}\)