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

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

=\(\frac{2\sqrt{a^2-b^2}}{2b}\):​\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)​

=\(\frac{\sqrt{a^2-b^2}}{b}\)*\(\frac{a^2+b^2}{\sqrt{a^2-b^2}}\)

=\(\frac{a^2+b^2}{b}\)

b/ Thế \(b=a-1\)thì ta có

\(P=\frac{a^2+\left(a-1\right)^2}{a-1}=\frac{2a^2-2a+1}{a-1}\)

\(\Leftrightarrow2a^2-\left(2+P\right)a+1+P=0\)

\(\Rightarrow\Delta_a=\left(2+P\right)^2-4.2.\left(1+P\right)\ge0\)

\(\Leftrightarrow P\ge2+2\sqrt{2}\)

\(A=\frac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\frac{2b}{b-a}.\)

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

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

\(=\frac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\frac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\frac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{4\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

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

\(=\left(\frac{\sqrt{a}^3+\sqrt{b}^3}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right)^2\)

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

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

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

1,

\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)

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

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

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

Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)

2, 

a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)

b, Ta có : \(B=\frac{2\sqrt{x}}{x-4}+\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)

\(=\frac{2\sqrt{x}}{x-4}+\frac{\sqrt{x}+2}{x-4}-\frac{\sqrt{x}-2}{x-4}\)

\(=\frac{2\sqrt{x}+2+2}{x-4}=\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{2}{\sqrt{x}-2}\)

c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)

\(a,Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\left(\frac{b}{a-\sqrt{a^2-b^2}}\right)\)

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

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

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

\(b.\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\frac{\sqrt{2}.\sqrt{b}}{2\sqrt{b}}=\frac{\sqrt{2}}{2}\)

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