\(D=a^{\dfrac{7}{2}}.a^{\dfrac{1}{3}}.a^{\dfrac{7}{4}}=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}=\sqrt[12]{a^{67}}\)

\(D=a^{\sqrt{2}-1}.a^{2\sqrt{2}}.a^{3-3\sqrt{2}}=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{3}}=a^2\)

\(D=\left(\sqrt{a}\right)^7\cdot\left(\sqrt[3]{a}\right)\left(\sqrt[4]{a}\right)^7\)

\(=a^{\dfrac{1}{2}\cdot7}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}\cdot7}\)

\(=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}\)

b: \(D=a^{\sqrt{2}-1}\cdot\left(a^2\right)^{\sqrt{2}}\cdot\left(a^3\right)^{1-\sqrt{2}}\)

\(=a^{\sqrt{2}-1}\cdot a^{2\sqrt{2}}\cdot a^{3-3\sqrt{2}}\)

\(=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{2}}=a^2\)

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

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

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

\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)

\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)

Bài 1:

a)  \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)

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

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

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

b)   \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)

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

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

\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)

c)  ĐK:  \(a\ge0;a\ne1\)

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

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

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

\(=1-a+a=1\)

Lời giải:

a. $=|3+\sqrt{2}|-|3-2\sqrt{2}|=(3+\sqrt{2})-(3-2\sqrt{2})$

$=3\sqrt{2}$

b. $=|\sqrt{7}-2\sqrt{2}|-|\sqrt{7}+2\sqrt{2}|$

$=(2\sqrt{2}-\sqrt{7})-(\sqrt{7}+2\sqrt{2})$

$=-2\sqrt{7}$

c.

$=|3+\sqrt{5}|+|3-\sqrt{5}|=(3+\sqrt{5})+(3-\sqrt{5})=6$

d.

$=|2-\sqrt{3}|-|2+\sqrt{3}|=(2-\sqrt{3})-(2+\sqrt{3})=-2\sqrt{3}$

\(1,ĐKx\ge5\)

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

\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)

\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)

\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)

2a,ĐK \(x\ge0;x\ne9\)

,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)

\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

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

a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)

\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)

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

=-a-1

b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)

\(=\left|3a-5\right|-2a+4\)

\(=5-3a-2a+4\)

=9-5a

c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)

\(=4a-3-\left|2a-1\right|\)

\(=4a-3-2a+1\)

\(=2a-2\)

d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)

\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)

\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)

\(=-a^2\)

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