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

\(=\dfrac{1}{6\sqrt{6}}-\dfrac{6}{\sqrt{6}}=-\dfrac{35}{6\sqrt{6}}\)

b)\(\left(\sqrt{6}+\sqrt{5}\right)^2+\left(\sqrt{6}-\sqrt{5}\right)^2\)

\(=6+2\sqrt{30}+5+6-2\sqrt{30}+5=22\)

2) Ta có: \(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)

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

\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)

\(=-11\)

3) Ta có: \(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)

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

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

\(=4\sqrt{6}-6\sqrt{2}+4\sqrt{5}-2\sqrt{15}\)

`e)(3/2sqrt6+2sqrt{2/3}-4sqrt{3/2})(3/2sqrt6+2sqrt{2/3}+4sqrt{3/2})`

`=(3/2sqrt6+2sqrt{2/3})^2-(4\sqrt{3/2})^2`

`=((3sqrt6)/2+(2sqrt2)/3)^2-16*3/2`

`=((9sqrt6)/6+(4sqrt6)/6)^2-24`

`=((13sqrt6)/6)^2-24`

`=13^2/6-24`

`=25/6`

Giải:

\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right).\left(3\sqrt{\dfrac{2}{3}}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)

\(=\left(\sqrt{\dfrac{27}{2}}+\sqrt{\dfrac{8}{3}}-\sqrt{24}\right).\left(\sqrt{6}-\sqrt{2}-\sqrt{6}\right).\left(-\sqrt{6}\right)\)

\(=\left(\dfrac{\sqrt{6}}{6}\right).\left(-\sqrt{2}\right).\left(-\sqrt{6}\right)\)

\(=\sqrt{2}\)

Vậy ...

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

\(A=\left[2-\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}\right]\left[2+\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}\right]\)

\(A=\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)\)

\(A=2^2-\left(\sqrt{5}\right)^2\)

\(A=4-5\)

\(A=-1\)

____

\(B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)

\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)\)

\(B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)\)

\(B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)\)

\(B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)\)

\(B=6-121\)

\(B=-115\)

6:ĐKXĐ: x>=0; x<>1/25

BPT=>\(\dfrac{3\sqrt{x}}{5\sqrt{x}-1}+3< =0\)

=>\(\dfrac{3\sqrt{x}+15\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{18\sqrt{x}-5}{5\sqrt{x}-1}< =0\)

=>\(\dfrac{1}{5}< \sqrt{x}< =\dfrac{5}{18}\)

=>\(\dfrac{1}{25}< x< =\dfrac{25}{324}\)

7:

ĐKXĐ: x>=0

BPT \(\Leftrightarrow\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}>\dfrac{8}{3}:\dfrac{8}{3}=1\)

=>\(\dfrac{\sqrt{x}+1}{2\sqrt{x}+3}-1>=0\)

=>\(\dfrac{\sqrt{x}+1-2\sqrt{x}-3}{2\sqrt{x}+3}>=0\)

=>\(-\sqrt{x}-2>=0\)(vô lý)

8:

ĐKXĐ: x>=0; x<>9/4

BPT \(\Leftrightarrow\dfrac{\sqrt{x}-2}{2\sqrt{x}-3}+4< 0\)

=>\(\dfrac{\sqrt{x}-2+8\sqrt{x}-12}{2\sqrt{x}-3}< 0\)

=>\(\dfrac{9\sqrt{x}-14}{2\sqrt{x}-3}< 0\)

TH1: 9căn x-14>0 và 2căn x-3<0

=>căn x>14/9 và căn x<3/2

=>14/9<căn x<3/2

=>196/81<x<9/4

TH2: 9căn x-14<0 và 2căn x-3>0

=>căn x>3/2 hoặc căn x<14/9

mà 3/2<14/9

nên trường hợp này Loại

9: 

ĐKXĐ: x>=0

\(BPT\Leftrightarrow\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}< =-\dfrac{1}{3}\)

=>\(\dfrac{2\sqrt{x}+3}{5\sqrt{x}+7}+\dfrac{1}{3}< =0\)

=>\(\dfrac{6\sqrt{x}+9+5\sqrt{x}+7}{3\left(5\sqrt{x}+7\right)}< =0\)

=>\(\dfrac{11\sqrt{x}+16}{3\left(5\sqrt{x}+7\right)}< =0\)(vô lý)

10: 

ĐKXĐ: x>=0; x<>1/49

\(BPT\Leftrightarrow\dfrac{6\sqrt{x}-2}{7\sqrt{x}-1}+6>0\)

=>\(\dfrac{6\sqrt{x}-2+42\sqrt{x}-6}{7\sqrt{x}-1}>0\)

=>\(\dfrac{48\sqrt{x}-8}{7\sqrt{x}-1}>0\)

=>\(\dfrac{6\sqrt{x}-1}{7\sqrt{x}-1}>0\)

TH1: 6căn x-1>0 và 7căn x-1>0

=>căn x>1/6 và căn x>1/7

=>căn x>1/6

=>x>1/36

TH2: 6căn x-1<0 và 7căn x-1<0

=>căn x<1/6 và căn x<1/7

=>căn x<1/7

=>0<=x<1/49

\(a,\dfrac{3}{\sqrt{7}-4}+\dfrac{4+\sqrt{7}}{3}\)

\(=\dfrac{9}{3\left(\sqrt{7}-4\right)}+\dfrac{\left(\sqrt{7}-4\right)\left(\sqrt{7}+4\right)}{3\left(\sqrt{7}-4\right)}\)

\(=\dfrac{9+7-16}{3\left(\sqrt{7}-4\right)}\)

\(=0\)

\(b,\left(\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}\right):\dfrac{1}{2\sqrt{3}}\)

\(=\left[\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\right]\cdot2\sqrt{3}\)

\(=\left(\sqrt{2}+\dfrac{\sqrt{3}-\sqrt{2}}{3-2}\right)\cdot2\sqrt{3}\)

\(=\left(\sqrt{2}+\sqrt{3}-\sqrt{2}\right)\cdot2\sqrt{3}\)

\(=\sqrt{3}\cdot2\sqrt{3}\)

\(=6\)

#\(Toru\)

Ta có: \(\left(\dfrac{3\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{3\sqrt{2}+2\sqrt{3}}{\sqrt{3}+\sqrt{2}}\right)\cdot\dfrac{5-2\sqrt{6}}{4}\)

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

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

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

\(=\dfrac{25-24}{4}=\dfrac{1}{4}\)