1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)

2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)

3: \(=\sqrt{3}+1-\sqrt{3}=1\)

\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)

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

\(=\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}-1}+\dfrac{\sqrt{6}\left(\sqrt{6}-1\right)}{\sqrt{6}}\)

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

\(=\sqrt{6}+\sqrt{6}-1\)

\(=2\sqrt{6}-1\)

=======================

\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)

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

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

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

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

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

\(=\dfrac{5\sqrt{3}}{-\sqrt{6}}=-\dfrac{5}{\sqrt{2}}\)

a.

\(\sqrt[3]{125}.\sqrt[3]{\frac{16}{10}}.\sqrt[3]{-0,5}=\sqrt[3]{125.\frac{16}{10}.(-0,5)}=\sqrt[3]{-100}\)

b.

\(=1+\frac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}=1+\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}=1+\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2})^3-1}=1+\sqrt[3]{2}-1=\sqrt[3]{2}\)

c.

\(\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}=\sqrt{3}+\sqrt[3]{(\sqrt{3}+1)^3}=\sqrt{3}+\sqrt{3}+1=2\sqrt{3}+1\)

d.

\(\frac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{\sqrt[3]{(\sqrt{3}+1)^3}}=\frac{(\sqrt{3}+1)^2}{\sqrt{3}+1}=\sqrt{3}+1\)

e.

Đặt \(\sqrt[3]{2+10\sqrt{\frac{1}{27}}}=a; \sqrt[3]{2-10\sqrt{\frac{1}{27}}}=b\)

Khi đó:

$a^3+b^3=4$

$ab=\frac{2}{3}$

$E^3=(a+b)^3=a^3+b^3+3ab(a+b)$
$E^3=4+2E$

$E^3-2E-4=0$
$E^2(E-2)+2E(E-2)+2(E-2)=0$

$(E-2)(E^2+2E+2)=0$

Dễ thấy $E^2+2E+2>0$ nên $E-2=0$

$\Leftrightarrow E=2$

a. \(\dfrac{\sqrt{2}.\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{7}.\left(\sqrt{3}+\sqrt{5}\right)}=\dfrac{\sqrt{2}}{\sqrt{7}}=\sqrt{\dfrac{2}{7}}\)

d. \(\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{5-2\sqrt{5}+1}}{\sqrt{5}-1}=\dfrac{\left(\sqrt{5}-1\right)^2}{\sqrt{5}-1}=\sqrt{5}-1\)

\(\sqrt{3-2\sqrt{2}}\)

d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)

\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)

\(=\dfrac{3}{x-y}\)

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

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

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

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

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

Câu 1,2 bạn đã đăng và có lời giải rồi

Câu 3:

\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)