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

câu 9 nhầm đề bài r bạn

\(3\sqrt{9a^6}-6a^3=3\left|3a^3\right|-6a^3\)

Xét \(a\ge0\Rightarrow\) biểu thức \(=9a^3-6a^3=3a^3\)

Xét \(a< 0\Rightarrow\) biểu thức \(=-9a^3-6a^3=-15a^3\)

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

\(=1-x+3x-1\left(\dfrac{1}{3}< x\le1\right)=2x\)

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

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

\(\left(\sqrt{10}+\sqrt{2}\right)\left(6-2\sqrt{5}\right)\sqrt{3+\sqrt{5}}=\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)^2\sqrt{2}.\sqrt{3+\sqrt{5}}\)

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

\(=\left(\sqrt{5}+1\right)^2\left(\sqrt{5}-1\right)^2=4^2=16\)

\(\sqrt{23-8\sqrt{7}}+\sqrt{8-2\sqrt{7}}=\sqrt{\left(2\sqrt{7}-4\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}\)

\(=2\sqrt{7}-4+\sqrt{7}-1=3\sqrt{7}-5\)

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)

\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)

\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)

\(=\sqrt{x-1}+1+1-\sqrt{x-1}=2\)

\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}\)

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

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

Xét \(x\ge8\Rightarrow\sqrt{x-4}\ge2\Rightarrow\)biểu thức \(=\sqrt{x-4}+2+\sqrt{x-4}-2\)

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

Xét \(x< 8\Rightarrow\sqrt{x-4}< 2\Rightarrow\) biểu thức \(=\sqrt{x-4}+2+2-\sqrt{x-4}=4\)

ta có: \(\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)=1\)

Đặt  : \(\sqrt{\left(5-2\sqrt{6}\right)^x}=a\) ; \(a\ge0\)

=> \(a+\frac{1}{a}=10\)

\(\Leftrightarrow a^2+1=10a\)

\(\Leftrightarrow a^2-10a+1=0\)

\(\Leftrightarrow a^2-2.5a+25-24=0\)

\(\Leftrightarrow\left(a-5\right)^2=24\)

\(\Leftrightarrow\orbr{\begin{cases}a-5=2\sqrt{6}\\a-5=-2\sqrt{6}\end{cases}\Rightarrow\orbr{\begin{cases}a=2\sqrt{6}+5\\a=5-2\sqrt{6}\end{cases}}}\)(TMĐK)

với \(a=5+2\sqrt{6}\)

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

\(\Rightarrow\frac{1}{\left(5+2\sqrt{6}\right)^x}=\left(5+2\sqrt{6}\right)^2\)

=> x= -2

với \(a=5-2\sqrt{6}\)

=> \(\sqrt{\left(5-2\sqrt{6}\right)^x}=5-2\sqrt{6}\) =>x=2

vậy x=2 hoặc x=-2

Nhận xét : \(\sqrt{\left(5-2\sqrt{6}\right)^x}.\sqrt{\left(5+2\sqrt{6}\right)^x}=1\)

Ta đặt \(\sqrt{\left(5-2\sqrt{6}\right)^x}=a\Rightarrow\sqrt{\left(5+2\sqrt{6}\right)^x}=\frac{1}{a}\)

Khi đó phương trình ban đầu trở thành :

\(a+\frac{1}{a}=10\Rightarrow a^2-10a+1=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=5+2\sqrt{6}\\a=5-2\sqrt{6}\end{cases}}\)

+) Với \(a=5+2\sqrt{6}\Rightarrow\sqrt{\left(5-2\sqrt{6}\right)^x}=5+2\sqrt{6}\)

\(\Leftrightarrow\left(5-2\sqrt{6}\right)^x=\left(5+2\sqrt{6}\right)^2=\left(\frac{1}{5-2\sqrt{6}}\right)^2\)

\(\Leftrightarrow x=-2\)

+) Với \(a=5-2\sqrt{6}\Rightarrow\sqrt{\left(5-2\sqrt{6}\right)^x}=5-2\sqrt{6}\)

\(\Leftrightarrow\left(5-2\sqrt{6}\right)^x=\left(5-2\sqrt{6}\right)^2\)

\(\Leftrightarrow x=2\)

Vậy \(x\in\left\{-2,2\right\}\) thỏa mãn đề.

\(pt\Leftrightarrow\left(5-2\sqrt{6}\right)^{\frac{x}{2}}+\left(5+2\sqrt{6}\right)^{\frac{x}{2}}=10\)

Thấy rằng \(5-2\sqrt{6}\) là nghịch đảo của \(5+2\sqrt{6}\), Vì vậy 

\(\left(5-2\sqrt{6}\right)^{\frac{x}{2}}\left(5+2\sqrt{6}\right)^{\frac{x}{2}}=1\)

Đặt \(\left(5-2\sqrt{6}\right)^{\frac{x}{2}}=t\) ta dc pt sau 

\(t+\frac{1}{t}=10\Rightarrow t^2-10t+1=0\Rightarrow t=5\pm2\sqrt{6}\)

Vì vậy \(t=5\pm2\sqrt{6}=\left(5-2\sqrt{6}\right)^{\pm1}=\left(5-2\sqrt{6}\right)^{\frac{x}{2}}\)

Suy ra \(\frac{x}{2}=\pm1\Rightarrow x=\pm2\) 

\(\left(5-2\sqrt{6}\right)^{\frac{x}{2}}+\left(5+2\sqrt{6}\right)^{\frac{x}{2}}=10\)

\(pt\Leftrightarrow\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^{2x}}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^{2x}}=10\)

\(\Leftrightarrow\left(\sqrt{3}-\sqrt{2}\right)^x+\left(\sqrt{3}+\sqrt{2}\right)^x=10\)

\(\Leftrightarrow\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^x}+\left(\sqrt{3}+\sqrt{2}\right)^x=10\)

\(\Leftrightarrow\frac{1}{t}+t=10\left(t=\left(\sqrt{3}+\sqrt{2}\right)^x\right)\)

\(\Leftrightarrow t^2-10t+1=0\)\(\Leftrightarrow t=5\pm2\sqrt{6}\)

\(\Rightarrow5\pm2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^x\)

\(\Leftrightarrow\left(\sqrt{3}+\sqrt{2}\right)^{\pm2}=\left(\sqrt{3}+\sqrt{2}\right)^x\)

\(\Rightarrow x=\pm2\). Vậy...