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

\(1,=20-7=13\\ b,=12-50=-38\\ c,=\sqrt{7}-2+\sqrt{7}+2=2\sqrt{7}\\ d,=\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}=2\sqrt{3}\\ e,=11+2\sqrt{30}\\ f,=8-2\sqrt{15}\\ g,=11+2\sqrt{6}\)

1) \(=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)

2) \(=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)

3) \(=\sqrt{7}-2+\sqrt{7}+2=2\sqrt[]{7}\)

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

5) \(=5+6-2\sqrt{5.6}=11-2\sqrt{30}\)

6) \(=3+5-2\sqrt{3.5}=8-4\sqrt{2}\)

7) \(=\left(2\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+2\sqrt{2\sqrt{2}.3}=11+2\sqrt{6\sqrt{2}}\)

`A=sqrt{(5-sqrt3)^2}+sqrt{(2-sqrt3)^2}`

`=5-sqrt3+2-sqrt3`

`=7-2sqrt3`

`B=sqrt{(3-sqrt2)^2}-sqrt{(1-sqrt2)^2}`

`=3-sqrt2-(sqrt2-1)`

`=4-2sqrt2`

`C=sqrt{(3+sqrt7)^2}-sqrt{(2-sqrt7)^2}`

`=3+sqrt7-(sqrt7-2)`

`=5`

`D=sqrt{4-2sqrt3}+sqrt{7+4sqrt3}`

`=sqrt{3-2sqrt3+1}+sqrt{4+2.2.sqrt3+3}`

`=sqrt{(sqrt3-1)^2}+sqrt{(2+sqrt3)^2}`

`=sqrt3-1+2+sqrt3=1+2sqrt3`

\(A=\left|5-\sqrt{3}\right|+\left|2-\sqrt{3}\right|=5-\sqrt{3}+2-\sqrt{3}=7-2\sqrt{3}\)

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

\(C=\left|3+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=3+\sqrt{7}-\sqrt{7}+2=5\)

\(D=\sqrt{3-2\sqrt{3}+1}+\sqrt{4+2.2\sqrt{3}+3}\)

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

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

a: Ta có: \(\sqrt{x}\left(\sqrt{x}-3\right)-5\left(\sqrt{x}+3\right)\)

\(=x-3\sqrt{x}-5\sqrt{x}-15\)

\(=x-8\sqrt{x}-15\)

b: Ta có: \(3\left(\sqrt{x}+2\right)+\left(\sqrt{x}+3\right)\left(2-\sqrt{x}\right)\)

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

\(=-x+2\sqrt{x}+12\)

c: Ta có: \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-5\left(\sqrt{x}-1\right)\)

\(=x-9-5\sqrt{x}+5\)

\(=x-5\sqrt{x}-4\)

d: Ta có: \(3\left(\sqrt{x}-2\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

\(=3\sqrt{x}-6-x+1\)

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

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.

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

`F=sqrt{(3-sqrt2)^2}+sqrt{(1-sqrt2)^2}``

`=3-sqrt2+sqrt2-1=2`

`G=sqrt{(5+sqrt7)^2}-sqrt{(2-sqrt7)^2}`

`=5+sqrt7-(sqrt7-2)`

`=5+sqrt7-sqrt7+2=2`

`H=sqrt{(3-sqrt{10})^2}+sqrt{(2-sqrt{10})^2}`

`=sqrt{10}-3+sqrt{10}-2`

`=2\sqrt{10}-5`

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

\(G=\left|5+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=5+\sqrt{7}-\sqrt{7}+2=7\)

\(H=\left|3-\sqrt{10}\right|+\left|2-\sqrt{10}\right|=\sqrt{10}-3+\sqrt{10}-2=2\sqrt{10}-5\)

từ dòng cuối là sai rồi bạn à

Bạn bỏ dòng cuối đi còn lại đúng rồi

Ở tử đặt nhân tử chung căn x chung  rồi lại đặt căn x +1 chung

Ở mẫu tách 3 căn x ra 2 căn x +căn x rồi đặt nhân tử 2 căn x ra 

rút gọn được \(\frac{3\sqrt{x}-5}{2\sqrt{x}+1}\)

cảm ơn bạn nha