a,\(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5=\left(\sqrt{\dfrac{25}{16}}-\dfrac{3}{4}\right):5=\left(\dfrac{5}{4}-\dfrac{3}{4}\right):5\)

\(=\dfrac{1}{2}:5=\dfrac{1}{10}\)

b,\(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2=\left[\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)\right]^2\)

\(=\left[3-4\right]^2=1\)

c,\(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)=11^2-\left(4\sqrt{3}\right)^2\)

\(=121-48=73\)

d,\(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)

\(=2-2\sqrt{2}+1-3+\dfrac{4\sqrt{2}}{5}+\sqrt{\dfrac{36}{25}.2}\)

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

\(=-2\sqrt{2}+\dfrac{10\sqrt{2}}{5}=-2\sqrt{2}+2\sqrt{2}=0\)

e,\(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{2021-2\sqrt{2021}.1+1}\)

\(=\left(1+\sqrt{2021}\right)\sqrt{\left(\sqrt{2021}-1\right)^2}\)

\(=\left(1+\sqrt{2021}\right)\left(\sqrt{2021}-1\right)\)

\(=\sqrt{2021}-1+\sqrt{2021^2}-\sqrt{2021}=2020\)

\(\sqrt{\dfrac{16}{49}}+\left(\dfrac{1}{2}\right)^3-\left|-\dfrac{4}{7}\right|-\dfrac{7}{8}\)

\(=\dfrac{4}{7}+\dfrac{1}{8}-\dfrac{4}{7}-\dfrac{7}{8}\)

\(=\dfrac{1}{8}-\dfrac{7}{8}=-\dfrac{6}{8}=-\dfrac{3}{4}\)

\(\left|\dfrac{1}{2}-\dfrac{3}{5}\right|\cdot\sqrt{9}+0,5\left(-2\dfrac{3}{5}\right)\)

\(=\left|\dfrac{5-6}{10}\right|\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{1}{10}\cdot3+\dfrac{1}{2}\cdot\dfrac{-13}{5}\)

\(=\dfrac{3}{10}-\dfrac{13}{10}=-\dfrac{10}{10}=-1\)

a) Ta có: \(\left(\dfrac{9}{25}-2\cdot18\right):\left(3\dfrac{4}{5}+0.2\right)\)

\(=\left(\dfrac{9}{25}-36\right):\left(\dfrac{19}{5}+\dfrac{1}{5}\right)\)

\(=\left(\dfrac{9}{25}-\dfrac{900}{25}\right):\dfrac{20}{5}\)

\(=\dfrac{-891}{25}\cdot\dfrac{1}{4}\)

\(=-\dfrac{891}{100}\)

b) Ta có: \(\dfrac{3}{8}\cdot19\dfrac{1}{3}+\dfrac{3}{8}\cdot33\dfrac{1}{3}\)

\(=\dfrac{3}{8}\cdot\dfrac{58}{3}+\dfrac{3}{8}\cdot\dfrac{100}{3}\)

\(=\dfrac{58}{8}+\dfrac{100}{8}\)

\(=\dfrac{158}{8}=\dfrac{79}{4}\)

c) Ta có: \(15\cdot\left(-\dfrac{2}{3}\right)^2-\dfrac{7}{3}\)

\(=15\cdot\dfrac{4}{9}-\dfrac{7}{3}\)

\(=\dfrac{20}{3}-\dfrac{7}{3}\)

\(=\dfrac{13}{3}\)

d) Ta có: \(\dfrac{1}{2}\sqrt{64}-\sqrt{\dfrac{4}{25}}+\left(-1\right)^{2007}\)

\(=\dfrac{1}{2}\cdot8-\dfrac{2}{5}-1\)

\(=4-1-\dfrac{2}{5}\)

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

\(=\dfrac{15}{5}-\dfrac{2}{5}=\dfrac{13}{5}\)

e) Ta có: \(\left(-\dfrac{5}{2}\right)^2:\left(-15\right)-\left(0.45+\dfrac{3}{4}\right)\cdot\left(-1\dfrac{5}{9}\right)\)

\(=\dfrac{25}{4}\cdot\dfrac{-1}{15}-\left(\dfrac{9}{20}+\dfrac{15}{20}\right)\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}-\dfrac{24}{20}\cdot\dfrac{-14}{9}\)

\(=\dfrac{-25}{60}+\dfrac{28}{15}\)

\(=\dfrac{-25}{60}+\dfrac{112}{60}\)

\(=\dfrac{87}{60}=\dfrac{29}{20}\)

f) Ta có: \(\left(-\dfrac{1}{3}\right)-\left(-\dfrac{3}{5}\right)^0+\left(1-\dfrac{1}{2}\right)^2:2\)

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

\(=\dfrac{-4}{3}+\dfrac{1}{4}\cdot\dfrac{1}{2}\)

\(=\dfrac{-4}{3}+\dfrac{1}{8}\)

\(=\dfrac{-32}{24}+\dfrac{3}{24}=\dfrac{-29}{24}\)

g) Ta có: \(\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{4}\right)^{20}\)

\(=\left(\dfrac{1}{2}\right)^{15}\cdot\left(\dfrac{1}{2}\right)^{40}\)

\(=\left(\dfrac{1}{2}\right)^{55}\)

\(=\dfrac{1}{2^{55}}\)

h) Ta có: \(\dfrac{5^4\cdot20}{25^5\cdot4^5}\)

\(=\dfrac{5^4\cdot5\cdot2^2}{5^{10}\cdot2^{10}}\)

\(=\dfrac{5^5}{5^{10}}\cdot\dfrac{2^2}{2^{10}}\)

\(=\dfrac{1}{5^5}\cdot\dfrac{1}{2^8}\)

\(=\dfrac{1}{800000}\)

1)(-1/2)^2:1/4-2.(-1/2)^3+căn 4

=1/4:1/4-2.-1/8+2

= 1-(-1/4)+2

=1+1/4+2=13/4

2) 3-(-6/7)^0+căn 9 :2

= 3-1+3:2

=3-1+3/2=7/2

3) (-2)^3+1/2:1/8-căn 25 + |-64|

= -8+4-5+64= 55

4) (-1/2)^4+|-2/3|-2007^0

= 1/16+2/3-1

= -13/48

5) = 178/495:623/495-17/60:119/120

= 2/7-2/7=0

6) [2^3.(-1/2)^3+1/2]+[25/22+6/25-3/22+19/25+1/2]

= [-1+1/2]+[(25/22-3/22)+(6/25+19/25)+1/2]

= -1/2+[1+1+1/2]

= -1/2+5/2=2

Mấy cái dấu chấm đó là  nhân nha bn!

a: \(=\dfrac{3}{4}-\dfrac{5}{6}+\dfrac{3}{2}=\dfrac{9-10+18}{12}=\dfrac{17}{12}\)

b: \(=\left(\dfrac{1}{9}+\dfrac{6}{9}\right)^2-\dfrac{1}{3}=\dfrac{49}{81}-\dfrac{27}{81}=\dfrac{22}{81}\)

c; \(=\dfrac{5}{11}\left(-\dfrac{3}{7}-\dfrac{5}{7}\right)+\dfrac{-8}{7}\cdot\dfrac{6}{11}=\dfrac{-8}{7}\left(\dfrac{5}{11}+\dfrac{6}{11}\right)=-\dfrac{8}{7}\)

d: \(=\dfrac{2^{26}}{2^{15}\cdot2^{12}}=\dfrac{1}{2}\)

a) \(4.\left(-\dfrac{1}{2}\right)^3-2.\left(-\dfrac{1}{2}\right)^2+3.\left(-\dfrac{1}{2}\right)+1\)

\(=4.\left(-\dfrac{1}{8}\right)-2.\dfrac{1}{4}+3.\left(-\dfrac{1}{2}\right)+1\)

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

\(=-\dfrac{3}{2}\)

b) \(8.\sqrt{9}-\sqrt{64}\)

\(=8.3-8\)

\(=24-8\)

\(=16\)

c) \(\sqrt{\dfrac{9}{16}}+\dfrac{25}{46}:\dfrac{5}{23}-\dfrac{7}{4}\)

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

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

\(=\dfrac{3}{2}\)

56:54=

d) Ta có: \(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)

\(=\dfrac{5\sqrt{x}-6-2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-9+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

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

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

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

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

\(=\dfrac{3x-9\sqrt{x}+3\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

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