a)

\(-12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)

\(=-12:\left(\dfrac{18}{24}-\dfrac{20}{24}\right)^2\)

\(=-12:\left(\dfrac{-1}{12}\right)^2\)

\(=-12:\dfrac{1}{144}\)

\(=-12\times\dfrac{144}{1}\)

\(=-1728\)

b)

\(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)

\(=\left(4\times\dfrac{3}{4}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)

\(=\left(3-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)

\(=\dfrac{5}{2}\times\dfrac{6}{5}-17\)

\(=3-17\)

\(=-14\)

a)\(=-12:\left(-\dfrac{1}{12}\right)^2\)
\(=-12:\dfrac{1}{144}\)\(=-12.144=-1728\)
b)\(=\left(8:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\left(6-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
\(=\dfrac{11}{2}.\dfrac{6}{5}-17=\dfrac{33}{5}-17=\dfrac{33}{5}-\dfrac{85}{5}=-\dfrac{2}{5}\)

   

a: Ta có: \(3\sqrt{2}\cdot5\sqrt{6}\cdot4\sqrt{12}\)

\(=\sqrt{18\cdot25\cdot6\cdot16\cdot12}\)

\(=\sqrt{518400}\)

=720

b: Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)^2+2\sqrt{14}\)

\(=9-2\sqrt{14}+2\sqrt{14}\)

=9

c: Ta có: \(\left(1+\sqrt{5}+\sqrt{6}\right)\left(1+\sqrt{5}-\sqrt{6}\right)\)

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

\(=2\sqrt{5}\)

Bài 2: 

a: A=(12,87+14,13)+(-14,7-37,3)

=27-52=-25

b: B=-1/3+2/5-2/3-3/5+1/5

=-1

a, = 3.25 + 15.4 - 16 : 2 = 75 + 60 - 8 = 135 - 8 = 127

a: =75+60-8=127

\(a,\dfrac{3}{5}-\dfrac{1}{2}\sqrt{1\dfrac{11}{25}}=\dfrac{3}{5}-\dfrac{1}{2}\sqrt{\dfrac{36}{25}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{\sqrt{6^2}}{\sqrt{5^2}}=\dfrac{3}{5}-\dfrac{1}{2}.\dfrac{6}{5}=\dfrac{3}{5}-\dfrac{6}{10}=\dfrac{3}{5}-\dfrac{3}{5}=0\)

\(b,\left(5+2\sqrt{6}\right)\left(5-2\sqrt{6}\right)=5^2-\left(2\sqrt{6}\right)^2=25-2^2.\sqrt{6^2}=25-4.6=25-24=1\)

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

\(d,\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\left(dk:x,y>0\right)\\ =\dfrac{\left(\sqrt{x^2}.\sqrt{y}+\sqrt{y^2}.\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\\ =\sqrt{x^2}-\sqrt{y^2}\\ =\left|x\right|-\left|y\right|\\ =x-y\)

4:

a: =4/15-2,9+11/15=1-2,9=-1,9

b: \(=-36,75+3,7-63,25+6,3=10-100=-90\)

c: \(=6,5+3,5-\dfrac{10}{17}-\dfrac{7}{17}=10-1=9\)

d: \(=\dfrac{13}{25}\left(-39,1-60,9\right)=\dfrac{13}{25}\left(-100\right)=-52\)

e: =-5/12-7/12-3,7-6,3=-1-10=-11

f: =2,8(-6/13-7/13)-7,2=-2,8-7,2=-10

cậu tách ra từng bài thôi nha.

\(a,\dfrac{2}{3}-\dfrac{5}{6}+\dfrac{3}{4}\)

\(=(\dfrac{2}{3}+\dfrac{3}{4})-\dfrac{5}{6}\)

\(=\dfrac{7}{12}\)

\(b,\dfrac{6}{7}.\dfrac{8}{13}+\dfrac{6}{13}.\dfrac{9}{7}-\dfrac{-4}{13}.\dfrac{6}{7}\)

\(=\dfrac{6}{7}.(\dfrac{8}{13}-\dfrac{4}{13})+(\dfrac{6}{13}.\dfrac{9}{7})\)

\(=(\dfrac{6}{7}.\dfrac{4}{13})+\dfrac{54}{91}\)

\(=\dfrac{24}{91}+\dfrac{54}{91}\)

\(=\dfrac{78}{91}=\dfrac{6}{7}\)

\(c,19.(-26)\)

\(=-494\)

a) `2/3 - 5/6 +3/4`

`=4/6-5/6+3/4`

`=-1/6+3/4`

`=-4/24+18/24`

`=14/24=7/12`

b) `6/7 .8/13 +6/13 .9/7 -4/13 .6/7`

`=6/7 .(8/13-4/13) + 6/13 .9/7`

`=6/7 . 4/13 + 6/13 . 9/7`

`= 6/13 .4/7 +6/13 .9/7`

`=6/13 . (4/7+9/7)`

`=6/13 . 13/7`

`=6/7`

c) `19.(-26)`

`=(1-20).(20+6)`

`=20+6-400-120`

`=-494`

a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)

\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)

\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)

b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)

c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

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