\(1)\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{1}{3}\)

\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)\)

\(=\frac{1}{3}.2\)

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

\(1)\)\(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{1}{3}\)

\(=\frac{1}{3}.\left(\frac{4}{5}+\frac{6}{5}-1\right)\)

\(=\frac{1}{3}.\left(\frac{4}{5}+\frac{6}{5}-\frac{5}{5}\right)\)

\(=\frac{1}{3}.1\)

\(=\frac{1}{3}\)

\(2)\)\(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}\)

\(=\frac{1}{7}.\frac{27}{26}-\frac{1}{7}.\frac{1}{26}\)

\(=\left(\frac{27}{26}-\frac{1}{26}\right).\frac{1}{7}\)

\(=1.\frac{1}{7}\)

\(=\frac{1}{7}\)

\(3)\)\(15\left(\frac{1}{5}-\frac{7}{15}+\frac{17}{3}\right)\)

\(=3-7+85\)

\(=81\)

\(4)\)\(38\left(2\frac{1}{19}-5\frac{7}{38}\right)\)

\(=38\left(\frac{39}{19}-\frac{197}{38}\right)\)

\(=78-197\)

\(=-119\)

\(5)\)\(75\left(\frac{-4}{25}+7\frac{1}{15}-5\frac{7}{75}\right)\)

\(=75\left(\frac{-4}{25}+\frac{106}{15}-\frac{382}{75}\right)\)

\(=-12+530-382\)

\(=136\)

Ddaeng

1, \(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{1}{3}\)

\(=\frac{1}{3}.\left(\frac{4}{5}+\frac{6}{5}\right)-\frac{1}{3}\)

\(=\frac{1}{3}.2-\frac{1}{3}\)

\(=\frac{2}{3}-\frac{1}{3}\)

\(=\frac{1}{3}\)

2, \(\frac{3}{7}.\frac{9}{26}-\frac{1}{14}.\frac{1}{13}\)

\(=\frac{1}{7}.3.\frac{9}{26}-\frac{1}{7}.\frac{1}{2}.\frac{1}{13}\)

\(=\frac{1}{7}.\frac{27}{26}-\frac{1}{7}.\frac{1}{26}\)

\(=\frac{1}{7}.\left(\frac{27}{26}-\frac{1}{26}\right)\)

\(=\frac{1}{7}.1\)

\(=\frac{1}{7}\)

3, \(15.\left(\frac{1}{5}-\frac{7}{15}+\frac{17}{3}\right)\)

\(=15.\frac{1}{5}-15.\frac{7}{15}+15.\frac{17}{3}\)

\(=3-7+85\)

\(=81\)

4, \(38.\left(2\frac{1}{19}+5\frac{7}{38}\right)\)

\(=38.\left(\frac{39}{19}+\frac{197}{38}\right)\)

\(=38.\frac{39}{19}+38.\frac{197}{38}\)

\(=78+197\)

\(=275\)

5, \(75.\left(\frac{-4}{25}+7\frac{1}{15}-5\frac{7}{75}\right)\)

\(=75.\left(\frac{-4}{25}+\frac{106}{15}-\frac{382}{75}\right)\)

\(=75.\frac{-4}{25}+75.\frac{106}{15}-75.\frac{382}{75}\)

\(=-12+530-382\)

\(=136\)

Giải:

a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\) 

=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\) 

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

=\(\dfrac{7}{5}+\dfrac{-1}{4}\) 

=\(\dfrac{23}{20}\) 

b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\) 

=\(\left[0+0\right]:\dfrac{2017}{2018}\) 

=0\(:\dfrac{2017}{2018}\) 

=0

c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)

=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) 

=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10

a: =25(15+45*3)

=25*150

=3750

b: \(=-10\left(25+75-50\right)=-10\cdot50=-500\)

c: =>3^x-2=27

=>x-2=3

=>x=5

d: =>2x-5=-4

=>2x=1

=>x=1/2

e: =>2(x-1)^2=32

=>(x-1)^2=16

=>x-1=4 hoặc x-1=-4

=>x=-3 hoặc x=5

f:  =>25(x+3)=75

=>x+3=3

=>x=0

Giải:

a) \(\left(9\dfrac{4}{9}+5\dfrac{2}{3}\right)-5\dfrac{1}{2}\) 

\(=\left(\dfrac{85}{9}+\dfrac{17}{3}\right)-\dfrac{11}{2}\) 

\(=\dfrac{136}{9}-\dfrac{11}{2}\) 

\(=\dfrac{173}{18}\) 

b) \(\dfrac{13}{9}.\dfrac{15}{4}-\dfrac{13}{9}.\dfrac{7}{4}-\dfrac{13}{9}.\dfrac{5}{4}\) 

\(=\dfrac{13}{9}.\left(\dfrac{15}{4}-\dfrac{7}{4}-\dfrac{5}{4}\right)\) 

\(=\dfrac{13}{9}.\dfrac{3}{4}\) 

\(=\dfrac{13}{12}\) 

c) \(\dfrac{2}{3}+\dfrac{5}{8}-\dfrac{-1}{3}+0,375\) 

\(=\left(\dfrac{2}{3}-\dfrac{-1}{3}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)\) 

\(=1+1\) 

\(=2\)

d) \(75\%-3\dfrac{1}{2}+1,5:\dfrac{10}{7}\) 

\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{3}{2}:\dfrac{10}{7}\) 

\(=\dfrac{3}{4}+\dfrac{7}{2}+\dfrac{21}{20}\) 

\(=\dfrac{53}{10}\) 

e) \(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\) 

\(=\dfrac{28}{15}.\dfrac{1}{4}.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\) 

\(=\dfrac{7}{5}+\dfrac{-47}{60}:\dfrac{47}{24}\) 

\(=\dfrac{7}{5}+\dfrac{-2}{5}\) 

\(=1\)

1) 58× 75+ 58× 50 -58 × 25

= 58( 75 + 50 - 25 )

= 58 . 100

= 5800

1/ 58 x 75 + 58 x 50 - 58 x 25

= 58 x ( 75 + 50 - 25 )

= 58 x 100

= 5800

2/= \(2\frac{3}{5}-\frac{3}{5}\cdot\frac{11}{84}\)

\(=\frac{13}{5}-\frac{3}{5}\cdot\frac{11}{84}\)

\(=\frac{13}{5}-\frac{11}{140}\)

\(=\frac{364-11}{140}\)

\(=\frac{353}{140}\)

3/  \(\frac{1}{3}-\frac{5}{4}\cdot\frac{4}{15}\)

\(=\frac{1}{3}-\frac{1}{3}\)

\(=0\)