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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(=3-7+85\)

\(=81\)

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

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

\(=78-197\)

\(=-119\)

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

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

\(=-12+530-382\)

\(=136\)

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

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

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

\(=-12+530-382\)

\(=136\)

1)

25.(75-45)-75.(45-25)

C1: =25.30-75.20

=750-1500

=-750

(-1).(-1)\(^2\).(-1)\(^3\).....(-1)\(^{100}\)

\(\Rightarrow\)(-1).1.(-1).1.....(-1).1

có tất cả 50 số -1

có tất cả 50 số 1

\(\Rightarrow\) \([\)(-1).50\(]\).\([\)1.50\(]\)

=-50.50=0

nhầm -50.50=-2500

Mọi người giúp mình với

câu này mình tính mãi nhưng ko biết làm

mình chịu rồi bạn đợi mấy bạn khác giải dùm nha

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

Bài 1: Thực hiện các phép tính sau:

\(a)\)Chưa rỏ đề

\(b)\)\(5025\div5-25\div5\)

\(=\)\(1005-5\)

\(=\)\(1000\)

\(c)\)\(218-180\div2\div9\)

\(=\)\(218-10\)

\(=\)\(208\)

\(d)\)\(\left(328-8\right)\div32\)

\(=\)\(320\div32\)

\(=\)\(10\)

Bài 1:

a) ( Tôi không nhìn rõ đầu bài )

b) 5025  :  5  -  25  :  5

=  ( 5025  -  25 )  :  5

=        5000   :   5

=       1000

c)   218  -  180  :  2  :  9

=     218   -    180  :  (  2  .  9 )

=      218    -    180   :  18

=       218    -    10

=       208

d)   (  328  -  8  )  :  32

=      320   :   32

=     10 

Bài 2: Thực hiện phép tính:

\(a)\)\(27.75+25.27-150\)

\(=\)\(27.\left(75+25\right)-150\)

\(=\)\(27.100-150\)

\(=\)\(2700-150\)

\(=\)\(2550\)

\(b)\)\(12\div\left\{400\div\left[500-\left(125+25.7\right)\right]\right\}\)

\(=\)\(12\div\left\{400\div\left[500-\left(125+175\right)\right]\right\}\)

\(=\)\(12\div\left\{400\div\left[500-300\right]\right\}\)

\(=\)\(12\div\left\{400\div200\right\}\)

\(=\)\(12\div2\)

\(=\)\(6\)

\(c)\)\(13.17-256\div16+14\div7-1\)

\(=\)\(221-16+2-1\)

\(=\)\(206\)

\(d)\)\(18\div3+182+3.\left(51\div17\right)\)

\(=\)\(6+182+3.3\)

\(=\)\(6+182+9\)

\(=\)\(197\)