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`@` `\text {Ans}`

`\downarrow`

`1)`

`5/7*37 13/23 - 51 13/23*5/7`

`= 5/7* (37 13/23 - 51 13/23)`

`= 5/7* (-14)`

`= -10`

`2)`

`-2/3 +1/3+0,5+2 1/2`

`= -2/3 + 1/3 + 1/2 + 5/2`

`= (-2/3+1/3) + (1/2+5/2)`

`= -1/3 + 3`

`=8/3`

`3)`

`-0,5+2/3+1/2`

`= -1/2 + 2/3 + 1/2`

`= (-1/2 + 1/2) + 2/3`

`= 2/3`

`4)`

`(8+2 1/3-3/5) -(5+0,4)-(3 1/2 -2)`

`= 8+ 7/3 - 3/5 - 5 - 0,4 - 7/2 + 2`

`= (8+2-5) + (-3/5 - 2/5) + (7/3 - 7/2)`

`= 5 - 1 - 7/6`

`= 4 - 7/6 = 17/6`

`5)`

`(2/9-7/12):3/4+(16/9-5/12):3/4`

`= (2/9 - 7/12) \times 4/3 + (16/9 - 5/12) \times 4/3`

`= 4/3 *(2/9 - 7/12 + 16/9 - 5/12)`

`= 4/3 * [(2/9 + 16/9) + (-7/12 - 5/12)]`

`= 4/3 * ( 2 - 1)`

`= 4/3 * 1 = 4/3`

`6)`

`-(2021.0,7+19,75) +0,7- (8-19,75)`

`= -2021*0,7 -19,75 + 0,7 - 8 + 19,75`

`= 0,7*(-2021 + 1) - 8`

`= -1414-8`

`= -1422`

`7)`

`15/34+7/21+19/34-20/15`

`= (15/34 + 19/34) + 7/21 - 20/15`

`= 1 + 7/21 - 20/15`

`= 4/3 - 20/15 =0`

`8)`

`2 5/6+1/6:(-5/8)`

`= 17/6 + (-4/15)`

`= 77/30`

`9)`

`(-2)^2 +2/9. (4/5-2/3)`

`= 4 + 2/9*2/15`

`= 4+4/135`

`= 544/135`

`10)`

`(-1/5+3/7):5/4+(-4/5+4/7):5/4`

`= (-1/5+3/7) * 4/5 + (-4/5+4/7) * 4/5`

`= 4/5*(-1/5 +3/7-4/5+4/7)`

`= 4/5*[(-1/5-4/5)+(3/7+4/7)]`

`= 4/5* (-1+1)`

`= 4/5*0=0`

`11)`

`2022,2021 . 1954,1945+ 2022,2021 . (-1954,1945)`

`= 2022,2021 * [1954,1945 + (-1954,1945)]`

`= 2022,2021*0 `

`= 0`

`12)`

`-5,2 .72 +69,1 +5,2 . (-28)+(-1,1)`

`= -5,2*72 + 69,1 - 5,2*28 - 1,1`

`= -5,2*(72+28) + (69,1 - 1,1)`

`= -5,2*100 + 68`

`= -520 + 68`

`= -452`

`13)`

`(7 -1/2-3/4) : (5-1/4-5/8)`

`= 23/4 \div 33/8`

`=46/33`

`14)`

`(8+ 2 1/3 -3/5) -(5+0,4) -( 3 1/3 - 2)`

`= 8+ 2 1/3 - 3/5 - 5 - 0,4 - 3 1/3 + 2`

`= (8+2-5) + (2 1/3 - 3 1/3) - (0,6 + 0,4) `

`= 5 - 1 - 1`

`= 3`

11 tháng 6 2023

help lười tính quá

 

\(\left(2+4+6+...+100\right).\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]:\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

Để í ngoặc \(\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]\)

\(\Leftrightarrow\left[\frac{6}{7}+-\frac{6}{7}\right]\)

\(\Leftrightarrow0\)

Vậy biểu thức \(\left(2+4+6+...+100\right).\left[\frac{3}{5}:0,7+3.\frac{-2}{7}\right]:\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)có giá trị bằng 0

8 tháng 4 2023

\(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{1\dfrac{1}{6}-\dfrac{7}{8}+0,7}\\ =\dfrac{2\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{\dfrac{2}{6}-\dfrac{2}{8}+\dfrac{2}{10}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\\ =\dfrac{2}{7}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}-\dfrac{1}{10}\right)}\\ =\dfrac{2}{7}-\dfrac{2}{7}=0\)

8 tháng 4 2023

phân số cuối là \(\dfrac{2}{7}-\dfrac{2\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}{7\left(\dfrac{1}{6}-\dfrac{1}{8}+\dfrac{1}{10}\right)}\) nha :vv

7 tháng 4 2020

a) \(\frac{2^7\cdot9^3}{6^5\cdot8^2}=\frac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\frac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\frac{3}{2^4}=\frac{3}{16}\)

c) \(\frac{5^4\cdot20^4}{25^4\cdot4^5}=\frac{5^4\cdot\left(2^2\cdot5\right)^4}{\left(5^2\right)^4\cdot\left(2^2\right)^5}=\frac{5^4\cdot2^8\cdot5^4}{5^8\cdot2^{10}}=\frac{1}{2^2}=\frac{1}{4}\)

d) \(\frac{\left(5^4\cdot20^4\right)^3}{125^4}=\frac{5^{12}\cdot20^{12}}{\left(5^3\right)^4}=\frac{5^{12}\cdot\left(2^2\cdot5\right)^{12}}{5^{12}}=2^{24}\cdot5^{12}\)

a: \(=\dfrac{9}{13}\cdot\dfrac{4}{5}=\dfrac{36}{65}\)

b: \(=\dfrac{-7}{10}:\dfrac{3}{2}=\dfrac{-7}{10}\cdot\dfrac{2}{3}=\dfrac{-14}{30}=-\dfrac{7}{15}\)

c: \(=\dfrac{7}{6}\left(3+\dfrac{1}{4}-\dfrac{1}{4}\right)=\dfrac{7}{6}\cdot3=\dfrac{7}{2}\)

28 tháng 7 2018

k mk đi

ai k mk 

mk sẽ k lại

thanks

tích mình với

ai tích mình

mình tích lại

thanks

30 tháng 7 2019

A=[2+4+6+...+100][3/5:0,7+3[-2/7]]:[1/2+1/4+1/6+...+1/100]

A=[2+4+6+...+100][6/7+[-6/7]]:[1/2+1/4+1/6+...+1/100]

A=[2+4+6+...+100][0]:[1/2+14+1/6+...+1/100]

A=0

CHỈ MK CÁCH VIẾT PHÂN SỐ ĐI

11 tháng 7 2023

1) \(5-\left(1+\dfrac{1}{3}\right):\left(1-\dfrac{1}{3}\right)\)

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

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

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

\(=5-2\)

\(=3\)

b) \(\left(1+\dfrac{2}{3}-\dfrac{5}{4}\right)-\left(1-\dfrac{5}{4}\right)+2022-\dfrac{2}{3}\)

\(=1+\dfrac{2}{3}-\dfrac{5}{4}-1+\dfrac{5}{4}++2022-\dfrac{2}{3}\)

\(=\left(1-1\right)+\left(\dfrac{2}{3}-\dfrac{2}{3}\right)+\left(-\dfrac{5}{4}+\dfrac{5}{4}\right)+2022\)

\(=0+0+0+2022\)

\(=2022\)

2) \(0,7^2\cdot x=0,49^2\)

\(\Rightarrow x=\dfrac{0,49^2}{0,7^2}\)

\(\Rightarrow x=\left(\dfrac{0,49}{0,7}\right)^2\)

\(\Rightarrow x=\left(0,7\right)^2\)

\(\Rightarrow x=0,49\)

b) \(x:\left(-0,5\right)^3=\left(0,5\right)^2\)

\(\Rightarrow x=\left(0,5\right)^2\cdot\left(-0,5\right)^3\)

\(\Rightarrow x=\left(-0,5\right)^5\)

\(\Rightarrow x=-\dfrac{1}{32}\)

2:

a: =>x*0,49=0,49^2

=>x=0,49

b: =>x=(0,5)^2*(-1)*(0,5)^3=-(0,5)^5