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24 tháng 8 2022

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

\(=\dfrac{-10}{30}+\dfrac{12}{30}-\dfrac{15}{30}=\dfrac{-13}{30}\)

27 tháng 1 2016

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27 tháng 1 2016

\(7832\)

15: A= 1/3-3/4+3/5+1/2007-1/36+1/15-2/9

Sửa đề: 

A=-3/4-2/9-1/36+1/3+3/5+1/15+1/2007

=-27/36-8/36-1/36+5/15+9/15+1/15+1/2007

=-1+1+1/2007=1/2007

16:

\(A=\dfrac{1}{3}+\dfrac{3}{5}+\dfrac{1}{15}-\dfrac{3}{4}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{64}\)

\(=\dfrac{5+9+1}{15}+\dfrac{-27-8-1}{36}+\dfrac{1}{64}\)

=1/64

17:

=1/2-1/2+2/3-2/3+3/4-3/4+4/5-4/5+5/6-5/6-6/7

=-6/7

`@` `\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

 

26 tháng 8 2019

a

\(A=1+3+3^2+3^3+....+3^{100}\)

\(3A=3+3^2+3^3+3^4+.....+3^{101}\)

\(2A=3^{101}-1\)

\(A=\frac{3^{101}-1}{2}\)

b

\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{99}}\)

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

\(B=1-\frac{1}{2^{99}}\)

c

\(C=5^{100}-5^{99}+5^{98}-5^{97}+....+5^2-5+1\)

\(5C=5^{101}-5^{100}+5^{99}-5^{98}+....+5^3-5^2+5\)

\(6C=5^{101}+1\)

\(C=\frac{5^{101}+1}{6}\)

\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\)

\(\Rightarrow\frac{1}{2}B=\)\(\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{100}\)

\(\Rightarrow B-\frac{1}{2}B=\left[\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+...+\left(\frac{1}{2}\right)^{99}\right]-\left[\left(\frac{1}{2}\right)+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{100}\right]\)

\(\Rightarrow\frac{1}{2}B=\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\Rightarrow B=\left[\frac{1}{2}-\left(\frac{1}{2}\right)^{100}\right].2\)

Đề bài yêu cầu gì?

AH
Akai Haruma
Giáo viên
25 tháng 10 2018

\(B=1+5+5^2+5^3+...+5^{2008}+5^{2009}\)

\(\Rightarrow 5B=5+5^2+5^3+5^4+...+5^{2009}+5^{2010}\)

Trừ theo vế:

\(5B-B=(5+5^2+5^3+5^4+...+5^{2009}+5^{2010})-(1+5+5^2+...+5^{2009})\)

\(4B=5^{2010}-1\)

\(B=\frac{5^{2010}-1}{4}\)

AH
Akai Haruma
Giáo viên
25 tháng 10 2018

\(S=\frac{3^0+1}{2}+\frac{3^1+1}{2}+\frac{3^2+1}{2}+..+\frac{3^{n-1}+1}{2}\)

\(=\frac{3^0+3^1+3^2+...+3^{n-1}}{2}+\frac{\underbrace{1+1+...+1}_{n}}{2}\)

\(=\frac{3^0+3^1+3^2+..+3^{n-1}}{2}+\frac{n}{2}\)

Đặt \(X=3^0+3^1+3^2+..+3^{n-1}\)

\(\Rightarrow 3X=3^1+3^2+3^3+...+3^{n}\)

Trừ theo vế:

\(3X-X=3^n-3^0=3^n-1\)

\(\Rightarrow X=\frac{3^n-1}{2}\). Do đó \(S=\frac{3^n-1}{4}+\frac{n}{2}\)