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BT
3
17 tháng 8 2023
a: =>x-2/5=3/4:1/3=3/4*3=9/4
=>x=9/4+2/5=45/20+8/20=53/20
b: =>x-2/3=7/3:4/5=7/3*5/4=35/12
=>x=35/12+2/3=43/12
c: 1/3(x-2/5)=4/5
=>x-2/5=4/5*3=12/5
=>x=12/5+2/5=14/5
d: =>2/3x-1/3-1/4x+1/10=7/3
=>5/12x-7/30=7/3
=>5/12x=7/3+7/30=77/30
=>x=77/30:5/12=154/25
e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)
=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)
=>x=19/7:23/28=76/23
f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5
=>13/12x=1/5+3/2+4/3+5/4=257/60
=>x=257/65
i: =>x^2-2/5x-x^2-2x+11/4=4/3
=>-12/5x=4/3-11/4=-17/12
=>x=17/12:12/5=85/144
11 tháng 8 2016
a)\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{3}{4}+...+\frac{1}{9}-\frac{1}{10}\)
= \(1+\left(\frac{-1}{2}+\frac{1}{2}\right)+\left(\frac{-1}{3}+\frac{1}{3}\right)+...+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{10}\)
= \(1-\frac{1}{10}\)
=\(\frac{9}{10}\)
b)\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
= \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
=\(1-\frac{1}{11}\)
= \(\frac{10}{11}\)
c) đặt A=\(\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}\)
\(\frac{1}{3}A\)=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)
\(\frac{2}{3}A\)=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(\frac{2}{3}A\)=\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(1+\left(\frac{-1}{3}+\frac{1}{3}\right)+\left(\frac{-1}{5}+\frac{1}{5}\right)+\left(\frac{-1}{7}+\frac{1}{7}\right)+\left(\frac{-1}{9}+\frac{1}{9}\right)-\frac{1}{11}\)
\(\frac{2}{3}A\)=\(\frac{10}{11}\)
A= \(\frac{10}{11}:\frac{2}{3}\)
A= \(\frac{10}{11}.\frac{3}{2}\)=\(\frac{15}{11}\)
d) giả tương tự câu c kết quả \(\frac{25}{11}\)
11 tháng 8 2016
tổng đặc biệt đó bạn
\(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(1-\frac{1}{10}=\frac{9}{10}\)
những câu sau cũng áp dụng như vậy nhé
NC
0
F
2
B
6 tháng 5 2021
A = \(\dfrac{1}{1+3}\) + \(\dfrac{1}{1+3+5}\) + \(\dfrac{1}{1+3+5+7}\) + ... + \(\dfrac{1}{1+3+5+7+...+2021}\)
\(\Leftrightarrow\) A = \(\dfrac{1}{\dfrac{\left(1+3\right).2}{2}}\) + \(\dfrac{1}{\dfrac{\left(1+5\right).3}{2}}\) + \(\dfrac{1}{\dfrac{\left(1+7\right).4}{2}}\) + ... + \(\dfrac{1}{\dfrac{\left(1+2021\right).1011}{2}}\)
= \(\dfrac{2}{2.4}\) + \(\dfrac{2}{3.6}\) + \(\dfrac{2}{4.8}\) + ... + \(\dfrac{2}{1011.2021}\)
= \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + \(\dfrac{1}{4.4}\) + ... + \(\dfrac{1}{2021.2021}\)
A < \(\dfrac{1}{4}\) + ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{2020.2021}\) )
< \(\dfrac{1}{4}\) + ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2020}\) - \(\dfrac{1}{2021}\) )
< \(\dfrac{1}{4}\) + ( \(\dfrac{1}{2}\) - \(\dfrac{1}{2021}\) ) < \(\dfrac{1}{4}\) + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)
Kiểu như vậy hả ?
29 tháng 6 2023
a: 14/5-7/5=7/5
b: 7/8-1/3+5/4
=21/24-8/24+30/24
=43/24
c; =7/6+5/6+2/15+13/15
=2+1
=3
d: =4*5/3*11=20/33
e: =2/9*1/6*1/4=2/9*1/24=1/108
2:
a: \(=\dfrac{3}{9}\cdot\dfrac{4}{4}\cdot\dfrac{5}{5}\cdot\dfrac{6}{6}\cdot\dfrac{7}{7}=\dfrac{1}{3}\)
b: \(=\dfrac{1}{6}\left(\dfrac{22}{3}-\dfrac{2}{3}\right)=\dfrac{10}{3}\cdot\dfrac{1}{6}=\dfrac{10}{18}=\dfrac{5}{9}\)
c; \(=\dfrac{1}{3}\left(9-\dfrac{2}{5}-\dfrac{3}{5}\right)=\dfrac{8}{3}\)
NH
0