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a: =2/5-3/5+3/7=3/7-1/5
=15/35-7/35
=8/35
b: =>5/7:x=4/3
=>x=5/7:4/3=5/7*3/4=15/28
c: =>x-1/3=15/8:4/5=15/8*5/4=75/32
=>x=75/32+1/3=257/96
d: =>2x+1/8=2/7
=>2x=9/56
=>x=9/112
e: =>2x=10/3-5/4-3/4=10/3-2=4/3
=>x=2/3
\(a,\dfrac{2}{5}+\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\\ =\dfrac{2}{5}+\dfrac{3}{7}-\dfrac{3}{5}\\=\left(\dfrac{2}{5}-\dfrac{3}{5}\right)+\dfrac{3}{7}\\ =-\dfrac{1}{5}+\dfrac{3}{7}\\ =-\dfrac{7}{35}+\dfrac{15}{35}\\ =\dfrac{8}{35}\\ b,1-\dfrac{5}{7}:x=-\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=1-\left(-\dfrac{1}{3}\right)\\ =>\dfrac{5}{7}:x=1+\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=\dfrac{3}{3}+\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=\dfrac{4}{3}\\ =>x=\dfrac{5}{7}:\dfrac{4}{3}\\ =>x=\dfrac{5}{7}.\dfrac{3}{4}\\ =>x=\dfrac{15}{28}\\ c,\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)=\dfrac{15}{8}\\ =>x-\dfrac{1}{3}=\dfrac{15}{8}:\dfrac{4}{5}\\ =>x-\dfrac{1}{3}=\dfrac{15}{8}.\dfrac{5}{4}\\ =>x-\dfrac{1}{3}=\dfrac{75}{32}\\ =>x=\dfrac{75}{32}+\dfrac{1}{3}\\ =>x=\dfrac{257}{96}\)
\(d,\dfrac{2}{3}:\left(2x+\dfrac{1}{8}\right)=\dfrac{7}{3}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{3}:\dfrac{7}{3}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{3}.\dfrac{3}{7}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{7}\\ =>2x=\dfrac{2}{7}-\dfrac{1}{8}\\ =>2x=\dfrac{16}{56}-\dfrac{7}{56}\\ =>2x=\dfrac{9}{56}\\ =>x=\dfrac{9}{56}:2\\ =>x=\dfrac{9}{112}\\ e,2x+\dfrac{3}{4}=\dfrac{10}{3}-\dfrac{5}{4}\\ =>e,2x+\dfrac{3}{4}=\dfrac{40}{12}-\dfrac{15}{12}\\ =>2x+\dfrac{3}{4}=\dfrac{25}{12}\\ =>2x=\dfrac{25}{12}-\dfrac{3}{4}\\ =>2x=\dfrac{25}{12}-\dfrac{9}{12}\\ =>2x=\dfrac{16}{12}\\ =>2x=\dfrac{4}{3}\\ =>x=\dfrac{4}{3}:2\\ =>x=\dfrac{4}{6}\\ =>x=\dfrac{2}{3}\)
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 \(\frac{25}{11}\)
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
\(\dfrac{x-1}{-10}=\dfrac{-7}{y}=\dfrac{z+5}{3}=\dfrac{-2}{4}=\dfrac{-1}{2}\)
=>x-1=5 và 7/y=1/2 và z+5=-3/2
=>x=6 và y=14 và z=-13/2
a) \(\dfrac{13}{20}+\dfrac{3}{5}+x=\dfrac{5}{6}\)
\(\Rightarrow\dfrac{5}{4}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{5}{4}\)
\(\Rightarrow x=\dfrac{-5}{12}\)
b) \(x+\dfrac{1}{3}=\dfrac{2}{5}-\dfrac{-1}{3}\)
\(\Rightarrow x+\dfrac{1}{3}=\dfrac{11}{15}\)
\(\Rightarrow x=\dfrac{11}{15}-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{5}\)
c)\(\dfrac{-5}{8}-x=\dfrac{-3}{20}-\dfrac{-1}{6}\)
\(\dfrac{-5}{8}-x=\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-5}{8}-\dfrac{1}{60}\)
\(\Rightarrow x=\dfrac{-77}{120}\)
d) \(\dfrac{3}{5}-x=\dfrac{1}{4}+\dfrac{7}{10}\)
\(\Rightarrow\dfrac{3}{5}-x=\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{3}{5}-\dfrac{19}{20}\)
\(\Rightarrow x=\dfrac{-7}{20}\)
e) \(\dfrac{-3}{7}-x=\dfrac{4}{5}+\dfrac{-2}{3}\)
\(\Rightarrow\dfrac{-3}{7}-x=\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-3}{7}-\dfrac{2}{15}\)
\(\Rightarrow x=\dfrac{-59}{105}\)
g) \(\dfrac{-5}{6}-x=\dfrac{7}{12}+\dfrac{-1}{3}\)
\(\Rightarrow\dfrac{-5}{6}-x=\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-5}{6}-\dfrac{1}{4}\)
\(\Rightarrow x=\dfrac{-13}{12}\)
a: Bạn ghi lại đề nha bạn
b: \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)
=>\(30x+60-6x+30-24x=100\)
=>\(\left(30x-6x-24x\right)+\left(60+30\right)=100\)
=>0x=100-90=10(vô lý)
c: \(\left(x-7\right)\left(x+3\right)< 0\)
TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)
=>-3<x<7
mà x nguyên
nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)
d: -1<2x-1<4
=>\(-1+1< 2x< 4+1\)
=>0<2x<5
=>0<x<2,5
mà x nguyên
nên \(x\in\left\{1;2\right\}\)
\(a.\frac{1}{7}\times\frac{-3}{8}+\frac{-13}{8}==\frac{-3}{56}+\frac{-13}{8}=\frac{-3}{56}+\frac{-91}{56}=\frac{-94}{56}=\frac{-47}{28}\)
\(b.\frac{3}{5}\times\frac{13}{40}-\frac{1}{10}\times\frac{16}{23}=\frac{39}{200}-\frac{8}{115}=\frac{577}{4600}\)
\(c.\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{1}{4}\right):\frac{3}{7}\)
\(=\left(\frac{-3}{4}+\frac{2}{5}\right)\times\frac{7}{3}+\left(\frac{3}{5}+\frac{1}{4}\right)\times\frac{7}{3}\)
\(=\frac{7}{3}\times\left(\frac{-3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{1}{4}\right)\)
\(=\frac{7}{3}\times\left[\left(\frac{-3}{4}+\frac{1}{4}\right)+\left(\frac{2}{5}+\frac{3}{5}\right)\right]\)
\(=\frac{7}{3}\times\left(\frac{-2}{4}+1\right)\)
\(=\frac{7}{3}\times\frac{1}{2}\)
\(=\frac{7}{6}\)
\(d.\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{6}-\frac{5}{12}\right)\)
\(=\frac{7}{8}:\frac{7}{72}+\frac{7}{8}:\left(\frac{-1}{4}\right)\)
\(=\frac{7}{8}\times\frac{72}{7}+\frac{7}{8}\times-4\)
\(=\frac{7}{8}\times\left(\frac{72}{7}+\left(-4\right)\right)\)
\(=\frac{7}{8}\times\frac{44}{7}\)
\(=\frac{11}{2}\)
a) \(\frac{3}{5}:\left(-\frac{1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(-\frac{1}{3}-1\frac{1}{15}\right)\)
\(=\frac{3}{5}:\left(-\frac{1}{15}-\frac{1}{6}-\frac{2}{6}-1+\frac{1}{15}\right)\)
\(=\frac{3}{5}:\left(-\frac{1}{2}-1\right)\)
\(=\frac{3}{5}:\left(-\frac{3}{2}\right)\)
\(=-\frac{2}{5}\)
b) \(\left(-\frac{3}{4}+\frac{5}{13}\right):\frac{2}{7}-\left(2\frac{1}{4}+\frac{8}{13}\right):\frac{2}{7}\)
\(=\left(-\frac{3}{4}+\frac{5}{13}-2+\frac{1}{4}+\frac{8}{13}\right):\frac{2}{7}\)
\(=\left(-\frac{1}{2}+1-2\right):\frac{2}{7}\)
\(=\left(-\frac{1}{2}-1\right):\frac{2}{7}\)
\(=-\frac{3}{2}:\frac{2}{7}\)
\(=-\frac{21}{4}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)
`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)
`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)
`b)`
\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)
`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)
`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)
`c)`
\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)
`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)
a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)
\(=\dfrac{1}{6}-\dfrac{1}{3}\)
\(=\dfrac{-1}{6}\)
b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)
\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)
\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)
\(=\dfrac{85}{36}+\dfrac{-7}{12}\)
\(=\dfrac{16}{9}\)
c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)
\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)
\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{6}{5}\)
\(=\dfrac{-3}{4}+\dfrac{6}{5}\)
\(=\dfrac{9}{20}\)
hey , đề bài sai ròi .
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 ?