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1.
a.
\(\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{7}\right)\)
\(=\frac{1}{3}+\frac{1}{5}-\frac{1}{7}\)
\(=\frac{35-21-15}{105}\)
\(=-\frac{1}{105}\)
b.
\(\frac{3}{5}-\left(\frac{3}{4}-\frac{1}{2}\right)\)
\(=\frac{3}{5}-\frac{3}{4}+\frac{1}{2}\)
\(=\frac{12-15+10}{20}\)
\(=\frac{7}{20}\)
c.
\(\frac{4}{7}-\left(\frac{2}{5}+\frac{1}{3}\right)\)
\(=\frac{4}{7}-\frac{2}{5}-\frac{1}{3}\)
\(=\frac{60-42-35}{105}\)
\(=-\frac{17}{105}\)
2.
a.
\(S=-\frac{1}{1\times2}-\frac{1}{2\times3}-\frac{1}{3\times4}-...-\frac{1}{\left(n-1\right)\times n}\)
\(S=-\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{\left(n-1\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
b.
\(S=-\frac{4}{1\times5}-\frac{4}{5\times9}-\frac{4}{9\times13}-...-\frac{4}{\left(n-4\right)\times n}\)
\(S=-\left(\frac{4}{1\times5}+\frac{4}{5\times9}+\frac{4}{9\times13}+...+\frac{4}{\left(n-4\right)\times n}\right)\)
\(S=-\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{n-4}-\frac{1}{n}\right)\)
\(S=-\left(1-\frac{1}{n}\right)\)
\(S=-1+\frac{1}{n}\)
Chúc bạn học tốt
Ta có:
(-3/2:3/-4)*(-9/2)-1/4<x/8<-1/2:3/4:1/8+1
Xét VT = (-3/2.-4/3).(-9/2)-1/4
= 2.-9/2-1/4
=-9-1/4=-37/4=--222/24
Xét VP = -1/2:3/4:1/8+1
=-1/2.4/3.8+1
=-16/3+1
=-13/3=-104/24
=>-222/24<x/8<-104/24=>-222/24<x.3/24<-104/24=>-222<x.3<-104
=>x.3={-221;-220;...;--105}Mà x.3 chia hết cho 3=>x.3 thuộc{-219;-216;...;-105}
=>x={-73;-72;.....-35}
Vậy ..........
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+\frac{2}{2018}+\frac{3}{2017}+...+\frac{2018}{2}+\frac{2019}{1}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{1}{2019}+1+\frac{2}{2018}+1+\frac{3}{2017}+1+...+\frac{2018}{2}+1+1}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{\frac{2020}{2019}+\frac{2020}{2018}+\frac{2020}{2017}+...+\frac{2020}{2}+\frac{2020}{2020}}\)
\(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}}{2020\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2020}\right)}\)
\(\frac{A}{B}=\frac{1}{2020}\)
\(2A=1+\frac{1}{2}+...+\frac{1}{2^{49}}\)
\(2A-A=1-\frac{1}{2^{50}}\)
\(A=1-\frac{1}{2^{50}}\)=> A bé hơn 1
tương tự nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\)
\(2A=2.\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{48}}+\frac{1}{2^{49}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{48}}+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{49}}+\frac{1}{2^{50}}\right)\)
\(A=1-\frac{1}{2^{50}}< 1\)
- \(\frac{11}{125}-\frac{17}{18}-\frac{5}{7}+\frac{4}{9}+\frac{17}{14}\)
\(=\left(-\frac{17}{18}+\frac{4}{9}\right)+\left(-\frac{5}{7}+\frac{17}{14}\right)+\frac{11}{125}\)
\(=-1+\frac{1}{2}+\frac{11}{125}\)
\(=-1+\frac{147}{125}\)
\(=\frac{22}{125}\)
2. \(1-\frac{1}{2}+2-\frac{2}{3}+3-\frac{3}{4}+4-\frac{1}{4}-3-\frac{1}{3}-2-\frac{1}{2}-1\)
\(=\left(1+2+3+4-3-2-1\right)\)\(+\left(-\frac{1}{2}-\frac{1}{2}\right)+\left(-\frac{2}{3}-\frac{1}{3}\right)+\left(-\frac{3}{4}-\frac{1}{4}\right)\)
\(=4-1-1-1\)
\(=1\)
a) \(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\Leftrightarrow\frac{3}{4}x=\frac{1}{2}\Leftrightarrow x=\frac{1}{2}\times\frac{4}{3}\Leftrightarrow x=\frac{2}{3}\)
b)\(1\frac{3}{4}x+1\frac{1}{2}=-\frac{4}{5}\Leftrightarrow\frac{7}{4}x+\frac{3}{2}=-\frac{4}{5}\Leftrightarrow\frac{7}{4}x=-\frac{23}{10}\)
\(\Leftrightarrow x=-\frac{23}{10}\times\frac{4}{7}\Leftrightarrow x=-\frac{46}{35}\)
c)\(\frac{3}{4}x+\frac{2}{5}x=1,2\Leftrightarrow x\left(\frac{3}{4}+\frac{2}{5}\right)=1,2\Leftrightarrow\frac{23}{20}x=1,2\)
\(\Leftrightarrow x=1,2\times\frac{20}{23}\Leftrightarrow x=\frac{24}{23}\)
d)\(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\Leftrightarrow\frac{1}{7x}=\frac{3}{14}-\frac{3}{7}\Leftrightarrow\frac{1}{7x}=-\frac{3}{14}\Leftrightarrow14=-3\times7x\)
\(\Leftrightarrow-21x=14\Leftrightarrow x=-\frac{2}{3}\)
e) \(-\frac{3}{4}-\left|\frac{4}{5}-x\right|=-1\Leftrightarrow\left|\frac{4}{5}-x\right|=-\frac{3}{4}+1\Leftrightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=-\frac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{11}{20}\\x=\frac{21}{20}\end{matrix}\right.\)
a, \(\frac{1}{4}+\frac{3}{4}x=\frac{3}{4}\\ \Rightarrow\frac{3}{4}x=\frac{1}{2}\\ \Rightarrow x=\frac{2}{3}\)
Vậy \(x=\frac{2}{3}\)
b, \(1\frac{3}{4}x+1\frac{1}{2}=\frac{-4}{5}\\ \frac{7}{4}x+\frac{3}{2}=\frac{-4}{5}\\ \Rightarrow\frac{7}{4}x=\frac{-23}{10}\\ \Rightarrow x=\frac{-46}{35}\)
Vậy \(x=\frac{-46}{35}\)
c, \(\frac{3}{4}x+\frac{2}{5}x=1,2\\ x\left(\frac{3}{4}+\frac{2}{5}\right)=\frac{6}{5}\\ x\cdot\frac{23}{20}=\frac{6}{5}\\ \Rightarrow x=\frac{24}{23}\)
Vậy \(x=\frac{24}{23}\)
d, \(\frac{3}{7}+\frac{1}{7}:x=\frac{3}{14}\\ \Rightarrow\frac{1}{7}:x=\frac{-3}{14}\\ \Rightarrow x=\frac{-2}{3}\)
Vậy \(x=\frac{-2}{3}\)
e, \(\frac{-3}{4}-\left|\frac{4}{5}-x\right|=-1\\ \Rightarrow\left|\frac{4}{5}-x\right|=\frac{1}{4}\\ \Rightarrow\left[{}\begin{matrix}\frac{4}{5}-x=\frac{1}{4}\\\frac{4}{5}-x=\frac{-1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{11}{20}\\x=\frac{21}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{11}{20};\frac{21}{20}\right\}\)
2: =>2x-1/4=5/6-1/2x
=>5/2x=5/6+1/4=13/12
=>x=13/30
3: =>3x-5/6=2/3-1/2x
=>3,5x=2/3+5/6=4/6+5/6=9/6=3,2
hay x=32/35