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A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).............\left(\frac{1}{9801}-1\right).\left(\frac{1}{10000}-1\right)\)
A=\(\left(\frac{1-4}{4}\right).\left(\frac{1-9}{9}\right).\left(\frac{1-16}{16}\right).............\left(\frac{1-9801}{9801}\right).\left(\frac{1-10000}{10000}\right)\)
A=\(\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....................\frac{-9800}{9801}.\frac{-9999}{10000}\)
A=\(\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}.....................\frac{-98.100}{99^2}.\frac{-99.101}{100^2}\)
A=\(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right)....................\left(-98\right).\left(-99\right)\right].\left(3.4.5............100.101\right)}{\left(2.3.4.........99.100\right).\left(2.3.4...............99.100\right)}\)
A=\(\frac{1.101}{100.2}\)=\(\frac{101}{200}\)
2
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.................+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}\)
\(\frac{1}{3.2}+\frac{1}{6.2}+\frac{1}{10.2}+.................+\frac{2}{2.x.\left(x+1\right)}=\frac{1}{2}.\frac{2015}{2017}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x+1}{2.\left(x+1\right)}-\frac{2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{\left(x+1\right)-2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x-1}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
=>\(\frac{x-1}{x+1}=\frac{2015}{2017}.\frac{1}{2}:\frac{1}{2}\)
\(\frac{x-1}{x+1}=\frac{2015}{2017}\)
=>x+1=2017
=>x=2018-1
=>x=2016
Vậy x=2016
Còn bài 3 em ko biết làm em ms lớp 6
Chúc anh học tốt
a)\(\left(\frac{-1}{3}\right)^3\cdot x=\frac{1}{81}\) \(< =>\frac{-1}{27}x=\frac{1}{81}\)\(< =>x=\frac{-1}{3}\)
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a) \(\left(x-\frac{2}{5}\right).\left(x+\frac{3}{7}\right)<0\)
\(\Rightarrow x-\frac{2}{5}<0\) hoặc \(x-\frac{2}{5}>0\)
\(x+\frac{3}{7}>0\) \(x+\frac{3}{7}<0\)
\(\Rightarrow x<\frac{2}{5}\) hoặc \(x>\frac{2}{5}\)
\(x>-\frac{3}{7}\) \(x<-\frac{3}{7}\)
\(\Rightarrow-\frac{3}{7} hoặc \(x\in rỗng\)
vậy \(-\frac{3}{7}
b) \(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)\le x\le\frac{1}{24}-\left(\frac{1}{8}-\frac{1}{3}\right)\)
\(\frac{-1}{12}\le x\le\frac{1}{4}\)
\(\frac{-1}{12}\le x\le\frac{3}{12}\)
\(\Rightarrow x=\frac{-1}{12};0;\frac{1}{12};\frac{2}{12};\frac{3}{12}\)
\(\left|5x-3\right|-3x=12\)
\(\Leftrightarrow\left|5x-3\right|=12+3x\)
\(\Leftrightarrow\hept{\begin{cases}5x-3=12+3x\\-\left(5x-3\right)=12+3x\end{cases}\Rightarrow\hept{\begin{cases}5x-3x=12+3\\-5x+3=12+3x\end{cases}\Rightarrow}\hept{\begin{cases}2x=15\\-5x-3x=12-3\end{cases}}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\-8x=9\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\x=\frac{-9}{8}\end{cases}}}\)
a)\(\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4\)
=> 2x + 7 = 4
2x = 4 - 7
2x = -3
x = -3 : 2
x = -1,5
Vậy x = -1,5
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\(\left(\frac{3}{4}\right)^{2x-1}=\left(\frac{3}{4}\right)^{5x-4}\)
\(\Rightarrow2x-1=5x-4\)
\(\Rightarrow2x-5x=1-4\)
\(\Rightarrow-3x=-3\)
\(\Rightarrow x=1\)
Vậy x = 1
vì (3/4)^2x-1=(3/4)^5x-4=> 2x-1=5x-4=> 2x=5x-3
3=5x-2x=3x=> x=1