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\(A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+....+\frac{10}{5^{10}}+\frac{11}{5^{11}}\)
\(\Rightarrow5A=1+\frac{2}{5}+\frac{3}{5^2}+....+\frac{10}{5^9}+\frac{11}{5^{10}}\)
\(\Rightarrow5A-A=\left(1+\frac{2}{5}+...+\frac{11}{5^{10}}\right)-\left(\frac{1}{5}+\frac{2}{5^2}+...+\frac{11}{5^{11}}\right)\)
\(\Rightarrow4A=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)(1)
Đặt \(B=1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)
\(\Rightarrow5B=5+1+\frac{1}{5}+...+\frac{1}{5^9}\)
\(\Rightarrow5B-B=\left(5+1+...+\frac{1}{5^9}\right)-\left(1+\frac{1}{5}+...+\frac{1}{5^{10}}\right)\)
\(\Rightarrow4B=5-\frac{1}{5^{10}}< 5\)
\(\Rightarrow B< \frac{5}{4}\)(2)
Thay (2) vào (1) \(\Rightarrow4A< \frac{5}{4}-\frac{11}{5^{11}}< \frac{5}{4}\)
\(\Rightarrow A< \frac{5}{16}\left(đpcm\right)\)
B. 1/3 - 1/3 - 3/5 +3/5 + 5/7 - 5/7 + 9/11 - 9/11 -11/13 + 11/ 13 + 7/9 + 13/15
= 0 -0-0-0-0+7/9 +13/15
= 74/45
a: =>x-8/5=1/20-1/10=-1/20
=>x=-0,05+1,6=1,55
b: =>x-3/2=4/3 hoặc x-3/2=-4/3
=>x=17/6 hoặc x=1/6
c: =>\(\left|x-\dfrac{1}{3}\right|=\dfrac{5}{2}-\dfrac{1}{4}+\dfrac{2}{3}=\dfrac{35}{12}\)
=>x-1/3=35/12 hoặc x-1/3=-35/12
=>x=39/12=13/4 hoặc x=-31/12
d: =>|x-5/8|=3/4
=>x-5/8=3/4 hoặc x-5/8=-3/4
=>x=11/8 hoặc x=-1/8
\(A=\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\Rightarrow5A=1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{11}{5^{10}}\)
\(\Rightarrow4A=5A-A=1+\left(\frac{1}{5}+\frac{1}{5^2}+\frac{...1}{5^{10}}\right)-\frac{11}{5^{11}}\)
\(< 1+\left(\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\right)< 1+\frac{1}{4}=\frac{5}{4}\)
\(\Rightarrow A< \frac{5}{4}:4=\frac{5}{16}\)
Lưu ý : \(M=\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\Rightarrow5M=1+\frac{1}{5}+...+\frac{1}{5^9}\Rightarrow4M=5M-M=1-\frac{1}{5^{10}}\)
\(\Rightarrow M=\frac{1}{4}-\frac{1}{5^{10}}:4< \frac{1}{4}\)
\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)
\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)
\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)
\(=-\frac{7}{9}.5\)
\(=-7\)
a)Bn Kaito Kid làm rùi!
B)Không viết lại đề
\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)
c)Không viết lại đề
\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)
\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)
\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)
D)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
\(\Leftrightarrow x:\frac{1}{45}=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{2}:\frac{1}{45}=\frac{45}{2}\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
c) \(\frac{4-3x}{2x+5}=0\Leftrightarrow4-3x=0\)
\(\Leftrightarrow3x=4\Rightarrow x=\frac{4}{3}\)
d) \(\left(x-2\right).\left(x+\frac{2}{3}\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{3}{2}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{3}{2}< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{3}{2}\end{matrix}\right.\end{matrix}\right.\)
Bài 2:
a) \(x:\left(\frac{2}{9}-\frac{1}{5}\right)=\frac{8}{16}\)
=> \(x:\frac{1}{45}=\frac{1}{2}\)
=> \(x=\frac{1}{2}.\frac{1}{45}\)
=> \(x=\frac{1}{90}\)
Vậy \(x=\frac{1}{90}.\)
b) \(\left(2x-1\right).\left(2x+3\right)=0\)
=> \(\left\{{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}2x=0+1=1\\2x=0-3=-3\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=1:2\\x=\left(-3\right):2\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{2};-\frac{3}{2}\right\}.\)
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