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Mình lộn
S = \(\frac{\left(1+90\right)\times90\div2}{4}\)
S = \(\frac{4095}{4}\)
\(\frac{\frac{2}{3}+\frac{1}{4}-\frac{3}{5}}{\frac{2}{3}-\frac{1}{4}+\frac{3}{5}}=\frac{\frac{2}{3}+\frac{1}{4}-\frac{3}{5}}{\frac{2}{3}-\left(\frac{1}{4}-\frac{3}{5}\right)}=\frac{\frac{2}{3}-\frac{7}{20}}{\frac{2}{3}+\frac{7}{20}}=\frac{\frac{19}{60}}{\frac{61}{60}}=\frac{19}{60}\times\frac{60}{61}=\frac{19}{61}\)
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~
ta thấy có thể giản ước 2/3 = 2/3 ,1/4=1/4,3/5=3/5
=> phép tính trên bằng 1
B=-4/5+4/52-4/53+...+4/5200
5B=-4+4/5-4/52+...+4/5201
5B+B=-4+4/5200
6B=-4x5200/5200+4/5200
6B=-4+4x5200/5200
Còn lại bạn tính nốt nha
a) \(3.\frac{5}{4}\)\(-\frac{3^2}{4}\)\(=\frac{3}{2}\)
b)\(\frac{-21}{10}\)\(+\frac{21}{10}\)\(-\frac{3}{4}\)\(-\frac{3}{4}\)\(=\left(\frac{-21}{10}+\frac{21}{10}\right)-\left(\frac{3}{4}+\frac{3}{4}\right)\)
\(=0-\frac{3}{2}\)\(=\frac{-3}{2}\)
c) \(\frac{3}{4}\)\(+\frac{9}{5}-\frac{3}{2}-1\)\(=\left(\frac{3}{4}-\frac{3}{2}\right)+\left(\frac{9}{5}-1\right)\)\(=\frac{-3}{4}\)\(+\frac{4}{5}\)\(=\frac{1}{20}\)
a, Ta có:
\(\frac{1}{2^3}< \frac{1}{1\cdot2\cdot3};\frac{1}{3^3}< \frac{1}{2\cdot3\cdot4};\frac{1}{4^3}< \frac{1}{3\cdot4\cdot5};...;\frac{1}{n^3}< \frac{1}{\left[n-1\right]n\left[n+1\right]}\)
\(\Rightarrow\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{3^3}+...+\frac{1}{n^3}< \frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}\)
Đặt \(A'=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}\)
\(\Rightarrow\frac{1}{2}A'=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{\left[n-1\right].n}-\frac{1}{n\left[n+1\right]}\)
\(\frac{1}{2}A'=\frac{1}{1\cdot2}-\frac{1}{n\left[n+1\right]}=\frac{1}{2}-\frac{1}{n\left[n+1\right]}=\frac{1}{4}-\frac{1}{2n\left[n+1\right]}< \frac{1}{4}\)
Vậy \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{\left[n-1\right]n\left[n+1\right]}< \frac{1}{4}\Leftrightarrow\frac{1}{2^3}+\frac{1}{3^3}+\frac{1}{4^3}+...+\frac{1}{n^3}< \frac{1}{4}\)
b,
\(C=\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}=1+\frac{1}{3}+1+\frac{1}{3^2}+1+\frac{1}{3^3}+...+1+\frac{1}{3^{98}}\)
\(=\left[1+1+1+...+1\right]+\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right]=98+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
Đặt \(C'=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)
\(\Rightarrow3C'=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{97}}\)
\(\Rightarrow3C'-C'=\left[1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{97}}\right]-\left[\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right]=1-\frac{1}{3^{98}}\)
\(\Rightarrow C'=\frac{1-\frac{1}{3^{98}}}{2}< 1\)
\(\Rightarrow98+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}< 98+1=99< 100\)
\(\Rightarrow\frac{4}{3}+\frac{10}{9}+\frac{28}{27}+...+\frac{3^{98}+1}{3^{98}}< 100\)
c,
\(D=\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{39}}\)
\(4D=5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}\)
\(4D-D=\left[5+\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}\right]-\left[\frac{5}{4}+\frac{5}{4^2}+...+\frac{5}{4^{38}}+\frac{5}{4^{39}}\right]\)
\(3D=5-\frac{5}{4^{39}}\Leftrightarrow D=\frac{5-\frac{5}{4^{39}}}{3}< \frac{5}{3}\)
Vậy:...........
AI THẤY ĐÚNG NHỚ ỦNG HỘ NHA
a ) \(\frac{5}{7}.\frac{-7}{9}-\frac{5}{7}.\frac{2}{9}-\frac{5}{9}\)
\(S=\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+....+\frac{90}{4}\)
\(=\frac{1+2+3+...+90}{4}\)
\(=\frac{\frac{90\left(90+1\right)}{2}}{4}\)
\(=\frac{4095}{4}\)
\(S=\frac{1}{4}+\frac{2}{4}+\frac{3}{4}+\frac{4}{4}+\frac{5}{4}+...+\frac{90}{4}\)
\(\Rightarrow S=\frac{1+2+3+4+5+...+90}{4}\)
=> S có 90 số hạng
\(\Rightarrow S=\frac{\left(90+1\right).90:2}{4}\)
\(S=\frac{4095}{4}\)