Tính A, biết: (tính nhanh, ko tính máy tính)
\(A=\frac{1}{2.32}+\frac{1}{3.33}+\frac{1}{4.34}\)
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KN
10 tháng 9 2019
\(A=\frac{1}{2.32}+\frac{1}{3.33}+...+\frac{1}{1973.2003}\)
\(=\frac{1}{30}\left(\frac{1}{2}-\frac{1}{32}+\frac{1}{3}-\frac{1}{33}+...+\frac{1}{1973}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1973}-\frac{1}{32}-\frac{1}{33}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}-\frac{1}{1974}-\frac{1}{1975}-...-\frac{1}{2003}\right)\)
\(B=\frac{1}{2.1974}+\frac{1}{3.1975}+...+\frac{1}{31.2003}\)
\(=\frac{1}{1972}\left(\frac{1}{2}-\frac{1}{1974}+\frac{1}{3}-\frac{1}{1975}+...+\frac{1}{31}-\frac{1}{2003}\right)\)
\(=\frac{1}{1972}\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}-\frac{1}{1974}-\frac{1}{1975}-...-\frac{1}{2003}\right)\)
Vậy \(\frac{A}{B}=\frac{1972}{30}\)
5 tháng 8 2019
\(P=...\)
\(=\frac{1}{30}\left(\frac{30}{2.32}+\frac{30}{3.33}+...+\frac{30}{1973.2003}\right)\)
\(=\frac{1}{30}\left(\frac{1}{2}-\frac{1}{32}+\frac{1}{3}-\frac{1}{33}+...+\frac{1}{1973}-\frac{1}{2003}\right)\)
\(=\frac{1}{30}\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1973}\right)-\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{2003}\right)\right]\)
\(=\frac{1}{30}\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}\right)-\left(\frac{1}{1974}+\frac{1}{1975}+...+\frac{1}{2003}\right)\right]\)
5 tháng 8 2019
\(Q=...\)
\(=\frac{1}{1972}\left(\frac{1972}{2.1974}+\frac{1972}{3.1975}+...+\frac{1}{31.2003}\right)\)
\(=\frac{1}{1972}\left(\frac{1}{2}-\frac{1}{1974}+\frac{1}{3}-\frac{1}{1975}+...+\frac{1}{31}-\frac{1}{2003}\right)\)
\(=\frac{1}{1972}\left[\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{31}\right)-\left(\frac{1}{1974}+\frac{1}{1975}+...+\frac{1}{2003}\right)\right]\)
DD
3
M
11 tháng 10 2017
\(\frac{33.44^2+55^3.33}{45.33^2-99.33^4}\)
\(=\frac{33.121.16+121.25.33}{5.9.33^2-11.9.33^2.33^2}\)
\(=\frac{33.121\left(16+25\right)}{1089.9.\left(5-11.1089\right)}\)
\(=\frac{3993.41}{9801.\left(-11974\right)}\)
\(=-\frac{163713}{117357174}\).
11 tháng 10 2017
\(\frac{33.44^2+55^3.33}{45.33^2-99.33^4}\)
=\(\frac{33.\left(4.11\right)^2+\left(5.11\right)^3.33}{9.5.33^2-9.11.33^4}\)
=\(\frac{33.16.11^2+125.11^3.33}{33^2.9.\left(5-11.33^2\right)}\)
= \(\frac{33.11^2.\left(16+125.11\right)}{33^2.9.\left(-11974\right)}\)
= \(\frac{132.1391}{33.9.\left(-107766\right)}\)
= \(\frac{183612}{-32006502}\)
mik ko chắc chắn lắm
24 tháng 4 2017
A= đã cho.
=>-A=1/20+1/30+1/42+1/56+1/72+1/90+1/110.
=>-A=1/4*5+1/5*6+1/6*7+1/7*8+1/8*9+1/9*10+1/10*11.
=>-A=1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11.
=>-A=1/4-1/11=7/44.
=>A=-7/44.
thay số 1/52 là 1/56 mới đúng,mk làm rồi.
tk mk nha các bn.
-chúc ai tk mk học giỏi-
25 tháng 4 2018
Ta có :
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
Vậy \(A=\frac{25}{17}\)
Chúc bạn học tốt ~
25 tháng 4 2018
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\frac{4}{21}\)
\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)
\(B=33\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(C=\frac{1}{2}.\frac{98}{99}\)
\(C=\frac{49}{99}\)
30A=30/2*32+30/3*33+30/4*34=1/2-1/32+1/3-1/33+1/4-1/34=99/100
A=3,3/100
frac2/3