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\(=\frac{23^{10}\left(1-23\right)}{23^{10}\left(23+21\right)}=\frac{-22}{44}=\frac{-1}{2}\)
\(\frac{23^{10}-23^{11}}{23^{11}+21.23^{10}}\)
\(=\frac{23^{10}.\left(1-23\right)}{23^{10}.\left(23+21\right)}\)
\(=\frac{-22}{44}=\frac{-22}{22.2}=\frac{-1}{2}\)
\(\frac{23^{10}-23^{11}}{23^{11}+21.23^{10}}=\frac{23^{10}\left(1-23\right)}{23^{10}\left(23+21\right)}=-\frac{22}{44}=-\frac{1}{2}\)
\(C=\dfrac{1}{3\cdot5}+\dfrac{1}{5\cdot7}+...+\dfrac{1}{37\cdot39}\)
\(2C=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{37\cdot39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{39}\)
\(2C=\dfrac{4}{13}\)
\(C=\dfrac{2}{13}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{37.39}\)
\(C=\frac{1}{2}.\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(C=\frac{1}{2}.\frac{12}{39}\)
\(C=\frac{4}{26}=\frac{2}{13}\)
\(A=\frac{1}{2}.\left(\frac{1}{3.5}+\frac{1}{5.7}+...\frac{1}{37.39}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{39}\right)=\frac{1}{2}.\frac{12}{39}=\frac{6}{39}\)
Ta đặt nhân tử chung nha :
\(A=\frac{1}{2}\left(\frac{1}{3.5}+\frac{1}{5.7}+.....+\frac{1}{37.39}\right)\)
\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{39}\right)\)
\(=\frac{1}{2}.\frac{12}{39}\)
\(=\frac{6}{39}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}\)
\(=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-\frac{2}{9}+...+\frac{2}{37}-\frac{2}{39}\)
\(=\frac{2}{3}-\frac{2}{39}\)
\(=\frac{8}{13}\)
Ta có:
\(\frac{2}{3.5}=\frac{1}{3}-\frac{1}{5}\)
\(\frac{2}{5.7}=\frac{1}{5}-\frac{1}{7}\)
\(\frac{2}{7.9}=\frac{1}{7}-\frac{1}{9}\)
\(......................................\)
\(\frac{2}{37.39}=\frac{1}{37}-\frac{1}{39}\)
nên \(C=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{37.39}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{37}-\frac{1}{39}\)
\(C=\frac{1}{3}-\frac{1}{39}=\frac{4}{13}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{21.23}\)
\(=5-\frac{5}{3}+\frac{5}{3}-\frac{5}{5}+\frac{5}{5}-\frac{5}{7}+...+\frac{5}{21}-\frac{5}{23}\)
\(=5\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{23}\right)\)
\(=5\left(1-\frac{1}{23}\right)\)
\(=5.\frac{22}{23}\)
\(=\frac{110}{23}\)
\(A=\frac{5}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{21}-\frac{1}{23}\right)\)
\(A=\frac{5}{2}.\left(1-\frac{1}{23}\right)\)
\(A=\frac{5}{2}.\frac{22}{23}\)
\(A=\frac{55}{23}\)
a, \(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{37.39}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\)
\(=\dfrac{1}{3}-\dfrac{1}{39}\)
\(=\dfrac{12}{39}\)
Vậy \(A=\dfrac{12}{39}\)
b,\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{73.76}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{73}-\dfrac{1}{76}\)
\(=1-\dfrac{1}{76}\)
\(=\dfrac{75}{76}\)
Vậy \(B=\dfrac{75}{76}\)
a) Ta có :
\(A=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+....................+\dfrac{2}{37.39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...................+\dfrac{1}{37}-\dfrac{1}{39}\)
\(A=\dfrac{1}{3}-\dfrac{1}{39}=\dfrac{4}{13}\)
b) Ta có :
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+..................+\dfrac{3}{73.76}\)
\(B=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+..................+\dfrac{1}{73}-\dfrac{1}{76}\)
\(B=1-\dfrac{1}{76}=\dfrac{75}{76}\)
~ Học tốt ~
62×21=1302
37×39=1443
11×39=429
21×23=483
=1443+1302−429−483=1833
=1833
xin 1 gp
1833
mik đang gấp nên ko kịp trình bày cho bạn nha!
sorry bạn!!!
Ta có: \(37\cdot39+62\cdot21-11\cdot39-21\cdot23\)
=\(39\left(37-11\right)+21\left(62-23\right)\)
\(=39\cdot26+21\cdot39=39\cdot\left(26+21\right)=39\cdot47=1833\)