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\(\dfrac{12,48:0,5.6,25.4.2}{2.3,14.1,25:0,25.10}\)
= \(\dfrac{12,48.2.5.4.2}{2.3,14.1.4.10}\)
= \(\dfrac{12,48.2}{2.3,14}\) = \(\dfrac{12,48}{3,14}\) = \(\dfrac{6,24}{1,57}\)
\(\frac{12,48:0,5.6,25.4.2}{2.3,12.1,25:0,25.10}\)= \(\frac{12,48.2.6.25.4.2}{2.3,12.1,25.4.10}\)= \(\frac{3,12.4.2.1,25.5.4.2}{2.3,12.1,25.4.5.2}\)= \(4\)
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 ~
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\frac{4}{21}\)
\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)
\(B=33\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(C=\frac{1}{2}.\frac{98}{99}\)
\(C=\frac{49}{99}\)
\(B1\)
\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{37}-\frac{1}{38}-\frac{1}{39}\)
\(=1-\frac{1}{39}\)
\(=\frac{38}{39}\)
\(B2\)
\(=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+.....+\frac{1}{99\cdot100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+......+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}\)
\(=\frac{25}{100}-\frac{1}{100}\)
\(=\frac{24}{100}\)
\(=\frac{6}{25}\)
Bài 1 :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}\)
\(=\frac{370}{741}\)
A=1+(1/6+1/12+1/20+...+1/90):2
A=1+(1/2-1/3+1/3-1/4+1/4-1/5+...+1/9-1/10):2
A=1+(1/2-1/10):2
A=1+2/5:2
A=1+1/5
A=6/5
Vậy A=6/5 nha bạn
Đúng 100%
k mk nha
Mk nhanh nhất
\(\frac{-2}{5}.\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{-2}{5}.\left(\frac{2}{17}+\frac{-2}{5}\right)\)
\(=\frac{-2}{5}.\frac{5}{17}-\frac{-2}{5}.\frac{3}{5}-\frac{-2}{5}.\frac{2}{17}-\frac{-2}{5}.\frac{-2}{5}\)
\(=\frac{-2}{5}.\left(\frac{5}{17}-\frac{2}{17}\right)-\frac{-2}{5}.\left(\frac{3}{5}+\frac{-2}{5}\right)\)
\(=\frac{-2}{5}.\frac{3}{17}-\frac{-2}{5}.\frac{1}{5}\)
\(=\frac{-2}{5}.\left(\frac{3}{17}-\frac{1}{5}\right)\)
\(=\frac{-2}{5}.\frac{-2}{85}\)
\(=\frac{4}{425}\)
\(\frac{-2}{5}.\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{-2}{5}.\left(\frac{2}{17}+\frac{-2}{5}\right)\)
= \(\frac{-2}{5}.\frac{-26}{85}-\frac{-2}{5}.\frac{-24}{85}\)
= \(\frac{-2}{5}.\left(\frac{-26}{85}-\frac{-24}{85}\right)\)
= \(\frac{-2}{5}.\frac{-2}{85}\)
= \(\frac{4}{425}\)
oi kho