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Đặt A = \(\frac{\frac{1}{2}}{1+2}+\frac{\frac{1}{2}}{1+2+3}+...+\frac{\frac{1}{2}}{1+2+3+....+100}\)
= \(\frac{1}{2}\left(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{100.101:2}\right)\)
= \(\frac{1}{2}\left(\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{100.101}\right)\)
= \(\frac{1}{2}.2\left(\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{100.101}\right)\)
= 1\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{100}-\frac{1}{101}\right)\)
= \(\frac{1}{2}-\frac{1}{101}=\frac{101}{202}-\frac{2}{202}=\frac{99}{202}\)
Từ 1 đến 100 có Số số hạng là :
( 100 -1 ) : 1 + 1 = 100 ( số )
=> TỪ 1/1 -> 100/100 có 100 số
TA có 1/1 + 2/2 + 3/3 + ... + 99/99 + 100/100
= 1 + 1 + 1 + ... + 1
=> có 100 số 1
= 100 x 1 = 100
Các bạn t i ck mk nếu đúng nha !
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+.....+\frac{3}{1+2+...+100}\)
\(=3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\)
\(=\frac{2}{2}.\left(3+\frac{3}{3}+\frac{3}{6}+...+\frac{3}{5050}\right)\)
\(=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)
\(=6.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\right)\)
\(=6.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=6.\left(1-\frac{1}{101}\right)\)
\(=6.\frac{100}{101}=\frac{600}{101}\)
Vậy \(A=\frac{600}{101}\)
\(A=3+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+...+100}\)
\(A=\frac{3.2}{2}+\frac{3.2}{\left(1+2\right).2}+\frac{3.2}{\left(1+2+3\right).2}+...+\frac{3.2}{\left(1+2+...+100\right).2}\)
\(A=\frac{6}{2}+\frac{6}{6}+\frac{6}{12}+...+\frac{6}{10100}\)
\(A=\frac{6}{1.2}+\frac{6}{2.3}+\frac{6}{3.4}+...+\frac{6}{100.101}\)
\(A=6\cdot\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(A=6\cdot\left(1-\frac{1}{101}\right)=6\cdot\frac{100}{101}=\frac{600}{101}\)
Vay A = ........
Ta có:2/2(1+2)+2/2(1+2+3)+2/2(1+2+3+4)+...+2/2(1+2+3+...+100)
=2/6+2/12+2/20+...+2/5050
=2/2.3+2/3.4+2/4.5+...+2/100.101
=2.(1/2.3+1/3.4+1/4.5+...+1/100.101)
=2.(1-1/2+1/3-1/4+1/4-1/5+...+1/100-1/101)
=2.(1-1/101)
=2.100/101
=200/101
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}+\frac{1}{3}\right)=\frac{4}{5}.\frac{3}{3}=\frac{4}{5}.1=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}+\frac{1}{6}\right)=\frac{3}{4}:\frac{2}{3}=\frac{9}{8}\)
\(\frac{2}{3}.\frac{4}{5}-\frac{1}{3}.\frac{4}{5}=\frac{4}{5}\left(\frac{2}{3}-\frac{1}{3}\right)=\frac{4}{5}.\frac{1}{3}=\frac{4}{15}\)
\(\frac{1}{2}:\frac{3}{4}-\frac{1}{6}:\frac{3}{4}=\frac{3}{4}:\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{3}{4}:\frac{1}{3}=\frac{9}{4}\)
\(\frac{2}{3}.\frac{4}{5}+\frac{1}{3}.\frac{4}{5}=\left(\frac{2}{3}+\frac{1}{3}\right).\frac{4}{5}=1.\frac{4}{5}=\frac{4}{5}\)
\(\frac{1}{2}:\frac{3}{4}+\frac{1}{6}:\frac{3}{4}=\frac{1}{2}.\frac{4}{3}+\frac{1}{6}.\frac{4}{3}=\left(\frac{1}{2}+\frac{1}{6}\right).\frac{4}{3}=\frac{2}{3}.\frac{4}{3}=\frac{8}{9}\)
c,d tương tự
\(\frac{3}{1}+\frac{3}{1+2}+\frac{3}{1+2+3}+...+\frac{3}{1+2+3+...+100}\)
\(=3\times\left(\frac{1}{0+1}+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\right)\)
\(=3\times\left(\frac{1}{\left(0+1\right)\times2:2}+\frac{1}{\left(1+2\right)\times2:2}+\frac{1}{\left(1+3\right)\times3:2}+...+\frac{1}{\left(1+100\right)\times100:2}\right)\)
\(=3\times\left(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+...+\frac{2}{100\times101}\right)\)
\(=3\times2\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{100\times101}\right)\)
\(=6\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\right)\)
\(=6\times\left(1-\frac{1}{101}\right)\)
\(=6\times\frac{100}{101}\)
\(=\frac{600}{101}\)
Ủng hộ mk nha ^_-