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Ta có: 1+1/2 +1/3 +...+1/98
=(1+1/98 )+(1/2 +1/97 )+(1/3 +1/96 )+...+(1/49 +1/50 )
=99/1.98 +99/2.97 +99/3.96 +...+99/49.50
=99(1/1.98 +1/2.97 +1/3.96 +...+1/49.50 )
⇒A=(1+1/2 +1/3 +...+1/98 ).2.3.4....98
=99(1/1.98 +1/2.97 +1/3.96 +...+1/49.50 ).2.3.4....98chia hết cho 99 (đpcm)
\(1:\frac{99}{100}:\frac{98}{99}:\frac{97}{98}:.........:\frac{2}{3}:\frac{1}{2}\)
\(=1.\frac{100}{99}.\frac{99}{98}.\frac{98}{97}......\frac{3}{2}.\frac{2}{1}\)
\(=\frac{1.100.99.98....3.2}{99.98.97......2.1}\)
\(=100\)
đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\)
\(\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}=\frac{100-1}{1}+\frac{100-2}{2}+...+\frac{100-99}{99}\)
\(=\frac{100}{1}-1+\frac{100}{2}-1+...+\frac{100}{99}-1=\left(\frac{100}{1}+\frac{100}{2}+...+\frac{100}{99}\right)-\left(1+1+...+1\right)\)
\(100+\left(\frac{100}{2}+\frac{100}{3}+...+\frac{100}{99}\right)-99=1+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)\)\(\Rightarrow\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}}=\frac{B}{100B}=\frac{1}{100}\)
a) \(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}.1\frac{1}{24}.........1\frac{1}{99}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}.\frac{25}{24}......\frac{100}{99}\)
\(=\frac{\left(2.2\right).\left(3.3\right).\left(4.4\right).\left(5.5\right)....\left(10.10\right)}{\left(1.3\right).\left(2.4\right).\left(3.5\right).\left(4.6\right).....\left(9.11\right)}\)
\(=\frac{\left(2.3.4.5...10\right).\left(23.4.5....10\right)}{\left(1.2.3.4...9\right).\left(3.4.5.6....11\right)}=\frac{10}{1}.\frac{2}{11}=\frac{20}{11}\)
b) \(\frac{99}{98}-\frac{98}{97}+\frac{1}{97.98}=\frac{99}{98}-\frac{98}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)-\left(\frac{98}{97}-\frac{1}{97}\right)\)
\(=\frac{98}{98}-\frac{97}{97}=1-1=0\)
đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\)
\(\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}=\frac{100-1}{1}+\frac{100-2}{2}+...+\frac{100-99}{99}=\frac{100}{1}-1+\frac{100}{2}-1+...+\frac{100}{99}-1\)
\(=\left(\frac{100}{1}+\frac{100}{2}+...+\frac{100}{99}\right)-\left(1+1+...+1\right)=100+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)-99\)
\(=1+100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}\right)=100\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)=100B\)
\(\Rightarrow\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+...+\frac{1}{99}}=\frac{B}{100B}=\frac{1}{100}\)
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\(\frac{99}{98}+\frac{96}{97}+\frac{1}{97.98}=\frac{99}{98}+\frac{96}{97}+\frac{1}{97}-\frac{1}{98}=\left(\frac{99}{98}-\frac{1}{98}\right)+\left(\frac{1}{97}+\frac{96}{97}\right)=1+1=2\)
= 2
mk bấm máy tính
đúng 10000000000000000000000000000000000000000000000000000000000000000000000%