Tính : ( 1 + 1/2) x ( 1 + 1/3) x ( 1 + 1/4) x ...... x ( 1+ 1/99)
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vì tử của tất cả các số là 1-1 mà 1-1=0
suy ra:=0+0+0+...+0 (100 số 0)
Suy ra:=0
vậy (1-1/1+2).(1-1/1+2+3).....(1-1/1+2+3+...+99+100)=0
M = \(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}\)
M = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
M = \(1-\dfrac{1}{100}\)
M = \(\dfrac{99}{100}\)
\(M=1\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+...+\dfrac{1}{99}\times\dfrac{1}{100}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(M=1-\dfrac{1}{100}\)
\(M=\dfrac{99}{100}\)
Bài 1:
a: Tổng là:
(-19+19)+(-18+18)+...+20=20
b: Tổng là:
-18+(-17+17)+...+0=-18
\(a,=\dfrac{1}{3}\times\left(\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{1}{3}\times\dfrac{5}{6}=\dfrac{5}{18}\\ b,=\dfrac{4}{5}\times\left(\dfrac{1}{2}-\dfrac{1}{3}\right)=\dfrac{4}{5}\times\dfrac{1}{6}=\dfrac{2}{15}\\ c,=456\times99-6\times99+456\\ =456\times\left(99+1\right)-594\\ =456\times100-594=45600-594=45006\\ d,=101\times\left(101-1\right)=101\times100=10100\)
Làm lại.
Giải:
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}...\frac{99}{100}\)
\(=\frac{1\times2\times3\times4\times...\times99}{2\times3\times4\times5\times6\times...\times100}\)
\(=\frac{1}{100}\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}...\frac{99}{100}\)
\(=\frac{1.2.3.4...99}{2.3.4.5.6...100}\)
\(=\frac{1}{100}\)
Ta có :
\(P=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(P=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}....\frac{9898}{99.100}\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}....\frac{98.101}{99.100}\)
\(P=\frac{1.2.3...98}{2.3.4....99}.\frac{4.5.6....101}{3.4.5...100}\)
\(P=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
Ủng hộ mk nha !!! ^_^
\(P=\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)
\(P=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)
\(P=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5...100}\)
\(P=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)....\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{100}{99}\)
\(=\frac{3.4.5...100}{2.3.4...99}=\frac{100}{2}=50\)