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1 ( 131,4 - 80,8 ) : 2,3 + 21,84 x 2 = 50,6 : 2,3 + 43,68 = 22 + 43,68= 65,68
1/2*3+1/3*4+1/4*5 + 1/5*6 + .... + 1/99 * 100
= 1/2 -1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 +..... + 1/99 - 1/100
= 1/2 - 1/100
= 49/100 nha bạn !
1/2x3+1/3x4+1/4x5+1/5x6+....+1/99x100
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+....+1/99-1/100
=1/2-1/100=49/100
1) =1/2 x 2/3 x 3/4 x 4/5 x .... x 2002/2003 x 2003/2004
=1/2004
2) 1/2 x X-3/4=5/6
1/2 x X =3/4+5/6
1/2 x X =19/12
X=19/6
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{2002}{2003}.\frac{2003}{2004}\)
\(=\frac{1.2.3...2002.2003}{2.3.4...2003.2004}=\frac{1}{2004}\)
\(\frac{1}{2}.x-\frac{3}{4}=\frac{5}{6}\)
\(\frac{1}{2}.x=\frac{5}{6}+\frac{3}{4}\)
\(\frac{1}{2}.x=\frac{10}{12}+\frac{9}{12}=\frac{19}{12}\)
\(x=\frac{19}{12}:\frac{1}{2}\)
\(x=\frac{19}{12}.2=\frac{19}{6}\)
B = \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x........x\left(1-\frac{1}{2003}\right)x\left(1-\frac{1}{2004}\right)\)
B = \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x.........x\frac{2002}{2003}x\frac{2003}{2004}\)
=> B = \(\frac{1}{2004}\)
\(B=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{2003}\right)\times\left(1-\frac{1}{2004}\right)\)
\(B=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2002}{2003}\times\frac{2003}{2004}\)
\(B=\frac{1\times2\times3\times...\times2002\times2003}{2\times3\times4\times...\times2003\times2004}\)
\(\Rightarrow B=\frac{1}{2004}\)
\(B=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{2003}\right)\times\left(1-\frac{1}{2004}\right)\)
\(\Rightarrow B=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2002}{2003}\times\frac{2003}{2004}\) (rút gọn từ trên tử xuống dưới mẫu nhé)
\(\Rightarrow B=\frac{1}{2004}\)
\(A=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2002}{2003}x\frac{2003}{2004}\)
\(A=\frac{1x\left(2x3x4x...x2002x2003\right)}{\left(2x3x4x...x2002x2003\right)x2004}=\frac{1}{2004}\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot....\cdot\left(1-\frac{1}{2003}\right)\cdot\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot....\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot4\cdot....\cdot2002\cdot2003}{2\cdot3\cdot4\cdot5\cdot....\cdot2003\cdot2004}\)
\(=\frac{1}{2004}\)
phân số thứ nhất bằng 0/2 =0
vậy tích đó chắc chắn bằng 0
chuẩn ko mn?