Tính: 3/4.8/9.15/16. ... .48/49
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\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{99.101}{100.100}\)
\(A=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
B = 3/4.8/9.15/16.....2499/2500
= 1.3/2^2 . 2.4/3^2 . 3.5/4^2 ..... 49.51/50^2
= 1.2.3.....49/1.2.3.....50 . 3.4.5....51/1.2.3.....50
= 1/50 . 51/2
= 51/100
Giải
B = 3/4.8/9.15/16.....2499/2500
B=1.3/2^2 . 2.4/3^2 . 3.5/4^2 ..... 49.51/50^2
B=1.2.3.....49/2.3.4.....50 . 3.4.5.....51/2.3.4.....50
B=1/50 . 51/2
B=51/100
\(M=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\frac{9999}{10000}\)
\(=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot\cdot\cdot\cdot\frac{99.101}{100.100}\)
\(=\frac{1}{2}\cdot\frac{101}{100}=\frac{101}{200}\)
Xét vế phải :
\(VP=\frac{99}{50}-\frac{97}{49}+...+\frac{7}{4}-\frac{5}{3}+\frac{3}{2}-1\)
\(=2.\left(\frac{99}{100}-\frac{97}{98}+...+\frac{7}{8}-\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(=2\left[\left(1-\frac{1}{100}\right)-\left(1-\frac{1}{98}\right)+...+\left(1-\frac{1}{4}\right)-\left(1-\frac{1}{2}\right)\right]\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{98}-\frac{1}{100}\right)\)
\(=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{49}+\frac{1}{50}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{50}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{25}+\frac{1}{26}+...+\frac{1}{50}\right)-\left(1+\frac{1}{2}+...+\frac{1}{25}\right)\)
\(=\frac{1}{26}+\frac{1}{27}+...+\frac{1}{49}+\frac{1}{50}=VT\Rightarrow\left(đpcm\right)\)
\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{99}{100}=\frac{1\cdot3}{2\cdot2}=\frac{2\cdot4}{3\cdot3}=\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{9\cdot11}{10\cdot10}=\frac{1}{2}\cdot\frac{11}{10}=\frac{11}{20}\)
gọi số phải tìm là A A
=(1.3).(2.4).(3.5)...(99.101)/
(2².3².4²...100²)
=(1.2.3...99).(3.4.5...101)/
[(1.2.3.4...100)(2.3.4...100)]
=101/(100.2)=101/200
= ( 1 x 3 / 2 x 2 ) . ( 2 x 4 / 3 x 3 ) . ( 3 x 5 / 4 x 4 ) . . . ( 6 x 8 / 7 x 7 ) ( mk viết thế cho rõ nhìn khi lm bỏ ngoặc cx dc )
= 1 x 3 x 2 x 4 x 3 x 4 x ... x 6 x 8 / 2 x 2 x 3 x 3 x 4 x 4 x ... x 7 x 7
= 1 x 8 / 2 x 7
= 8 / 14
= 4/7
Tk nha !!
Ta có :
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}........\frac{48}{49}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.......\frac{6.8}{7.7}\)
\(=\frac{\left(1.2.3.....6\right).\left(3.4.5.....8\right)}{\left(2.3.4......7\right).\left(2.3.4.....7\right)}\)
\(=\frac{1.2.3.....6}{2.3.4.....7}.\frac{3.4.5......8}{2.3.4......7}\)
\(=\frac{1}{7}.4\)
\(=\frac{4}{7}\)