1, Tính
b \(1\dfrac{1}{5}X1\dfrac{1}{6}X1\dfrac{1}{7}X...X1\dfrac{1}{1998}X1\dfrac{1}{1999}\)
Tính giúp mình , mình tick cho nhé !
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\(1\dfrac{1}{2}x1\dfrac{1}{3}x1\dfrac{1}{4}x1\dfrac{1}{5}x1\dfrac{1}{6}x1\dfrac{1}{7}x1\dfrac{1}{8}x1\dfrac{1}{9}\)
\(=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}x\dfrac{8}{7}x\dfrac{9}{8}x\dfrac{10}{9}\)
\(=x^7.\dfrac{3.4.5.6.7.8.9.10}{2.3.4.5.6.7.8.9}\)
\(=x^7.\dfrac{10}{2}\)
\(=5x^7\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{9}{8}\times\dfrac{10}{9}=\dfrac{10}{2}=5\)
\(...=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}....x\dfrac{1000}{999}\)
\(=\dfrac{1}{2}x\dfrac{1000}{1}=500\)
=3/2x4/3x5/4x....x1000/999
=1/2x1000=500
mình chưa chắc là đúng đâu nhé
\(=>C=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}.....\cdot\dfrac{101}{100}\)
\(C=\dfrac{3\cdot4\cdot5.......\cdot101}{2\cdot3\cdot4.........\cdot100}\)
\(C=\dfrac{101}{2}\)
1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\)...\(\times\)1\(\dfrac{1}{2005}\)
A = \(\dfrac{12+1}{12}\) \(\times\) \(\dfrac{13+1}{13}\) \(\times\) \(\dfrac{14+1}{14}\)\(\times\)...\(\times\) \(\dfrac{2006}{2005}\)
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\)...\(\times\) \(\dfrac{2006}{2005}\)
A = \(\dfrac{2006}{12}\)
A = \(\dfrac{1003}{6}\)
x1+x2=1/3; x1x2=-7/3
(x1-1)(x2-1)
=x1x2-(x1+x2)+1
=-7/3-1/3+1
=-8/3+1=-5/3
\(\dfrac{3}{x_1-2}+\dfrac{3}{x_2-2}=\dfrac{3x_2-6+3x_1-6}{\left(x_1-2\right)\left(x_2-2\right)}\)
\(=\dfrac{3\left(x_1+x_2\right)-12}{x_1x_2-2\left(x_1+x_2\right)+4}\)
\(=\dfrac{3\cdot\dfrac{1}{3}-12}{\dfrac{-7}{3}-2\cdot\dfrac{1}{3}+4}=-11\)
` a/`
` 2 - 1 5/6 + 2 2/3 = 2 - 11/6 - 8/3 = 1/6+ 8/3 = 1/6 + 16/6 = 17/6 `
`b/`
`5/9 xx ( 2 5/6 - 1 2/3 ) = 5/9 xx ( 17/6 - 5/3 ) = 5/9 xx 7/6 = 35/54 `
`c/`
` 1 1/3 : ( 2 + 1 1/6 : 2 5/6 ) `
`= 4/3 : ( 2 + 7/6 : 17/6 ) `
`= 4/3 : ( 2 + 7/6 xx 6/17 )`
`= 4/3 : ( 2 + 7/17 ) `
`= 4/3 : ( 34/17 + 7/17 ) `
`= 4/3 : 41/17 `
`= 4/3 xx 17/41 `
`= 68/123`
` d/`
` 2 3/5 : 3/4 xx 1 4/5 = 13/5 xx 4/3 xx 9/5 =52/15 xx 9/5 = 156/25`
\(=\dfrac{4}{5}+\dfrac{2}{3}x.\dfrac{55}{4}=\dfrac{4}{5}+\dfrac{55}{6}x\)
\(1\dfrac{1}{3}\times1\dfrac{1}{8}\times1\dfrac{1}{15}\times1\dfrac{1}{24}\times1\dfrac{1}{35}\)
= \(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}\times\dfrac{36}{35}\)
= \(\dfrac{4\times9\times16\times25\times36}{3\times8\times15\times24\times35}\)
= \(\dfrac{1\times2\times2\times25\times36}{1\times2\times15\times24\times35}\)
= \(\dfrac{4\times25\times36}{30\times24\times35}\)
= \(\dfrac{1\times25\times36}{30\times6\times35}=\dfrac{1}{7}\)
Sai rồi nhé! Từ dấu bằng thứ 2 xuống dấu bằng thứ 3 bạn làm sao đc z?
a: A=x1+x2=-5/2
b: \(=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-5}{2}:\left(-1\right)=\dfrac{5}{2}\)
c: \(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=\left(-\dfrac{5}{2}\right)^3-3\cdot\dfrac{-5}{2}\cdot\left(-1\right)\)
\(=-\dfrac{125}{8}-\dfrac{15}{2}=\dfrac{-185}{8}\)
e: \(E=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\sqrt{\left(-\dfrac{5}{2}\right)^2-4\cdot\left(-1\right)}=\sqrt{\dfrac{25}{4}+4}=\dfrac{\sqrt{41}}{2}\)
`1 1/5 xx 1 1/6 xx 1 1/7 xx .....xx 1 1/1998 xx 1 1/1999`
`=6/5 xx 7/6 xx 8/7 xx .......xx 1999/1998 xx 2000/19999`
`=(6xx 7xx8xx....xx1999xx2000)/(5xx6xx7xx....xx1998xx1999)`
`=2000/5`
`=400`
\(1\dfrac{1}{5}\cdot1\dfrac{1}{6}\) . \(1\dfrac{1}{7}\cdot\) .... . \(1\dfrac{1}{1998}.1\dfrac{1}{1999}\)
= \(\dfrac{6}{5}.\dfrac{7}{6}.\dfrac{8}{7}\) . ... . \(\dfrac{1999}{1998}.\dfrac{2000}{1999}\)
= \(\dfrac{2000}{5}=400\)