Tìm x:
(x -\(\dfrac{1}{3}\) ) x (\(\dfrac{2}{1x2}\)+ \(\dfrac{2}{2x3}\)+ \(\dfrac{2}{3x4}\) + … + \(\dfrac{2}{9x10}\)) = \(\dfrac{3}{4}\)
\(\dfrac{1}{1x2}\)+\(\dfrac{1}{2x3}\)+\(\dfrac{1}{3x4}\)+......+\(\dfrac{1}{9x10}\)
Tính bằng cách nhanh nhất
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{9\cdot10}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)
#kễnh
A = \(\dfrac{1}{2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+............+\dfrac{1}{9x10}\)
tính nhanh
=1-1/2+1/2-1/3+...+1/9-1/10
=1-1/10
=9/10
Tìm x biết:
\(\dfrac{x}{x+1}=\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{31x32}\)
Trả lời nhanh giúp mìn nhé
`x/(x+1)=1/(1xx2)+1/(2xx3)+1/(3xx4)+...+1/(31xx32)`
`=>x/(x+1)=1-1/2+1/2-1/3+1/3-1/4+...+1/31-1/32`
`=>x/(x+1)=1-1/32`
`=>x/(x+1)=31/32`
`=>32x=31(x+1)`
`=>32x=31x+31`
`=>32x-31x=31`
`=>x=31`
\(\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+........+\dfrac{1}{8x9}+\dfrac{1}{9x10}\right)xX=\dfrac{3}{4}\)
\(\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{9\times10}\right)\times x=\dfrac{3}{4}\)
\(\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\times x=\dfrac{3}{4}\)
\(\left(1-\dfrac{1}{10}\right)\times x=\dfrac{3}{4}\)
\(\dfrac{9}{10}\times x=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}\times\dfrac{10}{9}\)
\(x=\dfrac{5}{6}\)
Bài 3:(nâng cao) tính nhanh:
\(\dfrac{2}{1x2}\) + \(\dfrac{2}{2x3}\) + \(\dfrac{2}{3x4}\) + ........ +\(\dfrac{2}{18x19}\) + \(\dfrac{2}{19x20}\)
\(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{19\cdot20}\)
\(=2\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\right)\)
\(=2\cdot\left(1-\dfrac{1}{20}\right)\)
\(=2\cdot\dfrac{19}{20}\)
\(=\dfrac{19}{10}\)
C=\(\dfrac{2}{1x2}\)+\(\dfrac{2}{2x3}\)+\(\dfrac{2}{3x4}\)+...+\(\dfrac{2}{2018x2019}\)+\(\dfrac{2}{2019x2020}\)
\(C=\dfrac{2}{1\times2}+\dfrac{2}{2\times3}+...+\dfrac{2}{2019\times2020}\)
\(=2\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{2019\times2020}\right)\)
\(=2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\right)\)
\(=2\left(1-\dfrac{1}{2020}\right)=2.\dfrac{2019}{2020}=\dfrac{2019}{1010}\)
lớp 5 đây á
no no
đây ko phải lớp 5 mọi người nhỉ ?
BÀI 1: NHÂN ĐƠN THỨC VỚI ĐA THỨC
6) 5x +3 ( x2 -x - 1)
7) -\(\dfrac{2}{3}\)x ( -x4y2 -2x2 - 10y2)
8) \(\dfrac{2}{3}\)xy ( 3 x2y -3xy + y2)
9) (-2x).(3x2 - 2x +4)
10) 3x4 ( -2x3 + 5x2 - \(\dfrac{2}{3}\)x + \(\dfrac{1}{3}\))
9: \(\left(-2x\right)\left(3x^2-2x+4\right)=-6x^3+4x^2-8x\)
B= \(\dfrac{1}{1x2}\)+\(\dfrac{1}{2x3}\)+\(\dfrac{1}{3x4}\)+.....+\(\dfrac{1}{198x199}\)+\(\dfrac{1}{199x200}\)
\(B=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{199\times200}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)
\(=1-\dfrac{1}{200}=\dfrac{199}{200}\)
Tính
D = \(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+....+\dfrac{1}{2022x2023}\)