Tìm y biết: (y- \(\dfrac{1}{2}\)): (\(\dfrac{1}{2}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{12}\) + \(\dfrac{1}{20}\)+ ... + \(\dfrac{1}{90}\)) = \(\dfrac{1}{3}\)
K
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Những câu hỏi liên quan
15 tháng 6 2018
a, \(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)
\(\Rightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{100}\)
\(\Rightarrow\dfrac{1}{1}-\dfrac{1}{100}\)
\(\Rightarrow\dfrac{99}{100}\)
D
27 tháng 5 2022
`[-1]/2+[-1]/6+[-1]/12+[-1]/20+[-1]/30+[-1]/42+[-1]/56+[-1]/72+[-1]/90`
`=(-1)(1/2+1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90)`
`=(-1)(1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)`
`=(-1)(1-1/10)`
`=(-1). 9/10=-9/10`
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 5 2022
A = \(\dfrac{-1}{2}\) + \(\dfrac{-1}{6}\)+ \(\dfrac{-1}{12}\)+ \(\dfrac{-1}{20}\)+ \(\dfrac{-1}{30}\)+ \(\dfrac{-1}{42}\)+ \(\dfrac{-1}{56}\)+ \(\dfrac{-1}{72}\)+ \(\dfrac{-1}{90}\)
A = \(\dfrac{-1}{2}\) + \(\dfrac{-1}{2\times3}\)+ \(\dfrac{-1}{3\times4}\)+ \(\dfrac{-1}{4\times5}\)+ \(\dfrac{-1}{5\times6}\)+ \(\dfrac{-1}{6\times7}\)+ \(\dfrac{-1}{7\times8}\)+ \(\dfrac{-1}{8\times9}\)+ + \(\dfrac{-1}{9\times10}\)
A = - (\(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)- \(\dfrac{1}{5}\)+ \(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)+ \(\dfrac{1}{6}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{8}\)+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+ \(\dfrac{1}{9}\)-\(\dfrac{1}{10}\))
A = -(1-\(\dfrac{1}{10}\))
A = \(\dfrac{-9}{10}\)
31 tháng 7 2023
` a/`
` 3 1/5 : 2 1/3 : y = 12/7 `
` 48/35 : y = 12/7 `
` y = 48/35 : 12/7 `
` y = 48/35 xx 7/12 `
` y = 4/5`
Vậy ` y = 4/5`
`b/`
` 3 : y xx 3 1/2 = 2/3 xx 3/4 `
` 3 : y xx 7/2 = 1/2`
` 3 : y = 1/2 : 7/2 `
` 3 : y = 1/2 xx 2/7 `
` 3 : y = 1/7 `
` y = 3 : 1/7 `
` y = 3 xx 7`
` y = 21 `
Vậy ` y = 21 `
`c/`
` 3 2/3 - y + 1 3/4 = 2`
` 11/3 - y + 7/4 = 2 `
` 11/3 -y = 2 - 7/4`
` 11/3 - y = 1/4 `
` y = 11/3 - 1/4 `
` y = 41/12 `
Vậy ` y = 41/12`
2 tháng 8 2023
=9/10-(1/2+1/6+...+1/90)
=9/10-(1-1/2+1/2-1/3+...+1/9-1/10)
=9/10-9/10=0
17 tháng 1 2023
sau đây là phần chữa của mình:
\(=\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\)
= \(\dfrac{3}{10}\)
17 tháng 1 2023
= \(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
= \(\dfrac{1}{2}-\dfrac{1}{10}\)
= \(\dfrac{2}{5}\)
11 tháng 10 2023
b:
ĐKXĐ: x<>0
\(\dfrac{2}{x}+\dfrac{y}{3}=\dfrac{1}{6}\)
=>\(\dfrac{6+xy}{3x}=\dfrac{1}{6}\)
=>\(6\left(6+xy\right)=3x\)
=>\(x=2\left(6+xy\right)=12+2xy\)
=>\(x\left(1-2y\right)=12\)
mà x,y là các số nguyên
nên \(\left(x;1-2y\right)\in\left\{\left(12;1\right);\left(-12;-1\right);\left(4;3\right);\left(-4;-3\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(12;0\right);\left(-12;1\right);\left(4;-1\right);\left(-4;2\right)\right\}\)
c: ĐKXĐ: y<>-1
\(\dfrac{x}{3}+\dfrac{1}{y+1}=\dfrac{1}{6}\)
=>\(\dfrac{xy+x+3}{3\left(y+1\right)}=\dfrac{1}{6}\)
=>\(\dfrac{2\left(xy+x+3\right)}{6\left(y+1\right)}=\dfrac{y+1}{6\left(y+1\right)}\)
=>\(2xy+2x+6=y+1\)
=>\(2x\left(y+1\right)-\left(y+1\right)=-6\)
=>\(\left(2x-1\right)\left(y+1\right)=-6\)
mà x,y là các số nguyên
nên \(\left(2x-1;y+1\right)\in\left\{\left(1;-6\right);\left(-1;6\right);\left(3;-2\right);\left(-3;2\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(0;5\right);\left(2;-3\right);\left(-1;1\right)\right\}\)
14 tháng 5 2022
\(y\times\dfrac{1}{2}+y\times\dfrac{1}{6}+y\times\dfrac{1}{12}+y\times\dfrac{1}{20}=\dfrac{8}{5}\)
\(y\times\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}\right)=\dfrac{8}{5}\)
\(y\times\dfrac{4}{5}=\dfrac{8}{5}\)
\(\dfrac{8}{5}:\dfrac{4}{5}\)
\(y=\dfrac{8}{4}=2\)
(y - \(\dfrac{1}{2}\)) : \(\left(\dfrac{1}{2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\right)\)= \(\dfrac{1}{3}\)
(y\(-\dfrac{1}{2}\)): \(\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)= \(\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right):\left(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{10}\right)=\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right):\dfrac{3}{10}=\dfrac{1}{3}\)
\(\left(y-\dfrac{1}{2}\right)=\dfrac{1}{10}\)
y = \(\dfrac{3}{5}\)