\(A=\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{1x10}+\dfrac{1}{2x9}+\dfrac{1}{3x8}+...+\dfrac{1}{8x3}+\dfrac{1}{9x2}+\dfrac{1}{10x1}}\)
Những câu hỏi liên quan
\(\dfrac{1+3+6+10+...+45+55}{1x10+9x2+3x8+...9x2+10x1}\)
\(\dfrac{1+3+6+10+...+45+55}{1x10+2x9+3x8+...+9x2+10x1}\)
Mọi người bày tui cách giải bài này với. Nhanh nhé. Thanks 😘😘😘
Tính hợp lý
\(A= (\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\) B= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}})\)
8) Adfrac{9}{10}-dfrac{1}{90}-dfrac{1}{72}-dfrac{1}{56}-dfrac{1}{42}-dfrac{1}{30}-dfrac{1}{20}-dfrac{1}{12}-dfrac{1}{6}-dfrac{1}{2}
9) Bdfrac{1}{3}+dfrac{1}{3^2}+dfrac{1}{3^4}+...+dfrac{1}{3^{2014}}+dfrac{1}{3^{2015}}
10) Pdfrac{dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+...+dfrac{1}{2005}}{dfrac{2004}{1}+dfrac{2003}{2}+dfrac{2002}{3}+...+dfrac{1}{2004}}
Đọc tiếp
8) \(A=\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
9) \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{2014}}+\dfrac{1}{3^{2015}}\)
10) \(P=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2005}}{\dfrac{2004}{1}+\dfrac{2003}{2}+\dfrac{2002}{3}+...+\dfrac{1}{2004}}\)
8,A=\(\dfrac{9}{10}-\left(\dfrac{1}{10\times9}+\dfrac{1}{9\times8}+\dfrac{1}{8\times7}+...+\dfrac{1}{2\times1}\right)\)
=\(\dfrac{9}{10}-\left(\dfrac{1}{10}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{8}+...+\dfrac{1}{2}-1\right)\)
=\(\dfrac{9}{10}-\left(\dfrac{1}{10}-1\right)\)
=\(\dfrac{9}{10}-\dfrac{\left(-9\right)}{10}\)
=\(\dfrac{9}{5}\)
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Tính hợp lí:
A= \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
B= \(\dfrac{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}{\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}}\)
Cái này mk từng làm nhưng có chút sai sót vậy nên bn sữa cho mk chút nhé ! Thay vì N = ... thì bn thay bằng A = ... nha
Ta có :
N = 40 ( A = 40 )
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\((\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{9}+\dfrac{1}{10})\times x=\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}\)
`#iv`
`(1/2 +1/3 +1/4 +... +1/10)*x=1/9 + 2/8 + 3/7 +... +9/1`
`=>(1/2+1/3+1/4+...+1/10)*x=10*(1/10 + 1/9+1/8+1/7+...+1/2)`
`=>x=10*(1/10 + 1/9+1/8+1/7+...+1/2):(1/10 + 1/9+1/8+1/7+...+1/2)`
`=>x=10`
Vậy `x=10`
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Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
Bài 1:
\(a,\dfrac{1}{5}+\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}+...............+\dfrac{1}{1280}\)
\(b,\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+..............+\dfrac{1}{59049}\)
\(c,\dfrac{1}{2}\times3+\dfrac{1}{3}\times4+\dfrac{1}{4}\times5+\dfrac{1}{5}\times6\)
b)\(\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{9}+\dfrac{1}{10}\right)x=\dfrac{1}{9}+\dfrac{2}{8}+\dfrac{3}{7}+...+\dfrac{8}{2}+\dfrac{9}{1}\)