2A=1-1/2+1/2^2-...+1/2^98-1/2^99

=>3A=1-1/2^100

=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)

a)

`2/3+5/2-3/4`

`=10/4-3/4+2/3`

`=7/4+2/3`

`=21/12+8/12`

`=29/12`

b)

`2/5xx1/2:1/3`

`=2/10xx3/1`

`=6/10=3/5`

c)

`2/9:2/9xx1/3`

`=2/9xx9/2xx1/3`

`=1xx1/3`

`=1/3`

a, \(\dfrac{2}{3}\) + \(\dfrac{5}{2}\) - \(\dfrac{3}{4}\)

= \(\dfrac{8}{12}\) + \(\dfrac{30}{12}\) - \(\dfrac{9}{12}\)

= \(\dfrac{38-9}{12}\)

= \(\dfrac{29}{12}\)

b, \(\dfrac{2}{5}\) x \(\dfrac{1}{2}\) : \(\dfrac{1}{3}\)

= \(\dfrac{1}{5}\) x \(\dfrac{3}{1}\)

= \(\dfrac{3}{5}\)

c, \(\dfrac{2}{9}\) : \(\dfrac{2}{9}\) x \(\dfrac{1}{3}\)

= 1 x \(\dfrac{1}{3}\)

= \(\dfrac{1}{3}\)

\(A=\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+...+\frac{99^2}{98.100}\)

\(A=\frac{2.2}{1.3}+\frac{3.3}{2.4}+\frac{4.4}{3.5}+...+\frac{99.99}{98.100}\)

\(A=\frac{2}{1}+\frac{99}{100}\)

\(A=\frac{200}{100}+\frac{99}{100}=\frac{299}{100}\)

Hok tốt

Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)

=100

Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)

\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)

\(=\dfrac{8}{\dfrac{1}{5}}=40\)

\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)

`2/(1xx3)+2/(3xx5)+2/(5xx7)+...+2/(99xx101)` đề phải ntn chứ mà nhỉ

`=1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101`

`=1/1-1/101`

`=101/101-1/101`

`=100/101`

(Sửa phần 3 / 3 x 5 = 2 / 3 x 5)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{99\times101}\)

Ta có: \(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{99\times101}\right)\)

\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\times\left(1-\dfrac{1}{101}\right)\)

\(=2\times\dfrac{100}{101}\)

\(=\dfrac{200}{101}\)

Sửa bài ( dòng 3 đến hết bài )

... = \(2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{13\times15}+\dfrac{2}{15\times17}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}\)

\(=1-\dfrac{1}{17}\)

\(=\dfrac{16}{17}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{15\cdot17}\)

\(=2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{17}\)

\(=2-\dfrac{1}{17}\)

\(=\dfrac{33}{17}\)

N=1/2+1/2²+...+1/2¹⁰

2N=1+1/2+...+1/2⁹

2N-N=1-1/2¹⁰=1-1/1024=1023/1024

Giải:

N=1/2+1/2²+1/2³+...+1/2⁹+1/2¹⁰

2N=1+1/2+1/2²+...+1/2⁸+1/2⁹

2N-N=(1+1/2+1/2²+...+1/2⁸+1/2⁹)-(1/2+1/2²+1/2³+...+1/2⁹+1/2¹⁰)

N=1-1/2¹⁰=1023/1024

Chúc bạn học tốt!

\(A=1.3+2.4+3.5+.............+97.99+98.100\)
\(A=\left(2-1\right)\left(2+1\right)+\left(3-1\right)\left(3+1\right)+.............+\left(99-1\right)\left(99+1\right)\)
\(A=2^2-1+3^2-1+..............+99^2-1\)
\(A=1+2^2+3^2+............+99^2-99\)
Mà :
\(1+2+2^2+...........+n^2=\dfrac{\left(n+1\right)\left(n+2\right)}{6}\)
\(\Rightarrow A=\dfrac{99\left(99+1\right)\left(99+2\right)}{6}-99=\dfrac{99.100.101}{6}-99\)
\(A=166650-99=166551\)

~ Học tốt ~

Nhanh nha