Question

Khẩn cấp khẩn cấp,cứu vs

Tính nhanh

a)Cho A=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

b)Tính tổng B=\(\left(0,5\right)^2+1^2+\left(1,5\right)^2+2^2+...+\left(4,5\right)^2+5^2\)

Cho biết: \(1^2+2^2+3^2+...+10^2=385\)

Giúp mình gấp nha các bn,thks

 

Answer 1

a)A=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

      =\(\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)

      =\(\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

      =\(\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

      =\(\frac{1}{100}-\frac{99}{100}\)

      =\(\frac{-98}{100}=\frac{-49}{50}\)

Answer 2

a) $A= \backslash (\backslash frac\{1\}\{100\}-\backslash frac\{1\}\{100.99\}-\backslash frac\{1\}\{99.98\}-...-\backslash frac\{1\}\{3.2\}-\backslash frac\{1\}\{2.1\}\backslash )$ 

$A=$$\backslash (\backslash frac\{1\}\{100\}-\backslash left(\backslash frac\{1\}\{1.2\}+\backslash frac\{1\}\{2.3\}+...+\backslash frac\{1\}\{98.99\}+\backslash frac\{1\}\{99.100\}\backslash right)\backslash )$$\mathrm{(\; vận\; dụng quy tắc\; dấu\; ngoặc)}$\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

$\mathrm{\backslash (A=\backslash frac\{1\}\{100\}-\backslash left(1-\backslash frac\{1\}\{100\}\backslash right)\backslash ) = \backslash (\backslash frac\{1\}\{100\}-\backslash frac\{99\}\{100\}\backslash )}$ = \(\frac{-98}{100}\) = \(\frac{-49}{50}\)

Answer 3

$\frac{\mathrm{A=1/}}{100}-\frac{\mathrm{1/}}{100.99}-\frac{\mathrm{1/}}{\mathrm{99.98\; -...}}-\frac{\mathrm{1/}}{3.2}-\frac{\mathrm{1/}}{2.1}$

$A=\frac{\mathrm{1/}}{100}-(\frac{\mathrm{1/}}{1.2}+\frac{\mathrm{1/}}{2.3}+...+\frac{\mathrm{1/}}{98.99}+\frac{\mathrm{1/}}{99.100})$

 (vận dụng quy tắc dấu ngoặc)
$A=\frac{\mathrm{1/}}{100}-(1-\frac{\mathrm{1/}}{2}+\frac{\mathrm{1/}}{2}-\frac{\mathrm{1/}}{3}+...+\frac{\mathrm{1/}}{98}-\frac{\mathrm{1/}}{99}+\frac{\mathrm{1/}}{99}-$

$\frac{\mathrm{1/}}{\mathrm{100\; )}}$

 
$A=\frac{\mathrm{1/}}{100}-(1-\frac{\mathrm{1/}}{\mathrm{100)}}$ ) = $\frac{\mathrm{1/}}{100}-\frac{\mathrm{99/}}{100}=\frac{-\mathrm{98/}}{100}=\frac{-\mathrm{49/}}{50}$

/ là phần nha

Answer 4

xin lỗi mh chưa nghĩ ra