Question

5\(\dfrac{8}{17}\): x + (-\(\dfrac{1}{17}\)) : x + 3\(\dfrac{1}{17}\): 17\(\dfrac{1}{3}\)= \(\dfrac{4}{17}\)

\(\dfrac{1}{1.4}\)+ \(\dfrac{1}{4.7}\)+ \(\dfrac{1}{7.10}\)+ ... + \(\dfrac{1}{x.\left(x+3\right)}\)= \(\dfrac{6}{19}\)

So sánh hai biểu thức A và B, biết rằng:

A = \(\dfrac{2012}{2013}\)+ \(\dfrac{2013}{2014}\) và B =\(\dfrac{2012+2013}{2013+2014}\).

Answer 1

5\(\dfrac{8}{17}\):x + (-\(\dfrac{1}{17}\)) : x + 3\(\dfrac{1}{17}\) : 17\(\dfrac{1}{3}\)= \(\dfrac{4}{17}\)

\(\dfrac{93}{17}\).\(\dfrac{1}{x}\) + (-\(\dfrac{1}{17}\)) .\(\dfrac{1}{x}\) +\(\dfrac{3}{17}\)= \(\dfrac{4}{17}\)

\(\dfrac{1}{x}\).\(\dfrac{92}{17}\)=\(\dfrac{1}{17}\)

\(\dfrac{1}{x}\)=\(\dfrac{1}{17}\):\(\dfrac{92}{17}\) x= 92

Answer 2

\(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+...+\(\dfrac{1}{x.\left(x+3\right)}\)=\(\dfrac{6}{19}\)

3(\(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+...+\(\dfrac{1}{x.\left(x+3\right)}\))=3.\(\dfrac{6}{19}\) \(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{18}{19}\) 1-\(\dfrac{1}{x+3}\)=\(\dfrac{18}{19}\) \(\dfrac{1}{x+3}\)=\(\dfrac{1}{19}\) x+3 =19 x=19-3 x=17

Answer 3

Ta có:

B=\(\dfrac{2012+2013}{2013+2014}=\dfrac{2012}{2013+2014}+\dfrac{2013}{2013+2014}\)

Mà \(\dfrac{2012}{2013}>\dfrac{2012}{2013+2014}\)và \(\dfrac{2013}{2014}>\dfrac{2013}{2013+2014}\)

Vậy A>B