Đáp án:
Giải thích các bước giải:
Bài 1:
$\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1}{10}$ +..+ $\frac{1}{x.(x+1):2}$ 2.( $\frac{1}{6}$ + $\frac{1}{12}$ +....+ $\frac{1}{x.(x+1)}$ )= $\frac{2019}{2011}$
⇔ $\frac{2}{2}$. ($\frac{1}{3}$ + $\frac{1}{6}$ + $\frac{1}{10}$ +..+ $\frac{1}{x.(x+1):2}$ )= $\frac{2019}{2011}$
⇔ 2.( $\frac{1}{6}$ + $\frac{1}{12}$ +...+ $\frac{1}{x.(x+1)}$ )= $\frac{2019}{2011}$
⇔$\frac{1}{2.3}$ +$\frac{1}{3.4}$ +.....+ $\frac{1}{x.(x+1)}$ = $\frac{2019}{2011}$:2
⇔$\frac{1}{2}$ - $\frac{1}{3}$ +$\frac{1}{3}$ - $\frac{1}{4}$ +...+$\frac{1}{x}$ - $\frac{1}{x+1}$= $\frac{2019}{4022}$
⇔$\frac{1}{2}$ - $\frac{1}{x+1}$ = $\frac{2019}{4022}$
⇔$\frac{1}{x+1}$ = $\frac{1}{2}$ - $\frac{2019}{4022}$
⇔$\frac{1}{x+1}$=$\frac{-4}{2011}$
⇔2011= -4.(x+1)
⇔ x+1= $\frac{-2011}{4}$
⇔ x= $\frac{-2011}{4}$ -1 thành kết quả ∉ Z nên ko có x.
Vậy không có x thỏa mãn( Bạn làm lại bài này nhé, mình thấy hơi sai sai)
b, A= $\frac{2008}{2009}$ + $\frac{2009}{2010}$ + $\frac{2010}{2011}$
B=$\frac{2008+2009+ 2010}{2009+ 2010+ 2011}$ = $\frac{2008}{2009+ 2010+ 2011}$ + $\frac{2009}{2009+ 2010+ 2011}$ + $\frac{2010}{2009+ 2010+ 2011}$
Ta có: Vì: $\frac{2008}{2009}$ >$\frac{2008}{2009+ 2010+ 2011}$
$\frac{2009}{2010}$>$\frac{2009}{2009+ 2010+ 2011}$
$\frac{2010}{2011}$>$\frac{2010}{2009+ 2010+ 2011}$
Nên $\frac{2008}{2009}$ + $\frac{2009}{2010}$ + $\frac{2010}{2011}$> $\frac{2008}{2009+ 2010+ 2011}$ + $\frac{2009}{2009+ 2010+ 2011}$ + $\frac{2010}{2009+ 2010+ 2011}$ ⇒ A>B
Vậy A>B
Chúc bạn học tốt!
