toán

L

lan_phuong_000


Bài 1:\Leftrightarrow $\dfrac{-13}{12} < x \le 2$
Bài 2:
$= 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + ... + \dfrac{1}{2010} - \dfrac{1}{2011}$

$= \dfrac{2010}{2011}$
 
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T

tayhd20022001

Bài 1: Tìm x biết
$\dfrac{-2}{3}$ +$\dfrac{-5}{12}$ <x\leq4 -$\dfrac{1}{3}$:$\dfrac{1}{6}$
Giải
\Rightarrow Ta có :
\Leftrightarrow $\dfrac{-2}{3}$ +$\dfrac{-5}{12}$ <x\leq4 -$\dfrac{1}{3}$:$\dfrac{1}{6}$
\Rightarrow $\dfrac{-13}{12}$ <x\leq2=$\dfrac{24}{12}$
\Rightarrow x={-12;-11;...;24}
Bài 2:Tính tổng:
A=1/1.2+1/2.3+1/3.4+....+1/2010+2011
Ta có :
A=$\dfrac{1}{1.2}$+$\dfrac{1}{2.3}$+$\dfrac{1}{3.4}$+...+$\dfrac{1}{2010.2011}$
\Rightarrow A=1-$\dfrac{1}{2}$+$\dfrac{1}{2}$-$\dfrac{1}{3}$+...+$\dfrac{1}{2010}$-$\dfrac{1}{2011}$
\Rightarrow A=1-$\dfrac{1}{2011}$
\Rightarrow A=$\dfrac{2010}{2011}$
 
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