toán nâng cao lớp 5

T

thaonguyenkmhd

$1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+ \dfrac{2}{x(x+1)}=1+\dfrac{2009}{2010} \\ \leftrightarrow \dfrac{2}{1\times2}+\dfrac{2}{2\times3}+\dfrac{2}{3\times4}+ \dfrac{2}{4\times5}+...+\dfrac{2}{x(x+1)}=1+\dfrac{2009}{2010} \\ \leftrightarrow 2 \times ( \dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+ \dfrac{1}{4\times5}+...+\dfrac{1}{x(x+1)}=1+\dfrac{2009}{2010} \\ \leftrightarrow 2 \times (1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5} +...+\dfrac{1}{x}-\dfrac{1}{x+1})=1+\dfrac{2009}{2010} \\ \leftrightarrow 2 \times ( 1-\dfrac{1}{x+1})=1+\dfrac{2009}{2010}\\ \leftrightarrow 2-\dfrac{2}{x+1}=1+\dfrac{2009}{2010} \\ \leftrightarrow 2-1-\dfrac{2009}{2010}=\dfrac{2}{x+1} \\ \leftrightarrow \dfrac{1}{2010}=\dfrac{2}{x+1} \\ \leftrightarrow \dfrac{2}{4020}=\dfrac{2}{x+1} \\ \leftrightarrow x+1=4020 \\ \leftrightarrow x=4019$
 
V

vuhoanghuy71

1+13+16+110+...+2x(x+1)=1+20092010↔21×2+22×3+23×4+24×5+...+2x(x+1)=1+20092010↔2×(11×2+12×3+13×4+14×5+...+1x(x+1)=1+20092010↔2×(1−12+12−13+13−14+14−15+...+1x−1x+1)=1+20092010↔2×(1−1x+1)=1+20092010↔2−2x+1=1+20092010↔2−1−20092010=2x+1↔12010=2x+1↔24020=2x+1↔x+1=4020↔x=4019
 
C

chienhopnguyen

vuhoanghuy71 said:
1+13+16+110+...+2x(x+1)=1+20092010↔21×2+22×3+23×4+24×5+...+2x(x+1)=1+20092010↔2×(11×2+12×3+13×4+14×5+...+1x(x+1)=1+20092010↔2×(1−12+12−13+13−14+14−15+...+1x−1x+1)=1+20092010↔2×(1−1x+1)=1+20092010↔2−2x+1=1+20092010↔2−1−20092010=2x+1↔12010=2x+1↔24020=2x+1↔x+1=4020↔x=4019
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