1 .Tìm x biết
a. ( \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{97.100}\)) = \(\dfrac{0,33x}{2009}\)
b. 1 + \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1991}{1993}\)
c. \(\dfrac{1}{2013}x+1+\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{2012.2013}=2\)
d. 2x + \(\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}=10\)
2. Chứng minh rằng :
a. \(\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{99}}< \dfrac{1}{3}\)
b. \(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
1/
a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
Bài 1:
c: \(\Leftrightarrow x\cdot\dfrac{1}{2013}+1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2012}-\dfrac{1}{2013}=2\)
\(\Leftrightarrow x\cdot\dfrac{1}{2013}+2-\dfrac{1}{2013}=2\)
=>x*1/2013=1/2013
=>x=1
d: \(\Leftrightarrow2x+8+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{90}=10\)
=>2x+1/2-1/3+1/3-1/4+...+1/9-1/10=2
=>2x+2/5=2
=>x+1/5=1
=>x=4/5