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\(l=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+....+\dfrac{1}{97.100}\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{1}{3}-\dfrac{1}{300}< \dfrac{1}{3}\left(đpcm\right)\)
`#3107.101107`
1.
a)
`1/(1*4) + 1/(4*7) + 1/(7*10) + ... + 1/(100*103)`
`= 1/3 * (3/(1*4) + 3/(4*7) + 3/(7*10) + ... + 3/(100*103) )`
`= 1/3 * (1 - 1/4 + 1/4 - 1/7 + ... + 1/100 - 1/103)`
`= 1/3* (1 - 1/103)`
`= 1/3*102/103`
`= 34/103`
b)
`-1/3 + (-1/15) + (-1/35) + (-1/63) + ... + (-1/9999)`
`= - 1/3 - 1/15 - 1/35 - 1/63 - ... - 1/9999`
`= - (1/3 + 1/15 + 1/35 + ... + 1/9999)`
`= - (1/(1*3) + 1/(3*5) + 1/(5*7) + ... + 1/99*101)`
`= - 1/2 * (2/(1*3) + 2/(3*5) + 2/(5*7) + ... + 2/99*101)`
`= - 1/2* (1 - 1/3 + 1/3 - 1/5 + ... + 1/99 - 1/101)`
`= -1/2 * (1 - 1/101)`
`= -1/2*100/101`
`= -50/101`
2.
`3/(1*4) + 3/(4*7) + ... + 3/(94*97) + 3/(97*100)`
`= 1 - 1/4 + 1/4 - 1/7 + ... + 1/94 - 1/97 + 1/97 - 1/100`
`= 1-1/100`
`= 99/100`
Đặt \(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
\(3\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{\left(x+3\right)}\right)=3\cdot\frac{49}{148}\)
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{x\left(x+3\right)}=\frac{147}{148}\)
\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{147}{148}\)
\(1-\frac{1}{x-1}=\frac{147}{148}\)
\(\frac{1}{x-1}=1-\frac{147}{148}\)
\(\frac{1}{x-1}=\frac{1}{148}\)
\(\Rightarrow x-1=148\)
\(\Leftrightarrow x=148+1\)
\(\Leftrightarrow x=149\)
Vậy x=149
\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{49}{148}\)
\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{49}{148}\)
\(\Rightarrow\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{49}{148}\)
\(\Rightarrow\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{49}{148}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{49}{148}:\frac{1}{3}\)
\(\Rightarrow1-\frac{1}{x+3}=\frac{147}{148}\)
\(\Rightarrow\frac{1}{x+3}=1-\frac{147}{148}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{148}\)
\(\Rightarrow x+3=148\)
\(\Rightarrow x=148-3\)
\(\Rightarrow x=145\)
Vậy x = 145
_Chúc bạn học tốt_
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\)
\(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\)
\(\frac{\Rightarrow1}{a}=\frac{1}{b}=\frac{1}{c}\Rightarrow a=b=c\)
Thay vào M ta có
\(\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
P/s : hỏi từng câu thôi