A = 1.4 + 4.7 + 7.10 + ..... + 91.94 = ?
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9A = 1.4.[7+2] + 4.7. [10-1] + 7.10.[13-4] +...+ 91.94. [97-88]
= 1.4.7 + 1.2.4 + 4.7.10 - 1.4.7 + 7.10.13 - 4.7.10+...+ 91.94.97 - 88.91.94
= 1.2.4 + 91.94.97 = 8 +829738 = 829746 => A = 829746 : 9 = 92194
đúng cái nhé
\(\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{94.97}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)
mà \(\frac{96}{97}< 1\)
\(\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.07}< 1\)
vậy..................
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{91\cdot94}+\frac{3}{94\cdot97}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)
\(\Rightarrow\frac{96}{97}< 1\)
\(\Rightarrow\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)
Vậy \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)
Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)
\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\)
\(\Leftrightarrow1-\frac{1}{97}=\frac{96}{97}\)
Do \(\frac{96}{97}< 1\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}< 1\)
Vậy:.............................<1
Đặt 2/3 ra ngoài trong ngoặc còn :
1-1/4+1/4-1/7+...-1/97=96/97
Lấy 2/3 nhân với 96/97 sẽ ra đáp án nhé
\(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{91\cdot94}=\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{91\cdot94}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}\right)\)
\(=\frac{1}{3}\left[\left(1-\frac{1}{94}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{91}-\frac{1}{91}\right)\right]\)
\(=\frac{1}{3}\left[\left(\frac{94}{94}-\frac{1}{94}\right)+0+...+0\right]=\frac{1}{3}\cdot\frac{93}{94}=\frac{93}{282}\)
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)
⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)
Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.
Gọi biểu thức sau là A, ta có:
A=(5/1.4)+(5/4.7)+(5/7.10)+...+(5/91.94)
2A=(10/1.4)+(10/4.7)+(10/7.10)+...+(10/91.94)
2A=5/1-5/4+5/4-5/7+5/7-5/10+...+5/91-5/94
2A=5/1-5/4+5/4-5/7+5/7-5/10+...+5/91-5/94
2A=5/1-5/94
2A=465/94
=>A=465/94:2
=>A= tự tính nhé
\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{91.94}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}\right)\)
\(=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}\right)\)
\(=\frac{5}{3}.\left(1-\frac{1}{94}\right)=\frac{5}{3}.\frac{93}{94}=\frac{155}{94}\)
=\(\frac{31}{94}\)theo mình mới giải là vậy còn nếu không bạn có thể trình bày cách giải cho mình luôn được không?
`#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`