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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
\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{91.94}=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}\right)\)
\(=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}\right)\)
\(=\frac{5}{3}.\left(1-\frac{1}{94}\right)=\frac{5}{3}.\frac{93}{94}=\frac{155}{94}\)
Làm từng phần nha bạn
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{298\cdot301}+x=\frac{299}{301}\)
Đặt \(A+x=\frac{299}{301}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{298}-\frac{1}{301}\)
\(A=1-\frac{1}{301}\)
\(A=\frac{300}{301}\)
=> \(\frac{300}{301}+x=\frac{299}{301}\)
\(x=\frac{299-300}{301}\)
\(x=-\frac{1}{301}\)
\(A=5\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{301\cdot304}\right)\)
\(\frac{3A}{5}=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{301\cdot304}\)
\(\frac{3}{5}\cdot A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\)
\(\frac{3}{5}\cdot A=1-\frac{1}{304}\)
\(\frac{3}{5}\cdot A=\frac{303}{304}\)
\(A=\frac{505}{304}\)
Tính nhanh ;:
B = (1+2/1.4).(1+2/2.5).(1+2/3.6).(1+2/4.7)....(1+2/7.10
Nhớ giúp mình nhé cảm ơn nhiều
.................................
mik chỉ hỏi thui mà, mik có bắt bạn giúp âu
giúp mình với
5/1.4 + 5/4.7 + 5/7.10 + ... + 5/97.100
= 5/3 . (3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100)
= 5/3 . (1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/97 - 1/100)
= 5/3 . ( 1 - 1/100)
= 5/3 . 99/100
= 33/20
\(\dfrac{3}{2}A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4-1}{1.4}+\dfrac{7-4}{4.7}+\dfrac{10-7}{7.10}+...+\dfrac{97-94}{94.97}\)
\(\dfrac{3}{2}A=\dfrac{4}{1.4}-\dfrac{1}{1.4}+\dfrac{7}{4.7}-\dfrac{4}{4.7}+\dfrac{10}{7.10}-\dfrac{7}{7.10}+...+\dfrac{97}{94.97}-\dfrac{94}{94.97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}\)
\(\dfrac{3}{2}A=1-\dfrac{1}{97}=\dfrac{96}{97}\)
⇒ A = \(\dfrac{96}{97}:\dfrac{3}{2}=\dfrac{64}{97}\)
Câu B cách làm tương tự, thắc mắc gì bạn cứ hỏi nhé.
`S_1 = 5/(1.4) + 5/(4.7) +...+ 5/(97.100)`
`S_1 = 5 (1/(1.4) + 1/(4.7) +...+ 1/(97.100))`
`S_1 = 5/3 (3/(1.4) + 3/(4.7) +...+ 3/(97.100))`
`S_1 = 5/3 (1 - 1/4 + 1/4 - 1/7 + ...+ 1/97 - 1/100)`
`S_1 = 5/3 (1 - 1/100)`
`S_1 = 5/3 . 99/100`
`S_1 = 33/20`
Có : 3/5 A = 3/1.4 + 3/4.7 + 3/7.10 + ..... + 3/307.310
= 1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ........ + 1/307 - 1/310
= 1 - 1/310
= 309/310
=> A = 309/310 : 3/5 = 103/62
Tk mk nha
\(\frac{5}{1.4}+\frac{5}{4.7}+\frac{5}{7.10}+...+\frac{5}{307.310}\)
\(=\frac{5}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{307.310}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{307}-\frac{1}{310}\right)\)
\(=\frac{5}{3}\left(1-\frac{1}{310}\right)\)
\(=\frac{5}{3}.\frac{309}{310}=\frac{103}{62}\)