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\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{298\cdot301}+x=\frac{299}{301}\)

Đặt \(A+x=\frac{299}{301}\)

\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{298}-\frac{1}{301}\)

\(A=1-\frac{1}{301}\)

\(A=\frac{300}{301}\)

=> \(\frac{300}{301}+x=\frac{299}{301}\)

\(x=\frac{299-300}{301}\)

\(x=-\frac{1}{301}\)

\(A=5\cdot\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+...+\frac{1}{301\cdot304}\right)\)

\(\frac{3A}{5}=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{301\cdot304}\)

\(\frac{3}{5}\cdot A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{301}-\frac{1}{304}\)

\(\frac{3}{5}\cdot A=1-\frac{1}{304}\)

\(\frac{3}{5}\cdot A=\frac{303}{304}\)

\(A=\frac{505}{304}\)

3/1.4+3/4.7+3/7.10+......+3/40.43

=1/1-1/4+1/4-1/7+1/7-1/10+......+1/40-1/43

triệt tiêu hết cho nhau ta còn:

1/1-1/43=43/43-1/43=42/43

nhớ cho mình nhé

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{40.43}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)

\(=1-\frac{1}{43}\)

\(=\frac{42}{43}\)

Sao ở trên /3 xuống lại có 1 vậy??

\(3B=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}.\)

\(3B=\frac{4-1}{1.4}+\frac{7-4}{4.7}+\frac{10-7}{7.10}+...+\frac{103-100}{100.103}\)

\(3B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}=1-\frac{1}{103}=\frac{102}{103}\)

\(B=\frac{102}{3.103}=\frac{34}{103}\)

B=\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+......+\frac{1}{100.103}\)

B=\(\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{100.103}\right)\)

B=\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+......+\frac{1}{100}-\frac{1}{103}\right)\)

B=\(\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

B=\(\frac{1}{3}.\frac{102}{103}\)

B=\(\frac{34}{103}\)

\(\dfrac{x}{1.4}\)+\(\dfrac{x}{4.7}+\dfrac{x}{7.10}+\dfrac{x}{10.13}+\dfrac{x}{13.16}=\dfrac{5}{2}\) \(x.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)=\dfrac{5}{2}\) \(x.\left(\dfrac{1}{1}-\dfrac{1}{16}\right)=\dfrac{5}{2}\) \(x.\dfrac{15}{16}=\dfrac{5}{2}\) \(x=\dfrac{5}{2}:\dfrac{15}{16}\) \(x=\dfrac{80}{30}=\dfrac{8}{3}\) DAY LA BAI LAM CUA MK NHO TICK CHO MK NHA CAM ON BAN TRUOC

a: =1/2-1/3+1/3-1/4+...+1/99-1/100

=1/2-1/100=49/100

b; =5/3(1-1/4+1/4-1/7+...+1/100-1/103)

=5/3*102/103

=510/309=170/103

c: =1/2(1/3-1/5+1/5-1/7+...+1/49-1/51)

=1/2*16/51=8/51

More images for 1−14 +14 −17 +...+197 −1100 =0,99·x2009 100100 −1100 =0,99x2009 99100 =0,99x2009 =>0,99x*100=2009*9999x=2009*99=>x=2009Vậy x=2009 Đúng 4 Sai 0 Diana Andrea đã chọn câu trả lời này.Đỗ Lê Tú Linh 26/12/2015 lúc 22:10 Báo cáo sai phạm

ta nhân 3 cả hai vế, được : 

\(\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{102.105}\right)x=3\)

hay 

\(\left(\frac{4-1}{1.3}+\frac{7-4}{4.7}+...+\frac{105-102}{102.105}\right)x=3\) \(\Leftrightarrow\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+..+\frac{1}{102}-\frac{1}{105}\right)x=3\)

\(\Leftrightarrow\left(1-\frac{1}{105}\right)x=3\Leftrightarrow\frac{104}{105}.x=3\Leftrightarrow x=\frac{315}{104}\)

\(\text{a. Số số hạng của dãy là : }\left(2009-1\right)\div4+1=503\left(\text{số}\right)\)

\(\Rightarrow A=\frac{\left(2019+1\right)\times503}{2}=508030\)

\(\text{Vậy }A=508030\)

\(\text{b.Tương tự}\)

\(\text{c.Tương tự}\)

Kiệt à hình như cậu sai rồi 2009 chứ đâu phải 2019

Ở đoạn A = (2019-1)x503     = 508030

                               2

`2/(1.4) + 2/(4.7) + 2/(7.10) + 2/(10.13)`

`= 1/3.2(3/(1.4) + 3(4.7) + 3/(7.10) + 3/(10.13))`

`= 2/3 . (1 - 1/4 + 1/4 -1/7 + 1/7 - 1/10 + 1/10 - 1/13)`

`= 2/3 .(1 - 1/13)`

`= 2/3 . 12/13`

`= 8/13`