\(\text{Đặt: S= biểu thức cần tính}\)

\(\Rightarrow9S=1.4.7+4.7.9+......+19.22.9+4.2\)

\(\Rightarrow9S=1.4.7+4.7\left(10-1\right)+...+19.22\left(25-16\right)+8\)

\(\Rightarrow9S=19.22.25+8\Rightarrow S=1162\)

sai rồi 9S = 1.4.9 mà 

đề ???

Đăt A = 1.4 + 4.7 + 7.10 +  ....+ 19.22

=> 9A = 1.4.9 + 4.7.9 + 7.10.9 + ...+ 19.22.9

9A = 1.4.(7+2) + 4.7.(10-1) + 7.10.(13-4) + ...+ 19.22.(25-16)

9A = 1.4.7 + 4.2 + 4.7.10 - 1.4.7 + 7.10.13 - 4.7.10 + ....+ 19.22.25 - 16.19.22

9A = 4.2 + 19.22.25

A = 1 162

Dạng này có nhiều rồi, bạn tham khảo câu hỏi tương tự cũng được

\(C=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{22}.\)

\(C=\frac{1}{1}-\frac{1}{22}\)

\(C=\frac{21}{22}\)

\(C=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{19.22}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{22}\)

\(=1-\frac{1}{22}\)

\(=\frac{21}{22}\)

=>c= 1/1-1/4+1/4-1/7+....+1/19-1/22

=> c=  1-1/22

=> C= 21/22

 Các bạn giúp mk lên 1oosp nha.ai k mk mk k lại

Bài 1: Tính tổng S

\(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{19.22}\)

\(4S=\dfrac{4}{1.4}+\dfrac{4}{4.7}+\dfrac{4}{7.10}+...+\dfrac{4}{19.22}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{19}-\dfrac{1}{22}\)

\(=1-\dfrac{1}{22}\)

\(S=\dfrac{21}{22}.\dfrac{1}{4}=\dfrac{21}{88}\)

Ta có:
A  = 1/1.4 + 1/4.7 + 1/7.10 +...+ 1/16.19
3A= 1/3.(3/1.4 + 3/4.7 + 3/7.10 + ... + 3/16.19)
     = 1/3. (1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + ... + 1/16 - 1/19)
     = 1/3.(1 - 1/19)
     = 1/3. 18/19
    = 6/19

1.

E = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{4.7}\) + \(\dfrac{3}{7.10}\) + \(\dfrac{3}{10.13}\) + \(\dfrac{3}{13.16}\) + \(\dfrac{3}{16.19}\) + \(\dfrac{3}{19.22}\)

E = 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ... +\(\dfrac{1}{19}\) - \(\dfrac{1}{22}\)

E = 1 - \(\dfrac{1}{22}\)

E = \(\dfrac{21}{22}\)

2.

(x - 4)(x - 5) = 0

TH_1:

x - 4 = 0 => x = 4

TH_2:

x - 5 = 0 => x = 5

Vậy: x = 4 hoặc x = 5

\(A=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{61\cdot64}+\dfrac{3}{64\cdot67}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\)

\(A=1-\dfrac{1}{67}\) < 1

=> A<1

Ta có:

\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{61.64}+\dfrac{3}{64.67}\)

\(=3.\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{61}-\dfrac{1}{64}+\dfrac{1}{64}-\dfrac{1}{67}\right)\)

\(=3.\left(1-\dfrac{1}{67}\right)\)

\(=3.\dfrac{66}{67}\)

\(=\dfrac{198}{67}\)

Vì \(\dfrac{198}{67}\) có tử lớn hơn mẫu nên \(\dfrac{198}{67}>1\)

Vậy \(A>1\)

sửa bài:

...  \(=1-\dfrac{1}{67}\)

\(=\dfrac{66}{67}\)

Vì \(\dfrac{66}{67}\) có tử nhỏ hơn mẫu nên \(\dfrac{66}{67}< 1\)

Vậy \(A< 1\)

A= 2/1.4+2/4.7+2/7.10+...+2/97.100

= 2.(1/1.4+1/4.7+1/7.10+...+1/97.100)

= 2.(1/1-1/4+1/4-1/7+1/7-1/10+...+1/97-1/100)

= 2.(1/1-1/100)

= 2.(99/100)

=99/50

\(A=\dfrac{2}{1\cdot4}+\dfrac{2}{4\cdot7}+\dfrac{2}{7\cdot10}+...+\dfrac{2}{97\cdot100}\)

\(A=\dfrac{2}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{97\cdot100}\right)\)

\(A=\dfrac{2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(A=\dfrac{2}{3}\cdot\left(1-\dfrac{1}{100}\right)\)

\(A=\dfrac{2}{3}\cdot\dfrac{99}{100}\)

\(A=\dfrac{33}{50}\)

\(A=\dfrac{2}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{100}\right)=\dfrac{2}{3}\times\dfrac{99}{100}=\dfrac{33}{50}\)