ở câu 1 ở mỗi phẫn số chúng ta cộng thêm 1, tổng là ta cộng thêm 5. Lấy 5 + -5=0. Rồi ta được tất cả tử là x+200,đặt chung ra ngoài,từ đó tính x=-200

Câu 2: => 2/42 + 2/56 + 2/72 + ... + 2/x(x+1) = 2/9

=> 2/6*7+2/7*8+...+2/x(X+1) = 2/9

các bạn giúp mình với!!

tung từng vế một thôi

bạn nhác quá éo chịu suy nghĩ

bài này dễ vl

Bài 1:

a, \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2010}{2011}\)

\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(1-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(\frac{1}{5x+6}=1-\frac{2010}{2011}\)

\(\frac{1}{5x+6}=\frac{1}{2011}\)

=> 5x + 6 = 2011

    5x = 2011 - 6

    5x = 2005

    x = 2005 : 5

    x = 401

b, \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)

\(\frac{7}{x}=\frac{7}{15}\)

=> x = 15

c, ghi lại đề

d, ghi lại đề

Bài 2:

\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)

Tìm số tự nhiên x,y biết :x-3=y*(x+2)

|\(\frac32x\) + \(\frac12\)| = |4\(x\) - 1|

\(\left[\begin{array}{l}\frac32x+\frac12=-4x+1\\ \frac32x+\frac12=4x-1\end{array}\right.\)

\(\left[\begin{array}{l}\frac32x+4x=1-\frac12\\ \frac32x-4x=-1-\frac12\end{array}\right.\)

\(\left[\begin{array}{l}\frac{11}{2}x=\frac12\\ -\frac52x=-\frac32\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12:\frac{11}{2}\\ x=-\frac32:\frac{-5}{2}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac12\times\frac{2}{11}\\ x=-\frac32\times\frac{-2}{5}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{1}{11}\\ x=\frac35\end{array}\right.\)

Vậy \(x\in\) {\(\frac{1}{11};\frac35\)}

|\(\frac54x\) - \(\frac72\)| - |\(\frac58x\) + \(\frac35\)| = 0

|\(\frac54x\) - \(\frac72\)| = |\(\frac58x\) + \(\frac35\)|

\(\left[\begin{array}{l}\frac54x-\frac72=-\frac58x-\frac35\\ \frac54x-\frac72=\frac58x+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac54x+\frac58x=\frac72-\frac35\\ \frac54x-\frac58x=\frac72+\frac35\end{array}\right.\)

\(\left[\begin{array}{l}\frac{15}{8}x=\frac{29}{20}\\ \frac58x=\frac{41}{10}\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{29}{10}:\frac{15}{8}\\ x=\frac{41}{10}:\frac58\end{array}\right.\)

\(\left[\begin{array}{l}x=\frac{116}{75}\\ x=\frac{164}{25}\end{array}\right.\)

Vậy \(x\in\) {\(\frac{116}{75}\); \(\frac{164}{25}\)}

a) \(\frac{9}{20}\)                                                   c) \(\frac{-55}{4}\)                                                                                                                                 

b) \(\frac{116}{75}\)                                                d) \(\frac{-76}{45}\)

đúng hết đấy nhé mình tính kĩ lắm ko sai đâu

       chúc may mắn

3) 

3/5 + 3/7-3/11 /  4/5 + 4/7- 4/11

= 3.( 1/5 + 1/7 - 1/11)/4.(1/5+1/7-1/11)

= 3/4

1,

 ta có B = 196+197/197+198 = 196/(197+198) + 197/(197+198)

      196/197 > 196/197+198

      197/198 > 197/197+198

=> A>B

Ta có: \(B=\frac{1}{199}+\frac{2}{198}+...+\frac{199}{1}\)

\(=\frac{200-199}{199}+\frac{200-198}{198}+...+\frac{200-1}{1}\)

\(=\frac{200}{199}-\frac{199}{199}+\frac{200}{198}-\frac{198}{198}+...+\frac{200}{1}-\frac{1}{1}\)

\(=\left(\frac{200}{199}+\frac{200}{198}+...+\frac{200}{1}\right)-\left(\frac{199}{199}+\frac{198}{198}+...+\frac{1}{1}\right)\)

\(=200+200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)-199\)

\(=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)+\frac{200}{200}\)

\(=200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{200}+\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)

Ta có :

 \(B=\frac{1}{199}+\frac{2}{198}+....+\frac{198}{2}+\frac{199}{1}\)

 \(B=1+\frac{1}{199}+1+\frac{1}{198}+....+1+\frac{198}{2}\)

\(B=\frac{200}{199}+\frac{200}{198}+...+\frac{200}{2}\)

\(B=200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{200}}{200\left(\frac{1}{199}+\frac{1}{198}+...+\frac{1}{2}\right)}=\frac{1}{200}\)

Vậy \(\frac{A}{B}=\frac{1}{200}\)

a) = 3/3 x ( -24/54 +45/54 ) : 7/12

   = 1 x 21/54 x 12/7

   = 18/27 

( hiện tại mik chỉ lm đc thế này thui. thông cảm nk )

S = \(\frac12\times\frac13\) + \(\frac13\times\frac14\) + \(\frac14\times\frac15\) + \(\frac15\times\frac16\) + \(\frac17\times\frac18\) + \(\frac18\times\frac19\)

S = \(\frac12\) - \(\frac13\) + \(\frac13\) - \(\frac14\) + \(\frac14\) - \(\frac15\) + \(\frac15\) - \(\frac16\) + \(\frac17\) - \(\frac18\) + \(\frac18\) - \(\frac19\)

S = \(\frac12\) - \(\frac19\)

S = \(\frac{9}{18}-\frac{2}{18}\)

S = \(\frac{7}{18}\)