A = \(\left(\frac{7777}{8585}-\frac{77}{85}+\frac{7777}{16362}-\frac{77}{162}\right)\) . \(\frac{123498766}{987661234}\)

A = \(\left(\frac{7777}{8585}-\frac{7777}{8585}+\frac{7777}{16362}-\frac{7777}{16362}\right)\). \(\frac{123498766}{987661234}\)

A = 0 . \(\frac{123498766}{987661234}\) = 0

Câu 1 có vế sau = 0

Câu 2 có vế trước = 0

Một biểu thức nhân với 0 thì = 0, nên:

=> kết quả hai bài đều = 0

cả 2 câu đều bằng 0

à mình lộn đề bài là so sánh các phân sô sau nha mọi ng

ta thấy;

\(\frac{77}{76}+\frac{-1}{76}=1\)

\(\frac{84}{83}+\frac{-1}{83}=1\)

mà\(\frac{-1}{76}=\frac{1}{-76}>\frac{1}{-83}=\frac{-1}{83}\)

=>\(\frac{77}{76}< \frac{84}{83}\)

vậy...

b)ta thấy

\(\frac{47}{97}+\frac{50}{97}=1\)

\(\frac{63}{113}+\frac{50}{113}=1\)

mà \(\frac{50}{97}>\frac{50}{113}\)

=>\(\frac{47}{97}< \frac{63}{113}\)

vậy...

c cx làm tương tự z

bn gắng làm nha

a: x/5=32/80

nên x/5=2/5

hay x=2

13/x=26/30

nên 13/x=13/15

hay x=15

-x/7=22/-77

=>x/7=2/7

hay x=2

b: x/9=28/36

=>x/9=7/9

hay x=7

-10/x=50/55

=>-10/x=10/11

hay x=-11

a)

1) x/5 = 32/80

\(x : 5 = 32 : 80\) \(x = 5 \cdot \frac{32}{80}\) \(x = 5 \cdot \frac{2}{5}\) \(x = 2\)

2) 13/x = 26/30

\(13 : x = 26 : 30\) \(13 \cdot 30 = x \cdot 26\) \(390 = 26 x\) \(x = 390 : 26\) \(x = 15\)

3) -x/7 = 22/(-77)

\(- x : 7 = 22 : \left(\right. - 77 \left.\right)\) \(- x = 7 \cdot \frac{22}{- 77}\) \(- x = 7 \cdot \left(\right. - \frac{2}{7} \left.\right)\) \(- x = - 2\) \(x = 2\)

b)

1) x/9 = 28/36

\(x : 9 = 28 : 36\) \(x = 9 \cdot \frac{28}{36}\) \(x = 9 \cdot \frac{7}{9}\) \(x = 7\)

2) -10/x = 50/55

\(- 10 : x = 50 : 55\) \(- 10 \cdot 55 = 50 x\) \(- 550 = 50 x\) \(x = - 550 : 50\) \(x = - 11\)

3) 23/(-x) = 46/50

\(23 : \left(\right. - x \left.\right) = 46 : 50\) \(23 \cdot 50 = 46 \left(\right. - x \left.\right)\) \(1150 = - 46 x\) \(x = 1150 : \left(\right. - 46 \left.\right)\) \(x = - 25\)

✅ Kết quả cuối cùng

a)

1. x = 2
2. x = 15
3. x = 2

b)

1. x = 7
2. x = -11
3. x = -25

\(A=\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)

\(4A=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{15}+...+\frac{1}{n^2}-\frac{1}{4n}=\frac{56}{673}\)

\(\Rightarrow4A=\)

\(\frac{1}{21}+\frac{1}{77}+\frac{1}{165}+...+\frac{1}{n^2+4n}=\frac{56}{673}\)

\(\Rightarrow\frac{1}{3.7}+\frac{1}{7.11}+\frac{1}{11.15}+...+\frac{1}{n\left(n+4\right)}=\frac{56}{673}\)

\(\Rightarrow\frac{1}{4}\left(\frac{4}{3.7}+\frac{4}{7.11}+\frac{4}{11.15}+...+\frac{4}{n\left(n+4\right)}\right)=\frac{56}{673}\)

\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{n}-\frac{1}{n+4}\right)=\frac{56}{673}\)

\(\Rightarrow\frac{1}{4}\left(\frac{1}{3}-\frac{1}{n+4}\right)=\frac{56}{673}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{56}{673}:\frac{1}{4}\)

\(\Rightarrow\frac{1}{3}-\frac{1}{n+4}=\frac{224}{673}\)

\(\Rightarrow\frac{1}{n+4}=\frac{1}{3}-\frac{224}{673}\)

\(\Rightarrow\frac{1}{n+4}=\frac{1}{2019}\)

=> n + 4 = 2019 

     n = 2019 - 4

     n = 2015

\(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+...+\frac{2}{x\left(x+3\right)}=\frac{101}{770}\)
\(\Rightarrow\)\(\frac{3}{2}.\left(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+...+\frac{2}{x\left(x+3\right)}\right)=\frac{101}{770}\). 
\(\Rightarrow\)\(\frac{3}{40}+\frac{3}{88}+\frac{3}{154}+...+\frac{3}{x\left(x-3\right)}=\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x-1}\)
\(\Rightarrow\)\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\frac{308}{1540}-\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\frac{5}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\)\(\frac{1}{308}\)
\(\Rightarrow\)\(x+3=308\)
\(\Rightarrow\)\(x=308-3\)
\(\Rightarrow\)\(x=305\)