=\(\frac{3\left(\frac{1}{1}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{2}{4}+\frac{2}{6}+\frac{2}{8}}{5\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}\)

=\(\frac{3}{5}+\frac{2\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}{5\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}\right)}\)=\(\frac{3}{5}+\frac{2}{5}=\frac{5}{5}=1\)

Bằng 2/5

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}\)

=1

Ta có:

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left[\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right]}=\frac{1}{4}\)

Lại có:

\(\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{3\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}{4\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}=\frac{3}{4}\)

Vậy: 

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{1}{4}+\frac{3}{4}=1\)

Nhân vô rồi chuyển dấu lên và nhóm nhân -1ra ngoài rồi trg ngoặc là dãy có quy luật giải dãy đó r nhân phá ngoặc

đề học sinh giỏi hồi chiều ak!!!!!!!!! khó v:

a, => |5/3.x| = 1/6

=> 5/3.x = -1/6 hoặc 5/3.x = 1/6

=> x = -1/10 hoặc x = 1/10

Tk mk nha

\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)

<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)

<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)

<=> \(\left(x-2\frac{1}{5}\right)\) =    \(\frac{5}{6}\) : \(\frac{5}{12}\)

,<=> \(\left(x-2\frac{1}{5}\right)\)=   2 

<=. x = 2 + \(\frac{11}{5}\)

<=> x = \(\frac{21}{5}\)

\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)

\(\left(x+1\right)\left(x+3\right)=\left(0,5x+2\right)\left(2x+1\right)\)

\(x^2+4x+3=x^2+4,5x+2\)

\(x^2-x^2+4x-4,5x-2+3=0\)

\(1-0,5x=0\)

\(x=2\)

\(\left(\frac{1}{2}-\frac{3}{4}\right):\frac{1}{2}+\left(\frac{1}{2}\right)^2\)

\(=\left(\frac{2}{4}-\frac{3}{4}\right):\frac{1}{2}+\frac{1}{4}\)

\(=\frac{-1}{4}:\frac{1}{2}+\frac{1}{4}\)

\(=\frac{-1}{4}.\frac{2}{1}+\frac{1}{4}\)

\(=\frac{-1}{2}+\frac{1}{4}\)

\(=\frac{-2}{4}+\frac{1}{4}\)

\(=\frac{-1}{4}\)

1/2+2/3+3/4+4/5+5/6+6/7+7/8+8/9+9/10x9/10

=9/10x(1/2+2/3)+(3/4+4/5)+(5/6+6/7)+(7/8+8/9)

=9/10x(1/3+3/5+5/7+7/9)

9/10x(1/3+3/5)+(5/7+7/9)

=9/10x1/5+5/9

9/50+5/9

=10

Bn Long làm đúng rồi bn   nguyễn kim arica cứ làm theo cách đó là được .

Bn nào thấy đúng thì ủng hộ nha .

hình như đề bạn sai thì phải đúng ko biết ngỏ bạn làm

cái chỗ có 2 phân số 9/10 đấy