Áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{7x+2}{5x+7}=\frac{7x-1}{5x+1}=\frac{\left(7x+2\right)-\left(7x-1\right)}{\left(5x+7\right)-\left(5x+1\right)}=\frac{3}{6}=\frac{1}{2}\)

=> 2(7x + 2) = 5x + 7

14x + 4 = 5x + 7

14x - 5x = 7 - 4

9x = 3

x = 3:9

x = 0,(3)

\(\frac{x+1}{x-1}=\frac{7}{3}\)

=> \(3.\left(x+1\right)=7.\left(x-1\right)\)

=> \(3x+3=7x-7\)

=> \(3x+10=7x\)

=> \(4x=10\)

=> \(x=\frac{10}{4}=\frac{5}{2}\)

Vậy \(x=\frac{5}{2}\)

\(\frac{x+1}{x-1}=\frac{7}{3}\)

\(\Rightarrow3\left(x+1\right)=7\left(x-1\right)\)

\(\Leftrightarrow3x+3=7x-7\)

\(\Leftrightarrow-4x=-10\)

\(\Leftrightarrow x=\frac{5}{2}\)

~~~!!!

\(\frac{x+1}{x-1}=\frac{7}{3}\)

\(\Leftrightarrow3x+3=7x-7\)

\(\Leftrightarrow3x=7x-7-3\)

\(\Leftrightarrow-4x=-10\)

\(\Rightarrow x=\frac{5}{2}\)

a)\(\frac{1}{99.97}\)−\(\frac{1}{97.95}\)−\(\frac{1}{95.93}\)−…−\(\frac{1}{5.3}\)−\(\frac{1}{3.1}\)

=\(\frac{1}{99.97}\)−(\(\frac{1}{97.95}\)+\(\frac{1}{95.93}\)+…+\(\frac{1}{5.3}\)+\(\frac{1}{3.1}\))

=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).(\(\frac{1}{95}\)−\(\frac{1}{97}\)+\(\frac{1}{93}\)−\(\frac{1}{95}\)+…+\(\frac{1}{3}\)−\(\frac{1}{5}\)+1−\(\frac{1}{3}\))

=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).(1−\(\frac{1}{97}\))
=\(\frac{1}{99.97}\)−\(\frac{1}{2}\).\(\frac{96}{97}\)

=\(\frac{1}{99.97}\)−\(\frac{48}{97}\)

=\(\frac{1}{99.97}\)−\(\frac{48.99}{99.97}\)

=\(\frac{-4751}{9603}\)

ta có\(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-...-\frac{1}{1024}\)

\(=\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

tách

\(B=\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)

\(2B=\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)

\(2B-B=\frac{1}{2}-\frac{1}{1024}\)

thay vào B ta có 

\(\frac{1}{2}-\left(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{1024}=\frac{1}{1024}\)

\(A=\frac{1}{2}-\frac{1}{4}-\cdot\cdot\cdot-\frac{1}{1024}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\)

\(\Rightarrow2A=1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1-\frac{1}{2}-\cdot\cdot\cdot-\frac{1}{2^9}\right)-\left(\frac{1}{2}-\frac{1}{2^2}-\cdot\cdot\cdot-\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{1}{2}+\frac{1}{2^{10}}\)

\(\Rightarrow A=\frac{2^9+1}{2^{10}}\)

\(\Rightarrow A=\frac{513}{1024}\)

= \(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{18}\)\(\frac{1}{18}\)

= \(\frac{1}{3}-\frac{1}{18}\)

= \(\frac{5}{18}\)

kết quả là 439/2808

[{1/4-1/4}+ { 1/6-1/6}+{1/9-1/9}+{1/13-1/13}]*11/4=0

\(x\ne-\frac{7}{5};x\ne-\frac{1}{5}\)

Đề \(\Leftrightarrow\left(7x+2\right)\left(5x+1\right)=\left(7x-1\right)\left(5x+7\right)\)

\(\Leftrightarrow35x^2+7x+10x+2=35x^2+49x-5x-7\)

\(\Leftrightarrow17x-44x=-2-7\)

\(\Rightarrow-27x=-9\Rightarrow x=\frac{1}{3}\) (thỏa)

Vậy x = 1/3

Phân số thứ 20 à , hơi khó đó

Nhưng kết quả là:\(\frac{1}{1599}\) 

Kết quả đúng là \(\frac{1}{1599}\)

mình làm rùi 

ủng hộ nha

toan lop 6 day a co ma lop 5 y ngu hay sao ma hoi