a)Ta có:\(\dfrac{1717}{2929}=\dfrac{1717:101}{2929:101}=\dfrac{17}{29}\left(1\right)\)

.\(\dfrac{17171717}{29292929}=\dfrac{17171717:1010101}{29292929:1010101}=\dfrac{17}{29}\left(2\right)\)

Từ (1) và (2)=>\(\dfrac{1717}{2929}=\dfrac{17171717}{29292929}\)

b)Ta có \(\dfrac{3120-34}{4170-41}=\dfrac{3176}{4129}\left(1\right)\)

\(\dfrac{6420-68}{8340-82}=\dfrac{6352}{8258}=\dfrac{6352:2}{8258:2}=\dfrac{3176}{4129}\left(2\right)\)

Từ (1) và (2)=>đpcm

a) \(\frac{1717}{2929}=\frac{17.101}{29.101}=\frac{17}{29}\)

\(\frac{171717}{292929}=\frac{17.10101}{29.10101}=\frac{17}{29}\)

=> \(\frac{1717}{2929}=\frac{171717}{292929}\left(=\frac{17}{29}\right)\)

b) \(\frac{6420-68}{8340-82}=\frac{2\left(3210-34\right)}{2\left(4170-41\right)}=\frac{3210-34}{4170-41}\)

Vậy; \(\frac{3210-34}{4170-41}=\frac{6420-68}{8340-82}\)

a) Ta có :

\(\frac{1717}{2929}=\frac{17.101}{29.101}=\frac{17}{29}\)

\(\frac{171717}{292929}=\frac{17.10101}{29.10101}=\frac{17}{29}\)

Vì \(\frac{17}{29}=\frac{17}{29}\) nên \(\frac{1717}{2929}=\frac{171717}{292929}\)

b) Ta có:

\(\frac{6420-68}{8340-82}=\frac{2.3210-2.34}{2.4170-2.41}=\frac{2.\left(3210-34\right)}{2.\left(4170-41\right)}=\frac{3210-34}{4170-41}\)

Vì \(\frac{3210-34}{4170-41}=\frac{3210-34}{4170-41}\) nên \(\frac{3210-34}{4170-41}=\frac{6420-68}{8340-82}\)

\(\frac{4212}{14640}=\frac{4212:2}{14640:2}=\frac{2106}{7320}\)

\(\frac{6318}{21960}=\frac{6318:3}{21960:3}=\frac{2106}{7320}\)

Vậy\(\frac{2106}{7320}=\frac{4212}{14640}=\frac{6318}{21960}\)

tớ chỉ biết làm câu a thôi

a) 1717/2929 và 171717/292929

1717/2929 = 17x101/29x101    = 17/29                 và 17x10101/29x10101 = 17/29

Vậy hai phân số bằng nhau