Có: \(\dfrac{5454}{5757}-\dfrac{171717}{191919}.\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

=> \(\dfrac{54.101}{57.101}-\dfrac{17.10101}{19.10101}.\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

=> \(\dfrac{18}{19}-\dfrac{17}{19}.\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

=> \(\dfrac{21}{76}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

=> \(\dfrac{1}{4}:x=\dfrac{47}{380}\)

=> x = \(\dfrac{95}{47}\)

=>18/19-17/19*3/4+1/4:x=2/5

=>1/4:x=2/5-18/19+17/19*3/4=47/380

=>x=1/4:47/380=95/47

Môn Toán nha mọi người, mình nhầm!

\(\dfrac{5454}{5757}-\dfrac{171717}{191919}=\dfrac{1}{19}\)

\(\dfrac{2}{19}\)

Kết quả là:

 \(\frac{1}{19}\)

Ủng hộ mk nha các bn !!!

5454/5757=18/19

171717/191919=17/19

18/19-17/19=1/19

=54/57-17/19

=1/19

k nha

\(A=\frac{5454}{5757}-\frac{171717}{191919}=\frac{54}{57}-\frac{17}{19}=\frac{54}{57}-\frac{51}{57}=\frac{3}{57}=\frac{1}{19}\)

\(\frac{5454}{5757}-\frac{171717}{191919}=\frac{54x101}{57x101}-\frac{17x10101}{19x10101}=\frac{54}{57}-\frac{17}{19}=\frac{1}{19}\)

\(\frac{5454}{5757}-\frac{171717}{191919}=\frac{18.303}{19.303}-\frac{17.10101}{19.10101}=\frac{18}{19}-\frac{17}{19}=\frac{1}{19}\)

Ta có :

\(\frac{5454}{5757}=\frac{54.101}{57.101}=\frac{54}{57}\); \(\frac{171717}{191919}=\frac{17.10101}{19.10101}=\frac{17}{19}\)

Vậy \(\frac{54}{57}-\frac{17}{19}=\frac{1}{19}\)

\(\frac{5454}{5757}=\frac{54}{57}=\frac{18}{19};\frac{171717}{191919}=\frac{17}{19}\)

=> \(\frac{5454}{5757}=\frac{171717}{191919}=\frac{18}{19}-\frac{17}{19}=\frac{1}{19}\)

\(\dfrac{5454}{5757}-\dfrac{171717}{191919}=\dfrac{18\cdot3\cdot101}{19\cdot3\cdot101}-\dfrac{17\cdot10101}{19\cdot10101}=\dfrac{18}{19}-\dfrac{17}{19}=\dfrac{1}{19}\)

1) a) \(\frac{5454}{5757}-\frac{171717}{191919}=\frac{18\times3\times101}{19\times3\times101}-\frac{17\times10101}{19\times10101}=\frac{18}{19}-\frac{17}{19}=\frac{1}{19}\)

b) \(\frac{6}{5}\times\frac{7}{6}\times\frac{8}{7}\times....\times\frac{2021}{2020}=\frac{6\times7\times8\times...\times2021}{5\times6\times7\times...\times2020}=\frac{2021}{5}\)

2) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}=2\times\frac{1}{6}+2\times\frac{1}{12}+2\times\frac{1}{20}+...+2\times\frac{1}{90}\)

\(=2\times\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)

\(=2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{9\times10}\right)\)

\(=2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=2\times\left(\frac{1}{2}-\frac{1}{10}\right)=2\times\frac{2}{5}=\frac{4}{5}\)

b)Vì \(a-1< a+1\)

=> \(\frac{1}{a-1}>\frac{1}{a+1}\)

1/ a x b = 1/a - 1/b 

đk : b - a = 1

a)  Cách 1 :                                                          Cách 2

1 + 3 +5 + 7 + 9 + 11 + 13 + 15 + 17 + 19             1 + 3 +5 + 7 + 9 + 11 + 13 + 15 + 17 + 19

=(1 + 19) + (3 + 17) +.... + (9 + 11)                   Áp dụng công thức tính dãy số ta có :

= 20 + 20 + ... + 20                                           \(\frac{\left[\left(19-1\right):2+1\right].\left(19+1\right)}{2}=\frac{10.20}{2}=10.10=100\)

= 20 x 5 = 100

b) giống bài a     nhưng cách 1 làm dài lắm , mình sẽ làm cách 2

áp dụng công thức tính dãy số ta có:

\(\frac{\left[\left(200-4\right):4+1\right].\left(200+4\right)}{2}=\frac{50.204}{2}=50.102=5100\)

                         Tong :  \(\frac{\left[200+4\right].50}{2}\) =5100

a) Ta có C = 1 + 4 + 4² + ... + 4⁶

              4C = 4( 1 + 4 + 4² + ... + 4⁶ )

                    = 4 + 4² + 4³ + ... + 4⁷

b) Ta có C = 1 + 4 + 4² + ... + 4⁶

              4C = 4 + 4² + 4³ + ... + 4⁷

⇒ 4C - C = ( 4 + 4² + 4³ + ... + 4⁷ ) - ( 1 + 4 + 4² + ... + 4⁶ )

⇒ 3C = 4 + 4² + 4³ + ... + 4⁷ - 1 - 4 - 4² - ... - 4⁶

⇒ 3C = 4⁷ - 1

⇒ C = \(\dfrac{4^7-1}{3}\) ( đpcm )

vãi tui đang mắc ở bài này