Đáp án :

93/140 > 63/97 > 31/49

ko có cách giải , tự tìm hiểu

ủng hộ tớ nha

93/140 > 63/97 > 31/49

ai tk mình mình tk lại cho

\(\frac{31}{49}< \frac{62}{97}< \frac{93}{140}\)

38 + 93 = 131

28 + 94 = 122

     học tốt nha e và đừng quên k cho cj nha

\(38+93=131\)

\(28+94=122\)

hok tốt

38+93=131

28+94=122

hok tốt

\(39+93=132\)

\(93+38=131\)

học tốt

\(39+93=132\)

\(93+38=131\)

hok tốt

39 + 93 = 132

93 + 38 = 131

a. Ta có \(\frac{-29}{39}=-1+\frac{10}{39}\)

\(\frac{-39}{49}=-1+\frac{10}{49}\)
có -1 = -1 mà \(\frac{10}{39}>\frac{10}{49}\)do cùng tử mà mẫu càng lớn thì phân số đó bé hơn
vì vậy nên \(\frac{-29}{39}>\frac{-39}{49}\)

b. có \(\frac{20142014:\text{10001}}{20152015:\text{10001}}=\frac{2014}{2015}\)
vậy \(\frac{20142014}{20152015}=\frac{2014}{2015}\)

Chúc bạn học tốt

Ta có A = \(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)

= \(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)

= \(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)

= \(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\) 

B = \(\dfrac{\dfrac{2}{29}-\dfrac{2}{39}+\dfrac{2}{49}}{\dfrac{23}{29}-\dfrac{23}{39}+\dfrac{23}{49}}=\dfrac{2\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}{23\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}=\dfrac{2}{23}\) 

Lại có \(\dfrac{2}{23}>\dfrac{2}{24}=\dfrac{1}{12}\) hay A < B

Vậy A < B

Cảm ơn bạn nhé!

Rút gọn.

\(B=\dfrac{x^{39}x^{36}x^{33}...x^31}{x^{40}x^{38}x^{36}...x^21}=\dfrac{x^{\left(39+36+33+...+3\right)}}{x^{\left(40+38+36+...+2\right)}}\)

ta có: \(39+36+33+...+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)

\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)

=> \(B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)

Tương tự như B => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)

Ta có:

\(B=\dfrac{x^{\left(39+36+33+....+3\right)}}{x^{\left(40+38+36+....+2\right)}}\)

\(39+36+33+....+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)

\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)

\(\Rightarrow B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)

tương tự => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)