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a) 27/82 < 26/75 ( 2025/6250 < 2132\6250)
b) -49/78 > 64/ -95 ( - 3136/7410 > -4992/7410)
c) ta có: \(A=\frac{54.107-53}{53.107}=\frac{53.107+(107-53)}{53.107+54}=\frac{53.107+54}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}=\frac{134.269+\left(269-133\right)}{134.269+135}=\frac{134.269+136}{134.269+135}>1\)
\(\Rightarrow A< B\)
d) ta có: \(A=\frac{3^{10}+1}{3^9+1}=\frac{3.\left(3^9+1\right)-2}{3^9+1}=\frac{3.\left(3^9+1\right)}{3^9+1}-\frac{2}{3^9+1}=3-\frac{2}{3^9+1}\)
\(B=\frac{3^9+1}{3^8+1}=\frac{3.\left(3^8+1\right)-2}{3^8+1}=\frac{3.\left(3^8+1\right)}{3^8+1}-\frac{2}{3^8+1}=3-\frac{2}{3^8+1}\)
mà \(\frac{2}{3^9+1}< \frac{2}{3^8+1}\Rightarrow3-\frac{2}{3^9+1}< 3-\frac{2}{3^8+1}\)
=> A < B
B= 20^9+1/20^10+1
B= 20^9 +1 +19/ 20^10+1+19
B= 20^9 +20 /20^10+20
B= 20(20^8 +1) / 20(20^9+1)
B= 20^8+1 / 20^9+1 =A
=> A = B
Vậy...
b) C= 54.107- 53/ 53.107+ 54
C= (53+1)107-53 / 53.107 +54
C= 53.107+ 1.107 - 53/ 53.107 +54
C= 53.107 + 107 -53/ 53.107 +54
C= 53.107 + 54 / 53.107 + 54
C= 1
Vậy...
a, A = \(\frac{20^8+1}{20^9+1}\)