Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a.Vì \(\frac{17}{19}< 1\) và \(\frac{19}{17}>1\)
nên \(\frac{17}{19}< 1< \frac{19}{17}\)
hay \(\frac{17}{19}< \frac{19}{17}\)
b) \(\frac{15}{7}=2\frac{1}{7}\) và \(\frac{25}{12}=2\frac{1}{12}\)
Vì \(2\frac{1}{7}>2\frac{1}{12}\) nên \(\frac{15}{7}>\frac{25}{12}\)
\(A=\frac{54.107-53}{53.107+54}\)
\(\Leftrightarrow A=\frac{53.107+107-53}{53.107+54}\)
\(\Leftrightarrow A=\frac{53.107+54}{53.107+54}\)
\(\Leftrightarrow A=1\)
\(B=\frac{135.269-133}{134.269+135}\)
\(\Leftrightarrow B=\frac{134.269+269-133}{134.269+135}\)
\(\Leftrightarrow B=\frac{134.269+135}{134.269+135}\)
\(\Leftrightarrow B=1\)
Vì 1 = 1 nên A =B
A= (54.107-53)/(53.107+54)
= (53+1).107-53 / 53.107+54
=53.107+107-53 / 53.107+54
=53.107+54 / 54.107 + 54
=1
B= 135.269-133 / 134.269+135
= (134+1).269-133 / 134.269+135
= 134.269+269-133 / 134.269+135
=134.269+136 / 134.269+135
=134.269+135/ 134.269+135 + 1/134.269+135
=1 + 1/134.269+135 >1=A
\(A=\frac{54.107-53}{53.107+54}=1\)
\(B=\frac{135.269-133}{134.269+135}>1\)
\(A=\frac{54.107-53}{53.107+54}<\frac{135.269-133}{134.269+135}\)
\(A=\frac{54\cdot107-53}{53\cdot107+54}=\frac{\left(53+1\right)107-53}{53\cdot107+54}=\frac{53\cdot107+107-53}{53\cdot107+54}=\frac{53\cdot107+54}{53\cdot107+54}=1\)
\(B=\frac{135\cdot268-133}{134\cdot269+135}=\frac{\left(134+1\right)\cdot268-133}{134\cdot269+135}=\frac{134\cdot268+268-133}{34\cdot269+135}=\frac{134\cdot268+135}{134\cdot269+135}=1\)
Vì 1=1 nên A=B
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
\(a)\) Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(A=\frac{10^{11}-1}{10^{12}-1}< \frac{10^{11}-1+11}{10^{12}-1+11}=\frac{10^{11}+10}{10^{12}+10}=\frac{10\left(10^{10}+1\right)}{10\left(10^{11}+1\right)}=\frac{10^{10}+1}{10^{11}+1}=B\)
\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~