Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
đề có pải là A=\(\frac{19^{30}+5}{19^{31}+5}\) ; B=\(\frac{19^{31}+5}{19^{32}+5}\) PẢI KO BẠN
C = 1930+5/1931+5
=>19C = 1931+95/1931+5 = 1+ [90/1931+5]
D = 1931+5/1932+5
=>19D = 1932+95/1932+5 = 1 + [90/1932+5]
ma 90/1931+5 > 90/1932+5
=>19C > 19D
=>C > D
Ta có: \(A=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19A=\frac{19.\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
\(B=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19B=\frac{19.\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
Nên \(19A< 19B\Rightarrow A< B\)
Nhầm: Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow A>B\)
Ta có 1930<1931
\(\left(\frac{5}{19}\right)^{31}< \left(\frac{5}{19}\right)^{32}\)
5=5
công vế theo vế ta có
\(19^{30}+\left(\frac{5}{19}\right)^{31}+5< 19^{31}+\left(\frac{5}{19}\right)^{32}+5\)
Vậy A<B
\(a,\frac{-8}{15}=\frac{-8.12}{15.12}=\frac{-96}{180}\left(1\right)\)
\(\frac{7}{12}=\frac{7.15}{12.15}=\frac{105}{180}\left(2\right)\)
Từ 1 và 2 \(\Rightarrow\frac{-8}{15}< \frac{7}{12}\)
\(b,\frac{13}{19}=\frac{13.53}{19.53}=\frac{689}{1007}\left(1\right)\)
\(\frac{47}{53}=\frac{47.19}{53.19}=\frac{893}{1007}\left(2\right)\)
Từ 1 và 2 \(\Rightarrow\frac{13}{19}< \frac{47}{53}\)
Xét B = \(\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+14}{19^{32}+5+14}=\frac{19^{31}.19}{19^{32}.19}=\frac{19\left(19^{30}+1\right)}{19\left(19^{31}+1\right)}=\frac{19^{30}+1}{19^{31}+1}< \frac{19^{30}+5}{19^{31}+5}=A\)Vậy A > B
\(\frac{19^{30}+5}{19^{31}+5}\)và \(\frac{19^{31}+5}{19^{32}+5}\)
Xét biểu thức \(\frac{19^{30}+5}{19^{31}+5}\)là A và biểu thức \(\frac{19^{31}+5}{19^{32}+5}\)là B
\(B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+14}{19^{32}+5+14}\)
\(=\)\(\frac{19^{31}.19}{19^{32}.19}\)\(=\)\(\frac{19.\left(19^{30}+1\right)}{19.\left(19^{31}+1\right)}\)
\(=\)\(\frac{19^{30}+1}{19^{31}+1}\)
Mà \(\frac{19^{30}+1}{19^{31}+1}< \frac{19^{30}+5}{19^{31}+5}\)
Nên \(A>B\)
Ta có : \(A=\frac{19^{30}+15}{19^{31}+15}\)
\(\Rightarrow19A=\frac{19^{31}+285}{19^{31}+15}=\frac{19^{31}+15+270}{19^{31}+15}=1+\frac{270}{19^{31}+15}\)
Lại có \(B=\frac{19^{31}+15}{19^{32}+15}\)
\(\Rightarrow19B=\frac{19^{32}+285}{19^{32}+15}=\frac{19^{32}+15+270}{19^{32}+15}=1+\frac{270}{19^{32}+15}\)
Vì \(\frac{270}{19^{32}+15}< \frac{270}{19^{31}+15}\Rightarrow1+\frac{270}{19^{32}+5}< 1+\frac{270}{19^{31}+15}\Rightarrow19B< 19A\Rightarrow B< A\)