Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(10A=\frac{10\left(10^{29}+10^{10}\right)}{10^{30}+10^{10}}=\frac{10^{30}+10^{11}}{10^{30}+10^{10}}=1+\frac{10^{11}-10^{10}}{10^{30}+10^{10}}\)
\(10B=\frac{10\left(10^{30}+10^{10}\right)}{10^{31}+10^{10}}=\frac{10^{31}+10^{11}}{10^{31}+10^{10}}=1+\frac{10^{11}-10^{10}}{10^{31}+10^{10}}\)
\(10^{30}+10^{10}< 10^{31}+10^{10}\Rightarrow\frac{10^{11}-10^{10}}{10^{30}+10^{10}}>\frac{10^{11}-10^{10}}{10^{31}+10^{10}}\)
\(\Rightarrow10A=1+\frac{10^{11}-10^{10}}{10^{30}+10^{10}}>10B=1+\frac{10^{11}-10^{10}}{10^{31}+10^{10}}\)
\(\Rightarrow A>B\)
a, Ta có : \(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}=\frac{29}{87}>\frac{29}{88}\)
\(\Rightarrow\frac{13}{38}>\frac{29}{88}\Rightarrow\frac{-13}{38}< \frac{29}{-88}\)
b, Ta có: \(3^{301}>3^{300}=\left(3^3\right)^{100}=27^{100}\left(1\right)\)
\(5^{199}< 5^{200}=\left(5^2\right)^{100}=25^{100}\left(2\right)\)
Do \(25^{100}< 27^{100}\Rightarrow5^{200}< 3^{300}\)\(\left(3\right)\)
Từ \(\left(1\right),\left(2\right),\left(3\right)\Rightarrow5^{199}< 5^{200}< 3^{300}< 3^{301}\Rightarrow5^{199}< 3^{301}\)
c, Ta có: \(\frac{10^{2018}+5}{10^{2018}-8}=\frac{10^{2018}-8+13}{10^{2018}-8}=1+\frac{13}{10^{2018}-8}\)
\(\frac{10^{2019}+5}{10^{2019}-8}=\frac{10^{2019}-8+13}{10^{2019}-8}=1+\frac{13}{10^{2019}-8}\)
Do \(\frac{13}{10^{2018}-8}>\frac{13}{10^{2019}-8}\Rightarrow1+\frac{13}{10^{2018}-8}>1+\frac{13}{10^{2019}-8}\Rightarrow\frac{10^{2018}+5}{10^{2018}-8}>\frac{10^{2019}+5}{10^{2019}-8}\)
-29.(85-47)-85.(47-29)
=-29.38-85.18
=-1102-1503
=-2605
-15.(23-29)+23-(15-29)
=-15.(-6)+23-(-14)
=90+23+14
=113+14
=127
-13.(55-29)+23.(15-29)
=-13.26+23.(-14)
=-338+(-322)
=-660
-10.(2+4)-5.(4-3)
=-10.6-5.1
=-60-5
=-65
Ta có 22/37 < 29/37 và 29/37 < 29/33
=> 22/37< 29/37 < 29/33
\(\frac{1}{2}B=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{15.16}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{15}-\frac{1}{16}=\frac{1}{4}-\frac{1}{16}=\frac{3}{16}\)
=>\(B=\frac{3}{16}:\frac{1}{2}=\frac{3}{8}\)
\(C=\left(\frac{3}{29}-\frac{1}{5}\right)\cdot\frac{29}{3}=1-\frac{1}{5}\cdot\frac{29}{3}=1-\frac{29}{15}=-\frac{14}{15}\)
Ta có :
\(A=\frac{10^{29}+5}{10^{29}-2}\)\(=\frac{10^{29}-2+7}{10^{29}-2}\)\(=\frac{10^{29}-2}{10^{29}-2}+\frac{7}{10^{29}-2}\)\(=1+\frac{7}{10^{29}-2}\)
\(B=\frac{10^{29}}{10^{29}-7}=\frac{10^{29}-7+7}{10^{29}-7}=\frac{10^{29}-7}{10^{29}-7}+\frac{7}{10^{29}-7}=1+\frac{7}{10^{29}-7}\)
Vì \(\frac{7}{10^{29}-2}< \frac{7}{10^{29}-7}\Leftrightarrow A< B\)