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b: \(2^{91}=\left(2^{13}\right)^7\)
\(5^{35}=\left(5^5\right)^7\)
mà \(2^{13}>5^5\)
nên \(2^{91}>5^{35}\)
Ta có :
\(B=\frac{2017+2018}{2018+2019}=\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
Vì :
\(\frac{2017}{2018}>\frac{2017}{2018+2019}\)
\(\frac{2018}{2019}>\frac{2018}{2018+2019}\)
\(\Rightarrow\)\(\frac{2017}{2018}+\frac{2018}{2019}>\frac{2017+2018}{2018+2019}\) hay \(A>B\)
Vậy \(A>B\)
Chúc bạn học tốt ~
Bài toán : So sánh A và B
\(A=\frac{2018^{100}}{1+2018+2018^2+...+2018^{100}}\)
+) Ta có \(\frac{1}{A}=\frac{1+2018+2018^2+...+2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{2018}{2018^{100}}+\frac{2018^2}{2018^{100}}+...+\frac{2018^{100}}{2018^{100}}\)
\(=\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1\)
\(B=\frac{2019^{100}}{1+2019+2019^2+...+2019^{100}}\)
+) Ta có \(\frac{1}{B}=\frac{1+2019+2019^2+...+2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{2019}{2019^{100}}+\frac{2019^2}{2019^{100}}+...+\frac{2019^{100}}{2019^{100}}\)
\(=\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
+) \(\frac{1}{2018^{100}}>\frac{1}{2019^{100}}\)
\(\frac{1}{2018^{99}}>\frac{1}{2019^{99}}\)
.....................................
\(1=1\)
\(\Rightarrow\frac{1}{2018^{100}}+\frac{1}{2018^{99}}+\frac{1}{2018^{98}}+...+1>\frac{1}{2019^{100}}+\frac{1}{2019^{99}}+\frac{1}{2019^{98}}+...+1\)
\(\Rightarrow\frac{1}{A}>\frac{1}{B}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
Ta có :
\(\frac{2017}{2018}>\frac{2017}{2018+2019}\)
\(\frac{2018}{2019}>\frac{2018}{2018+2019}\)
\(\Rightarrow\frac{2017}{2018}+\frac{2018}{2019}>\frac{2017}{2018+2019}+\frac{2018}{2018+2019}\)
\(\Rightarrow\frac{2017}{2018}+\frac{2018}{2019}>\frac{2017+2018}{2018+2019}\)
\(\Rightarrow A>B\)
Chúc bạn học tốt !!!!
Vì \(\frac{2017}{2018}>\frac{2017}{2018+2019}\)
Vì \(\frac{2018}{2019}>\frac{2018}{2018+2019}\)
\(\Rightarrow\frac{2017}{2018}+\frac{2018}{2019}>\frac{2017+2018}{2018+2019}\)
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 \(\frac{13}{10^{2018}-8}>\frac{13}{10^{2019}-8}\Rightarrow1+\frac{13}{10^{2018}-8}>1+\frac{13}{10^{2019}-8}\Rightarrow\frac{10^{2018}+5}{10^{2018}-8}>\frac{10^{2019}+5}{10^{2019}-8}\)
b. 291 >290 =(25)18=3218
535 <536 =(52)18=2518
Vì 3218> 2518 => 290> 536 => 291> 535
Tiếp sức cho Quân Tạ Minh câu a) nè! Áp dụng BĐT tam giác vô giải tuyệt cú mèo!!!
a) \(2018^{2019}+2018^{2018}\) và \(2019^{2018}\)
Áp dụng BĐT tam giác: a + b > c ( a,b là hai cạnh của tam giác, c là cạnh còn lại.) Thế \(2018^{2019}=a;2018^{2018}=b;2019^{2018}=c\) theo BĐT tam giác,ta có: \(2018^{2019}+2018^{2018}>2019^{2018}\)