So sánh
\(\frac{3}{\frac{9}{11}}\); \(\frac{\frac{3}{9}}{11}\);\(\frac{3}{9+11}\)và\(\frac{3}{9\times11}\).
Cho \(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
So sánh S với 10
Ta có :
\(S=\frac{3}{2}+\frac{4}{3}+\frac{5}{4}+\frac{6}{5}+\frac{7}{6}+\frac{8}{7}+\frac{9}{8}+\frac{10}{9}+\frac{11}{10}+\frac{12}{11}\)
\(S=\frac{2+1}{2}+\frac{3+1}{3}+\frac{4+1}{4}+...+\frac{11+1}{11}\)
\(S=\left(1+\frac{1}{2}\right)+\left(1+\frac{1}{3}\right)+\left(1+\frac{1}{4}\right)+...+\left(1+\frac{1}{11}\right)\)
\(S=\left(1+1+1+...+1\right)+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)\)
\(S=10+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{11}\right)>10\)
\(\Rightarrow\)\(S>10\)
Vậy \(S>10\)
Chúc bạn học tốt ~
So sánh:
\(\frac{9}{10};\frac{3}{4};\frac{11}{14};\frac{13}{18}\)
so sánh
\(\frac{5}{9}< 9-\sqrt{\frac{11}{2}}< 3\)
Phương ngân là bồ tao ai đụng vào tao đánh chết
đây là chỗ để hỏi đáp học tập không phải chỗ nói tào lao
A=\(\frac{3^3}{1}-\frac{5^3}{3}+\frac{7^3}{6}-\frac{9^3}{10}+\frac{11^3}{15}-\frac{13^3}{21}+...+\frac{1993^3}{4950}\). So sánh A và B=814
Tính A= \(\left[\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right]:\frac{2014}{2015}\)
So sánh 199110 với 99612
tính b
b=\(\frac{0,275-0,5+\frac{3}{11}}{0,625+0,5+\frac{5}{11}}:\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
So sánh
A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^9}+.....+\frac{1}{3^{33}}với\frac{1}{2}\)
So sánh: \(\frac{3^{10}+1}{3^9+1}\)và \(\frac{3^{11}+1}{3^{10}+1}\)
Đặt A= \(\frac{3^{10}+1}{3^9+1}\) đặt B= \(\frac{3^{11}+1}{3^{10}+1}\)
Vì B<1 => B< \(\frac{3^{11}+1+2}{3^{10}+1+2}\) = \(\frac{3^{11}+3}{3^{10}+3}\) = \(\frac{3\cdot\left(3^{10}+1\right)}{3\cdot\left(3^9+1\right)}\) = \(\frac{3^{10}+1}{3^9+1}\) = A
Vậy B<A
Ta có :
\(\frac{3^{11}+1}{3^{10}+1}>1\) nên \(\frac{3^{11}+1}{3^{10}+1}>\frac{3^{11}+1+2}{3^{10}+1+2}=\frac{3^{11}+3}{3^{10}+3}=\frac{3\left(3^{10}+1\right)}{3\left(3^9+1\right)}=\frac{3^{10}+1}{3^9+1}\)
Vậy \(\frac{3^{11}+1}{3^{10}+1}>\frac{3^{10}+1}{3^9+1}\)
So sánh
\(\left(\frac{9}{11}-0.81\right)^{2005}......\frac{1}{10^{4010}}\)
\(\left(\frac{9}{11}-0,81\right)^{2005}=\left(\frac{9}{1100}\right)^{2005}=0,00\left(81\right)^{2005}\)
\(\frac{1}{10^{4010}}=\frac{1}{100^{2005}}=\left(\frac{1}{100}\right)^{2005}=0,01^{2005}\)
Vì 0,00(81)<0,01 nên \(\left(\frac{9}{11}-0,81\right)^{2005}< \frac{1}{10^{4010}}\)
So sánh phân số trung gian \(\frac{11}{14}và\frac{9}{11}\)
phân só trung gian là 9/14
Ta có: 11/14>9/14 và 9/14<9/11
=> 11/14<9/11