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T
1 tháng 7 2021
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\text{ }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
11 tháng 8 2016
So sánh:
\(P=\frac{4}{7}+5+\frac{3}{7^2}+\frac{5}{7^3}+\frac{6}{7^4}\)
\(Q=\frac{5}{7^4}+5+\frac{6}{7^2}+\frac{4}{7}+\frac{5}{7^3}\)
Ta có : \(P=\left\{\frac{4}{7}+5+\frac{5}{7^3}\right\}+\left\{\frac{3}{7^2}+\frac{6}{7^4}\right\}\)
\(Q=\left\{\frac{4}{7}+5+\frac{5}{7^3}\right\}+\left\{\frac{5}{7^4}+\frac{6}{7^2}\right\}\)
So sánh : \(\frac{3}{7^2}+\frac{6}{7^4}\)và \(\frac{5}{7^4}+\frac{6}{7^2}\)
Ta có : \(\frac{3}{7^2}+\frac{6}{7^4}=\frac{49.3}{7^4}+\frac{6}{7^4}\)
\(\frac{5}{7^4}+\frac{6}{7^2}=\frac{5}{7^4}+\frac{49.6}{7^4}\)
Vì 49.3 + 6 < 49.6 + 5 nên Q > P.
17 tháng 8 2018
a)
\(A=\left(\frac{19}{24}-\frac{7}{24}\right)-\left(\frac{1}{2}+\frac{1}{3}\right)\)
\(A=\frac{1}{2}-\frac{1}{2}+\frac{1}{3}\)
\(A=\frac{1}{3}\)
\(B=\left(\frac{7}{12}-\frac{5}{12}\right)+\left(\frac{5}{6}+\frac{1}{4}-\frac{3}{7}\right)\)
\(B=\left(\frac{1}{6}+\frac{5}{6}\right)+\frac{1}{4}-\frac{3}{7}\)
\(B=\frac{5}{4}-\frac{3}{7}\)
\(B=\frac{23}{28}\)
17 tháng 8 2018
b)
\(x=A-B\)
\(x=\frac{1}{3}-\frac{23}{28}\)
\(x=\frac{-41}{84}\)
T
1 tháng 7 2021
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B