so sánh :\(\frac{3}{11}\)và \(\frac{5}{11}\)
Những câu hỏi liên quan
So sánh A và B nếu :
\(A=-\frac{1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)va \(B=\frac{1}{2022}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
So sánh A và B
\(A=-\frac{1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)
\(B=-\frac{1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\text{ }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
So sánh \(A\) và \(B\),biết:
\(A=\frac{-1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)
\(B=\frac{-1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
So sánh A và B biết:
A = \(-\frac{1}{2013}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)
B = \(-\frac{1}{2013}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
\(A=\left(-\frac{1}{2013}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\right)-\frac{4}{11^4};B=\left(-\frac{1}{2013}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{3}{11^2}\right)-\frac{4}{11^2}\)
Vì 114 > 112 nên \(\frac{4}{11^4}-\frac{4}{11^2}\) => A > B
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Bài 1 ; So sánh
\(A=\frac{-1}{2011}-\frac{3}{11^2}-\frac{5}{11^3}-\frac{7}{11^4}\)
\(B=\frac{1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
Mình cần gấp lắm ạ , Ai làm đúng và nhanh nhất mình tick cho
\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)
\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)
\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)
=> A > B
Vậy A > B
So sánh các phân số.
a) $\frac{2}{3}$ và $\frac{{11}}{{18}}$
b) $\frac{{36}}{{63}}$ và $\frac{5}{7}$
c) $\frac{{55}}{{110}}$ và $\frac{4}{8}$
a) $\frac{2}{3} = \frac{{2 \times 6}}{{3 \times 6}} = \frac{{12}}{{18}}$
Ta có $\frac{{12}}{{18}} > \frac{{11}}{{18}}$ nên $\frac{2}{3} > \frac{{11}}{{18}}$
b) $\frac{{36}}{{63}} = \frac{{36:9}}{{63:9}} = \frac{4}{7}$
Ta có $\frac{4}{7} < \frac{5}{7}$ nên $\frac{{36}}{{63}}$ < $\frac{5}{7}$
c)
$\frac{{55}}{{110}} = \frac{{55:55}}{{110:55}} = \frac{1}{2}$ ; $\frac{4}{8} = \frac{1}{2}$
Vậy $\frac{{55}}{{110}}$ = $\frac{4}{8}$
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So sánh: \(\frac{5^{11}+1}{5^{11}+2}\) và \(\frac{5^{11}+2}{5^{11}+3}\)
So sánh
A=\(\frac{2006}{2007}-\frac{2007}{2008}+\frac{2008}{2009}-\frac{2009}{2010}\)
B=\(-\frac{1}{2006.2007}-\frac{1}{2008.2009}\)
So sánh
B=\(-\frac{1}{2011}-\frac{7}{11^2}-\frac{5}{11^3}-\frac{3}{11^4}\)
A=\(-\frac{1}{2011}-\frac{3}{11^2}-\frac{5}{11^4}\)
a) So sánh \(\frac{{ - 11}}{5}\) với \(\frac{{ - 7}}{4}\) bằng cách viết –2 ở dạng phân số có mẫu số thích hợp.
Từ đó suy ra kết quả so sánh \(\frac{{ - 11}}{5}\) với \(\frac{{ - 7}}{4}\).
b) So sánh \(\frac{{2020}}{{ - 2021}}\) với \(\frac{{ - 2022}}{{2021}}\).
a) Ta có: \( - 2 = \frac{{ - 2}}{1} = \frac{{ - 40}}{{20}}\)
\(\frac{{ - 11}}{5} = \frac{{ - 44}}{{20}} < \frac{{ - 40}}{{20}}\) nên \(\frac{{ - 11}}{5} < -2\).
\(\frac{{ - 7}}{4} = \frac{{ - 7.5}}{{4.5}} = \frac{{ - 35}}{{20}} > \frac{{ - 40}}{{20}}\) nên \(\frac{{ - 7}}{4} > -2\)
Vậy \(\frac{{ - 11}}{5} < \frac{{ - 7}}{4}\).
b) Ta có: \(\frac{{2020}}{{ - 2021}} = \frac{{ - 2020}}{{2021}} > \frac{{ - 2022}}{{2021}}\)
Vậy \(\frac{{2020}}{{ - 2021}} > \frac{{ - 2022}}{{2021}}\)
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