\(\frac{-4}{11}.\frac{-3}{11}.\frac{-2}{11}...........\frac{3}{11}.\frac{4}{11}\)
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Tính giá trị biểu thức \(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}+\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
=3/4-3/5+3/7+3/13 / 11/4-11/5+11/7+11/13 + 3/4-3/5+3/7+3/13 / 11/4-11/5+11/7+11/13
=3.1/4-3.1/5+3.1/7+3.1/13 / 11.1/4-11.1/5+11.1/7+11.1/13 + 3.1/4-3.1/5+3.1/7+3.1/13 / 11. 1/4-11.1/5+11.1/7+11.1/13
=3.(1/4-1/5+1/7+1/13) / 11.(1/4-1/5+1/7+1/13) + 3.(1/4-1/5+1/7+1/13) / 11.(1/4-1/5+1/7+1/13)
=3/11+3/11
=6/11
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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
\(5\frac{6}{11}x+8\frac{9}{11}x+2\frac{3}{11}=3\frac{4}{11}x-\frac{8}{11}\)
\(5\frac{6}{11}x+8\frac{9}{11}x+2\frac{3}{11}=3\frac{4}{11}x-\frac{8}{11}\)
\(\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)
\(\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)
= \(\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}\)
= \(\frac{3}{11}\)
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\(\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}=\frac{3.\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11.\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)
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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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\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{3}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
Ta có:\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{3}-\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)
\(\Rightarrow P=\frac{3.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3}{11}\)
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\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{3}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(\Rightarrow P=\frac{4419}{21208}\)
tíc mình nha
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