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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 ~
\(A=\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\\ \Leftrightarrow A=\dfrac{1}{3}-\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{9}{11}-\dfrac{11}{13}\\ \Leftrightarrow A=\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-\left(\dfrac{3}{5}-\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)-\left(\dfrac{7}{9}-\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)-\dfrac{11}{13}\\ \Leftrightarrow A=0-0+0-0+0-\dfrac{11}{13}\\ \Leftrightarrow A=\dfrac{-11}{13}\)
a=(1/3-1/3)+(3/5-3/5)+(5/7-5/7)+(9/11-9/11)+(11/13-11/13)+13/15
a=13/15
= 1/3 - 1/3 + 5/7 - 5/7 - 7/9 + 7/9 +9/11 - 9/11 -11/13 + 11/13 +13/15
= 0 + 0 - 0 + 0 -0 + 13/15
= 0 + 13/15
= 13/15
\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
\(=\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{3}{5}-\frac{3}{5}\right)+\left(\frac{5}{7}-\frac{5}{7}\right)+\left(\frac{7}{9}-\frac{7}{9}\right)+\left(\frac{9}{11}-\frac{9}{11}\right)+\left(\frac{11}{13}-\frac{11}{13}\right)+\frac{13}{15}\)
\(=0+0+0+0+0+0+\frac{13}{15}\)
\(=\frac{13}{15}\)
\(=\left(\frac{135}{11}-\frac{58}{11}\right)+\left(\frac{13}{4}+\frac{5}{4}\right)-\frac{6}{13}\)
\(=7+\frac{9}{2}-\frac{6}{13}\)
\(=\frac{23}{2}-\frac{6}{13}\)
\(=\frac{287}{26}.\)
Chúc bạn học tốt!
\(12\frac{3}{11}-\frac{6}{13}+3,25-5\frac{3}{11}-4\frac{7}{13}+\sqrt{1\frac{9}{16}}\)
\(=12+\frac{3}{11}-\frac{6}{13}+3,25-5-\frac{3}{11}-4-\frac{7}{13}+\sqrt{\frac{25}{16}}\)
\(=\left(12-5-4\right)+\left(\frac{3}{11}-\frac{3}{11}\right)+\left(\frac{-6}{13}-\frac{7}{13}\right)+\frac{5}{4}+3,25\)
\(=3+0+\left(-1\right)+1,25+3,25=2+4,5=6,5\)
Chắc bạn gõ nhầm số hạng thứ 3 phải là +5/7.
Tổng này đối xứng qua 13/15. Các đối xứng trái dấu nên Tổng = 13/15
Ta có : \(A-1=\frac{9^{11}+1}{9^{11}-7}-1=\frac{8}{9^{11}-7}\) ; \(B-1=\frac{9^{12}+3}{9^{12}-5}-1=\frac{8}{9^{12}-5}\)
Cần so sánh : \(9^{11}-7\) và \(9^{12}-5\)
Ta viết : \(9^{12}-5=9^{11}.9-5=9^{11}.\left(1+8\right)-5=\left(9^{11}-7\right)+\left(8.9^{11}+2\right)\)
Xét : \(\left(9^{12}-5\right)-\left(9^{11}-7\right)=\left(9^{11}-7\right)+\left(8.9^{11}+2\right)-\left(9^{11}-7\right)=8.9^{11}+2>0\)
\(\Rightarrow9^{12}-5>9^{11}-7\)
Do đó : \(B-1>A-1\Rightarrow B< A\)