A=\(1\dfrac{1}{2}x1\dfrac{1}{3}x1\dfrac{1}{4}x1\dfrac{1}{5}x......x1\dfrac{1}{2021}x1\dfrac{1}{2023}.TìmA\)
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\(1\dfrac{1}{2}x1\dfrac{1}{3}x1\dfrac{1}{4}x1\dfrac{1}{5}x1\dfrac{1}{6}x1\dfrac{1}{7}x1\dfrac{1}{8}x1\dfrac{1}{9}\)
\(=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}x\dfrac{8}{7}x\dfrac{9}{8}x\dfrac{10}{9}\)
\(=x^7.\dfrac{3.4.5.6.7.8.9.10}{2.3.4.5.6.7.8.9}\)
\(=x^7.\dfrac{10}{2}\)
\(=5x^7\)
\(=\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times...\times\dfrac{9}{8}\times\dfrac{10}{9}=\dfrac{10}{2}=5\)
\(...=\dfrac{3}{2}x\dfrac{4}{3}x\dfrac{5}{4}x\dfrac{6}{5}x\dfrac{7}{6}....x\dfrac{1000}{999}\)
\(=\dfrac{1}{2}x\dfrac{1000}{1}=500\)
=3/2x4/3x5/4x....x1000/999
=1/2x1000=500
mình chưa chắc là đúng đâu nhé
\(=>C=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}.....\cdot\dfrac{101}{100}\)
\(C=\dfrac{3\cdot4\cdot5.......\cdot101}{2\cdot3\cdot4.........\cdot100}\)
\(C=\dfrac{101}{2}\)
\(1\dfrac{1}{3}\times1\dfrac{1}{8}\times1\dfrac{1}{15}\times1\dfrac{1}{24}\times1\dfrac{1}{35}\)
= \(\dfrac{4}{3}\times\dfrac{9}{8}\times\dfrac{16}{15}\times\dfrac{25}{24}\times\dfrac{36}{35}\)
= \(\dfrac{4\times9\times16\times25\times36}{3\times8\times15\times24\times35}\)
= \(\dfrac{1\times2\times2\times25\times36}{1\times2\times15\times24\times35}\)
= \(\dfrac{4\times25\times36}{30\times24\times35}\)
= \(\dfrac{1\times25\times36}{30\times6\times35}=\dfrac{1}{7}\)
Sai rồi nhé! Từ dấu bằng thứ 2 xuống dấu bằng thứ 3 bạn làm sao đc z?
a: \(1\dfrac{1}{2}\cdot1\dfrac{1}{3}\cdot1\dfrac{1}{4}\)
\(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\)
\(=\dfrac{5}{2}\)
b: \(1\dfrac{1}{2}:1\dfrac{1}{3}:1\dfrac{1}{4}\)
\(=\dfrac{3}{2}:\dfrac{4}{3}:\dfrac{5}{4}\)
\(=\dfrac{3}{2}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}=\dfrac{9}{10}\)
1\(\dfrac{1}{12}\) \(\times\) 1\(\dfrac{1}{13}\) \(\times\) 1\(\dfrac{1}{14}\) \(\times\)...\(\times\)1\(\dfrac{1}{2005}\)
A = \(\dfrac{12+1}{12}\) \(\times\) \(\dfrac{13+1}{13}\) \(\times\) \(\dfrac{14+1}{14}\)\(\times\)...\(\times\) \(\dfrac{2006}{2005}\)
A = \(\dfrac{13}{12}\) \(\times\) \(\dfrac{14}{13}\) \(\times\) \(\dfrac{15}{14}\) \(\times\)...\(\times\) \(\dfrac{2006}{2005}\)
A = \(\dfrac{2006}{12}\)
A = \(\dfrac{1003}{6}\)
= 13/5 x 1/4 x 3/2
= 13 x 1 x 3/ 5 x 4 x 2
= 39/40
Ta có: \(\Delta=4m^2+4m-11\)
Để phương trình có 2 nghiệm phân biệt \(\Leftrightarrow4m^2+4m-11>0\)
Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m+3\\x_1x_2=2m+5\end{matrix}\right.\)
Để phương trình có 2 nghiệm dương phân biệt
\(\Leftrightarrow\left\{{}\begin{matrix}4m^2+4m-11>0\\2m+3>0\\2m+5>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m< \dfrac{-1-2\sqrt{3}}{2}\\m>\dfrac{-1+2\sqrt{3}}{2}\end{matrix}\right.\\m>-\dfrac{3}{2}\\m>-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow m>\dfrac{-1+2\sqrt{3}}{2}\)
Mặt khác: \(\dfrac{1}{\sqrt{x_1}}+\dfrac{1}{\sqrt{x_2}}=\dfrac{4}{3}\)
\(\Rightarrow\dfrac{x_1+x_2+2\sqrt{x_1x_2}}{x_1x_2}=\dfrac{16}{9}\) \(\Rightarrow\dfrac{2m+3+2\sqrt{2m+5}}{2m+5}=\dfrac{16}{9}\)
\(\Rightarrow18m+27+18\sqrt{2m+5}=32m+80\)
\(\Leftrightarrow14m-53=18\sqrt{2m+5}\)
\(\Rightarrow\) ...
a: A=x1+x2=-5/2
b: \(=\dfrac{x_1+x_2}{x_1x_2}=\dfrac{-5}{2}:\left(-1\right)=\dfrac{5}{2}\)
c: \(=\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)\)
\(=\left(-\dfrac{5}{2}\right)^3-3\cdot\dfrac{-5}{2}\cdot\left(-1\right)\)
\(=-\dfrac{125}{8}-\dfrac{15}{2}=\dfrac{-185}{8}\)
e: \(E=\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\)
\(=\sqrt{\left(-\dfrac{5}{2}\right)^2-4\cdot\left(-1\right)}=\sqrt{\dfrac{25}{4}+4}=\dfrac{\sqrt{41}}{2}\)
Ta có:
\(A=1\dfrac{1}{2}\times1\dfrac{1}{3}\times1\dfrac{1}{4}\times1\dfrac{1}{5}\times...\times1\dfrac{1}{2021}\times1\dfrac{1}{2022}\\ =\dfrac{3}{2}\times\dfrac{4}{3}\times\dfrac{5}{4}\times\dfrac{6}{5}\times...\times\dfrac{2022}{2021}\times\dfrac{2023}{2022}\\ =\dfrac{2023\times\left(3\times4\times5\times...\times2022\right)}{2\times\left(3\times4\times5\times...\times2022\right)}=\dfrac{2023}{2}=1011\dfrac{1}{2}\)
Đs....