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\(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-12}{13}.\dfrac{4}{17}\)
= \(\dfrac{-4}{13}.\dfrac{5}{17}+\dfrac{-4}{13}.\dfrac{12}{17}\)
= \(\dfrac{-4}{13}.\left(\dfrac{5}{17}+\dfrac{12}{17}\right)\)
= \(\dfrac{-4}{13}.\dfrac{17}{17}\)
= \(\dfrac{-4}{13}.1\)
= \(\dfrac{-4}{13}\)
= \(\dfrac{-4.5-12.4}{13.17}\)
=\(\dfrac{-4\left(5+12\right)}{13.17}\)
=\(\dfrac{-4.17}{13.17}\)
=\(\dfrac{-4}{13}\)
(x + 1)4 = (x + 1)3
⇒ (x + 1)4 - (x + 1)3 = 0
⇒ (x + 1)3 . (x + 1 - 1) = 0
⇒ (x + 1)3 . x = 0
⇒ \(\left[{}\begin{matrix}\left(x+1\right)^3=0\\x=0\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x+1=0\\x=0\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-1\\x=0\end{matrix}\right.\)
Vậy \(x\in\left\{0;-1\right\}\)
\(A=1+2+2^2+...+2^{2017}\)
\(\Rightarrow A=\dfrac{2^{2017+1}-1}{2-1}\)
\(\Rightarrow A=2^{2018}-1\)
mà \(B=2^{2018}\)
\(\Rightarrow A-B=2^{2018}-1-2^{2018}\)
\(\Rightarrow A-B=-1\)
\(2A=2+2^2+2^3+...+2^{2018}\)
\(\Rightarrow A=2A-A=2^{2018}-1\)
\(\Rightarrow A-B=2^{2018}-1-2^{2018}=-1\)
a) \(S_{xq}=\left(a+b\right).2.h\)
mà \(\left\{{}\begin{matrix}S_{xq}=120\left(cm^2\right)\\h=60\left(cm\right)\end{matrix}\right.\)
\(\Rightarrow120\left(a+b\right)=120\)
\(\Rightarrow a+b=1\)
\(\Rightarrow\left(a+b\right)^2=1\)
\(\Rightarrow a^2+b^2+2ab=1\)
mà \(a^2+b^2\ge2ab\) (do \(\left(a-b\right)^2=a^2+b^2-2ab\ge0,\forall ab>0\))
\(\Rightarrow4ab\le1\)
\(\Rightarrow ab\le\dfrac{1}{4}\left(1\right)\)
Để thể tích hình hộp chữ nhật có thể tích lớn nhất khi :
\(\left(ab\right)max\left(V=abh;h=60cm\right)\)
\(\left(1\right)\Rightarrow\left(ab\right)max=\dfrac{1}{4}\)
Vậy \(ab=\dfrac{1}{4}\) thỏa mãn đề bài
A = \(\dfrac{2}{5.7}\) + \(\dfrac{5}{7.12}\) + \(\dfrac{7}{12.19}\) + \(\dfrac{9}{19.28}\) + \(\dfrac{11}{28.39}\) + \(\dfrac{1}{30.40}\)
A = \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{12}\) + \(\dfrac{1}{12}\) - \(\dfrac{1}{19}\) + \(\dfrac{1}{19}\) - \(\dfrac{1}{28}\) + \(\dfrac{1}{28}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)
A = \(\dfrac{1}{5}\) - \(\dfrac{1}{39}\) + \(\dfrac{1}{1200}\)
A = \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\)
B = \(\dfrac{1}{20}\) + \(\dfrac{1}{44}\) + \(\dfrac{1}{77}\) + \(\dfrac{1}{119}\) + \(\dfrac{1}{170}\)
B = 2 \(\times\) ( \(\dfrac{1}{2.20}\) + \(\dfrac{1}{2.44}\) + \(\dfrac{1}{2.77}\) + \(\dfrac{1}{2.119}\) + \(\dfrac{1}{2.170}\))
B = 2 \(\times\) ( \(\dfrac{1}{40}\) + \(\dfrac{1}{88}\) + \(\dfrac{1}{154}\) + \(\dfrac{1}{238}\) + \(\dfrac{1}{340}\))
B = 2 \(\times\) ( \(\dfrac{1}{5.8}\) + \(\dfrac{1}{8.11}\) + \(\dfrac{1}{11.14}\) + \(\dfrac{1}{14.17}\) + \(\dfrac{1}{17.20}\))
B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{3}{5.8}\) + \(\dfrac{3}{8.11}\)+ \(\dfrac{3}{11.14}\) + \(\dfrac{3}{14.17}\) + \(\dfrac{3}{17.20}\))
B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\) - \(\dfrac{1}{14}\) + \(\dfrac{1}{14}\) - \(\dfrac{1}{17}\) + \(\dfrac{1}{17}\) - \(\dfrac{1}{20}\))
B = \(\dfrac{2}{3}\) \(\times\) ( \(\dfrac{1}{5}\) - \(\dfrac{1}{20}\))
B = \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{20}\)
B = \(\dfrac{1}{10}\) = \(\dfrac{34}{340}\) < \(\dfrac{34}{195}\) + \(\dfrac{1}{1200}\)
Vậy A > B