giúp mình với ạ
-1/21 - 1/28 =
/ nghĩa là phần
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a) Đặt A= \(\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{36}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{72}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\dfrac{1}{2}\)A=\(\dfrac{1}{2}-\dfrac{1}{9}\)
\(\dfrac{1}{2}\)A=\(\dfrac{7}{18}\)
A=\(\dfrac{7}{9}\)
Ta có:\(\frac{x+21}{1998}-\frac{x+25}{1994}+\frac{x+63}{652}-\frac{x+75}{648}=0\)
\(\Rightarrow\frac{x+21}{1998}-\frac{x+25}{1994}+\frac{x+63}{652}-\frac{x+75}{648}+4-4=0\)
\(\Rightarrow\frac{x+21}{1998}+1-\frac{x+25}{1994}-1+\frac{x+63}{652}+3-\frac{x+75}{648}-3=0\)
\(\Rightarrow\frac{x+2019}{1998}-\frac{x+2019}{1994}-\frac{x+2019}{652}+\frac{x+2019}{648}=0\)
\(\Rightarrow\left(x+2019\right)\times\left(\frac{1}{1998}-\frac{1}{1994}+\frac{1}{652}-\frac{1}{648}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2019=0\\\frac{1}{1998}-\frac{1}{1994}+\frac{1}{652}-\frac{1}{648}=0\end{cases}}\)
Vì \(\frac{1}{1998}-\frac{1}{1994}+\frac{1}{652}-\frac{1}{648}\ne0\)
\(\Rightarrow x+2019=0\)
\(\Rightarrow x=-2019\)
Vậy \(x=-2019\)
\(A=\dfrac{2x-4\sqrt{x}+2-\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{x-4}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
\(=\dfrac{2x-4\sqrt{x}+2-2x+4\sqrt{x}+\sqrt{x}-2}{\sqrt{x}+2}\cdot\dfrac{1}{\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{x}+2}\)
=14/15*20/21*...*209/210
\(=\dfrac{4\cdot7}{5\cdot6}\cdot\dfrac{5\cdot8}{6\cdot7}\cdot...\cdot\dfrac{19\cdot22}{20\cdot21}\)
\(=\dfrac{4\cdot5\cdot6\cdot...\cdot19}{5\cdot6\cdot7\cdot...\cdot20}\cdot\dfrac{7\cdot8\cdot9\cdot...\cdot22}{6\cdot7\cdot8\cdot...\cdot21}=\dfrac{11}{15}\)
=-4/84-3/84=-7/84=-1/12
\(-\dfrac{1}{21}-\dfrac{1}{28}\)
\(=-\dfrac{28}{588}-\dfrac{21}{588}\)
\(=-\dfrac{49}{588}=-\dfrac{1}{12}\)