Ax1= 1+2+3+4+...=50 giúp mình với
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Ta có:
\(\dfrac{-2}{3}=\dfrac{-2.20}{3.20}=\dfrac{-40}{60}\)
\(-0,4=\dfrac{-2}{5}=\dfrac{-2.12}{5.12}=\dfrac{-24}{60}\)
\(\dfrac{-5}{4}=\dfrac{-5.15}{4.15}=\dfrac{-75}{60}\)
\(-1,2=\dfrac{-6}{5}=\dfrac{-6.12}{5.12}=\dfrac{-72}{60}\)
Vì \(\dfrac{-75}{60}< \dfrac{-72}{60}< \dfrac{-40}{60}< \dfrac{-24}{60}\)
nên \(\dfrac{-5}{4}< -1,2< \dfrac{-2}{3}< -0,4\)
Sắp xếp theo thứ tự tăng dần:
\(\dfrac{-5}{4}< -1,2< \dfrac{-2}{3}< -0,4\)
\(#Wendy.Dang\)
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\(\dfrac{2011}{2010}< \dfrac{2011+1}{2010+1}=\dfrac{2012}{2011}\)
\(\Rightarrow\dfrac{2011}{2010}< \dfrac{2012}{2011}\)
\(\dfrac{2012}{2011}và\dfrac{2011}{2010}\\ \dfrac{2012}{2011}-1=\dfrac{1}{2011}\\ \dfrac{2011}{2010}-1=\dfrac{1}{2010}\\ \)
Vì \(\dfrac{1}{2011}< \dfrac{1}{2010}\\ \Rightarrow\dfrac{2012}{2011}< \dfrac{2011}{2010}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Các đơn thức là :
\(\left(1-\dfrac{1}{\sqrt[]{3}}\right)x^2;x^2.\dfrac{7}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Các đơn thức là :
\(\dfrac{x^2y}{2};\dfrac{x}{-5^2};\dfrac{-4}{5}\)
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\(\dfrac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\left(1\right)\)
\(Đkxđ:x\ne2009;x\ne2010\)
Đặt \(t=x-2010\left(t\ne0\right)\)
\(\Rightarrow2009-x=-\left(t+1\right)\)
\(\left(1\right)\Leftrightarrow\dfrac{\left(t+1\right)^2-\left(t+1\right)t+t^2}{\left(t+1\right)^2+\left(t+1\right)t+t^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{t^2+2t+1-t^2-t+t^2}{t^2+2t+1+t^2+t+t^2}=\dfrac{19}{49}\)
\(\Leftrightarrow\dfrac{t^2+t+1}{3t^2+3t+1}=\dfrac{19}{49}\)
\(\Leftrightarrow49t^2+49t+49=57t^2+57t+19\)
\(\Leftrightarrow8t^2+8t-30=0\)
\(\Leftrightarrow4t^2+4t-15=0\)
\(\Leftrightarrow\left(4t^2+4t+1\right)-16=0\)
\(\Leftrightarrow\left(2t+1\right)^2=16=4^2\)
\(\Leftrightarrow\left[{}\begin{matrix}2t+1=4\\2t+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{3}{2}\\t=-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2010=\dfrac{3}{2}\\x-2010=-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4023}{2}\\x=\dfrac{4015}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{6}+\dfrac{7}{12}-\dfrac{9}{20}+\dfrac{11}{30}-\dfrac{13}{42}\)
\(=\dfrac{1}{2.3}+\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}-\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\dfrac{1}{5}+\dfrac{1}{6}-\left(\dfrac{1}{6}+\dfrac{1}{7}\right)\)
\(=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)
Ax1 = 1+2+3+4+...+50
A có 50 số hạng nên A = (1 + 50).50:2 = 1275