Ax1 = 1+2+3+4+...+50

 A có 50 số hạng nên A = (1 + 50).50:2 = 1275

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:

\(\dfrac{-5}{4}< -1,2< \dfrac{-2}{3}< -0,4\)

\(#Wendy.Dang\)

Câu hỏi của bạn mà sao ...

\(\dfrac{2011}{2010}< \dfrac{2011+1}{2010+1}=\dfrac{2012}{2011}\)

\(\Rightarrow\dfrac{2011}{2010}< \dfrac{2012}{2011}\)

\(\dfrac{2012}{2011}và\dfrac{2011}{2010}\\ \dfrac{2012}{2011}-1=\dfrac{1}{2011}\\ \dfrac{2011}{2010}-1=\dfrac{1}{2010}\\ \)

Vì \(\dfrac{1}{2011}< \dfrac{1}{2010}\\ \Rightarrow\dfrac{2012}{2011}< \dfrac{2011}{2010}\)

Các đơn thức là :

\(\left(1-\dfrac{1}{\sqrt[]{3}}\right)x^2;x^2.\dfrac{7}{2}\)

Các đơn thức là :

\(\dfrac{x^2y}{2};\dfrac{x}{-5^2};\dfrac{-4}{5}\)

\(\dfrac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\left(1\right)\)

\(Đkxđ:x\ne2009;x\ne2010\)

Đặt \(t=x-2010\left(t\ne0\right)\)

\(\Rightarrow2009-x=-\left(t+1\right)\)

\(\left(1\right)\Leftrightarrow\dfrac{\left(t+1\right)^2-\left(t+1\right)t+t^2}{\left(t+1\right)^2+\left(t+1\right)t+t^2}=\dfrac{19}{49}\)

\(\Leftrightarrow\dfrac{t^2+2t+1-t^2-t+t^2}{t^2+2t+1+t^2+t+t^2}=\dfrac{19}{49}\)

\(\Leftrightarrow\dfrac{t^2+t+1}{3t^2+3t+1}=\dfrac{19}{49}\)

\(\Leftrightarrow49t^2+49t+49=57t^2+57t+19\)

\(\Leftrightarrow8t^2+8t-30=0\)

\(\Leftrightarrow4t^2+4t-15=0\)

\(\Leftrightarrow\left(4t^2+4t+1\right)-16=0\)

\(\Leftrightarrow\left(2t+1\right)^2=16=4^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2t+1=4\\2t+1=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{3}{2}\\t=-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-2010=\dfrac{3}{2}\\x-2010=-\dfrac{5}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4023}{2}\\x=\dfrac{4015}{2}\end{matrix}\right.\)

\(\dfrac{1}{6}+\dfrac{7}{12}-\dfrac{9}{20}+\dfrac{11}{30}-\dfrac{13}{42}\)

\(=\dfrac{1}{2.3}+\dfrac{7}{3.4}-\dfrac{9}{4.5}+\dfrac{11}{5.6}-\dfrac{13}{6.7}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{4}-\left(\dfrac{1}{4}+\dfrac{1}{5}\right)+\dfrac{1}{5}+\dfrac{1}{6}-\left(\dfrac{1}{6}+\dfrac{1}{7}\right)\)

\(=\dfrac{1}{2}-\dfrac{1}{7}=\dfrac{5}{14}\)