\(\dfrac{1}{1.3}+\dfrac{1}{2.3}+\dfrac{1}{2.5}+\dfrac{1}{3.5}+\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.9}\) 

\(=\dfrac{1}{1.3}+\dfrac{1}{3.2}+\dfrac{1}{2.5}+\dfrac{1}{5.3}+\dfrac{1}{3.7}+\dfrac{1}{7.4}+\dfrac{1}{4.9}\) 

\(=\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right):\dfrac{1}{2}\) 

\(=\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right):\dfrac{1}{2}\) 

\(=\left(\dfrac{1}{2}-\dfrac{1}{9}\right):\dfrac{1}{2}\) 

\(=\dfrac{7}{18}:\dfrac{1}{2}\) 

\(=\dfrac{7}{9}\)

Giải:

Diện tích lớp học HCN là:

       \(11.5,5=60,5\left(m^2\right)\) 

Đổi \(60,5m^2=0,605dam^2\) 

Vì \(0,70>0,605\) nên diện tích ngôi nhà lớn hơn diện tích lớp học

Chúc bạn học tốt!

Giải:

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

\(TH1:x\ge0\) 

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

  \(x-\dfrac{5}{6}.x=\dfrac{5}{6}\)   

\(x.\left(1-\dfrac{5}{6}\right)=\dfrac{5}{6}\) 

         \(x.\dfrac{1}{6}=\dfrac{5}{6}\) 

              \(x=\dfrac{5}{6}:\dfrac{1}{6}\) 

              \(x=5\) 

\(TH2:x\le0\) 

\(\left|x\right|-\dfrac{5}{6}.x=\dfrac{4}{9}.\dfrac{15}{8}\) 

\(-x-\dfrac{5}{6}.x=\dfrac{5}{6}\) 

\(x.\left(-1-\dfrac{5}{6}\right)=\dfrac{5}{6}\) 

     \(x.\dfrac{-11}{6}=\dfrac{5}{6}\) 

               \(x=\dfrac{5}{6}:\dfrac{-11}{6}\) 

               \(x=\dfrac{-5}{11}\) 

Vậy \(x\in\left\{\dfrac{-5}{11};5\right\}\)

Giải:

\(\dfrac{4}{1.4}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\) 

\(=1+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\) 

\(=1+\dfrac{1}{5}-\dfrac{1}{2005}\) 

\(=1+\dfrac{80}{401}\) 

\(=\dfrac{481}{401}\)

\(\left(x+1\right)^2-\left(x+1\right).\left(x-1\right)=0\) 

\(\Rightarrow\left(x+1\right).\left(x+1\right)-\left(x+1\right).\left(x-1\right)=0\) 

\(\Rightarrow\left(x+1\right).\left[\left(x+1\right)-\left(x-1\right)\right]=0\) 

\(\Rightarrow\left(x+1\right).\left[x+1-x+1\right]=0\) 

\(\Rightarrow\left(x+1\right).\left[\left(x-x\right)+\left(1+1\right)\right]=0\) 

\(\Rightarrow\left(x+1\right).\left[0+2\right]=0\) 

\(\Rightarrow\left(x+1\right).2=0\) 

\(\Rightarrow x+1=0:2\) 

\(\Rightarrow x+1=0\) 

\(\Rightarrow x=0-1\) 

\(\Rightarrow x=-1\)

Giải:

 

1) Vì +) Om; Ot cùng ∈ 1 nửa mặt phẳng bờ xy

         +) \(x\widehat{O}m< x\widehat{O}t\left(130^o< 150^o\right)\) 

⇒Om nằm giữa Ox và Ot

\(\Rightarrow x\widehat{O}m+m\widehat{O}t=x\widehat{O}t\) 

      \(130^o+m\widehat{O}t=150^o\) 

                 \(m\widehat{O}t=150^o-130^o\) 

                 \(m\widehat{O}t=20^o\) 

2) Vì \(x\widehat{O}y\) là góc bẹt

\(\Rightarrow x\widehat{O}y=180^o\) 

\(\Rightarrow x\widehat{O}t+t\widehat{O}y=180^o\) (2 góc kề bù)

    \(150^o+t\widehat{O}y=180^o\) 

               \(t\widehat{O}y=180^o-150^o\) 

               \(t\widehat{O}y=30^o\) 

Vì \(x\widehat{O}t>t\widehat{O}y\left(150^o>30^o\right)\) 

⇒Oy không phải là tia p/g của \(x\widehat{O}t\)

\(x+6=\dfrac{3}{2}\) 

       \(x=\dfrac{3}{2}-6\) 

       \(x=-\dfrac{9}{2}\)

\(\dfrac{2}{3}+\dfrac{1}{3}.\left(-\dfrac{4}{9}+\dfrac{5}{6}\right):\dfrac{7}{12}\) 

\(=\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{7}{18}:\dfrac{7}{12}\) 

\(=\dfrac{2}{3}+\dfrac{2}{9}\) 

\(=\dfrac{8}{9}\)

Giải:

a) \(x:-\dfrac{3}{5}=-\dfrac{10}{21}\) 

             \(x=-\dfrac{10}{21}.-\dfrac{3}{5}\) 

             \(x=\dfrac{2}{7}\) 

b) \(-\dfrac{2}{3}x+\dfrac{1}{5}=\dfrac{3}{10}\) 

            \(-\dfrac{2}{3}x=\dfrac{3}{10}-\dfrac{1}{5}\) 

            \(-\dfrac{2}{3}x=\dfrac{1}{10}\) 

                  \(x=\dfrac{1}{10}:-\dfrac{2}{3}\) 

                  \(x=-\dfrac{3}{20}\) 

c) \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\) 

             \(\left(x+\dfrac{1}{5}\right)^2=\dfrac{26}{25}-\dfrac{17}{25}\) 

             \(\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\) 

\(\Rightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{5}\right)^2=\left(\dfrac{3}{5}\right)^2\\\left(x+\dfrac{1}{5}\right)^2=\left(-\dfrac{3}{5}\right)^2\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\) 

d) \(\left(x+\dfrac{1}{2}\right).\left(\dfrac{2}{3}-2x\right)=0\) 

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

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