Ta có :

\(\left(\dfrac{3}{4}.x-\dfrac{9}{16}\right).\left(1,5+\dfrac{-3}{5}:x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}.x-\dfrac{9}{16}=0\\1,5+\dfrac{-3}{5}:x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}.x=\dfrac{9}{16}\\\dfrac{3}{5}:x=\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{16}:\dfrac{3}{4}\\x=\dfrac{3}{5}:\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{2}{5}\end{matrix}\right.\)

1,8 . x - 30% . x = -4,5

\(\Rightarrow\dfrac{18}{10}.x-\dfrac{3}{10}.x=\dfrac{-9}{2}\)

\(\Rightarrow\left(\dfrac{18}{10}-\dfrac{3}{10}\right).x=\dfrac{-9}{10}\)

\(\Rightarrow5.x=\dfrac{-9}{2}\)

\(\Rightarrow x=\dfrac{-9}{10}\)

\(\dfrac{x+1}{65}+1+\dfrac{x+3}{63}+1=\dfrac{x+5}{61}+1+\dfrac{x+7}{59}+1\)

\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{63}-\dfrac{1}{61}-\dfrac{1}{59}\ne0\right)=0\Leftrightarrow x=-66\)

Ta có :

\(250.\left(\dfrac{1}{5}-\dfrac{1}{6}\right)^2\)\(=250.\left(\dfrac{1}{30}\right)^2\)\(=250.\dfrac{1}{900}=\dfrac{5}{18}\)

Lại có : 

\(250.\left(\dfrac{1}{5}\right)^2-\dfrac{1}{6}\)\(=250.\left(\dfrac{1}{25}\right)^2-\dfrac{1}{6}\)\(=250.\dfrac{1}{625}-\dfrac{1}{6}\)\(=\dfrac{2}{5}-\dfrac{1}{6}=\dfrac{7}{30}\)

Vì \(\dfrac{5}{18}>\dfrac{5}{20}=\dfrac{1}{4}=\dfrac{7}{28}>\dfrac{7}{30}\) nên \(250.\left(\dfrac{1}{5}-\dfrac{1}{6}\right)^2>250.\left(\dfrac{1}{5}\right)^2-\dfrac{1}{6}\)

\(\dfrac{-19}{27}< \dfrac{-16}{27}< \dfrac{-16}{29}\)

1+2+...+2306

Ta có: (2306-1) : 1 + 1= 2306 số hạng

Có số cặp tổng là: 2306 : 2= 1153 cặp

Mỗi cặp có giá trị là: 2306 + 1=  2307 đơn vị

Tổng dãy số đó là: 1153 x 2307= 2659971

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