\(2+\dfrac{3}{2}+\dfrac{3}{6}+\dfrac{3}{12}+...+\dfrac{3}{90}\)

\(=2+3\times\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{90}\right)\)

\(=2+3\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+...+\dfrac{1}{9\times10}\right)\)

\(=2+3\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=2+3\times\left(1-\dfrac{1}{10}\right)\)

\(=2+3\times\dfrac{9}{10}\)

\(=2+\dfrac{27}{10}\)

\(=\dfrac{47}{10}\)

\(\left(27-x\right)\cdot\left(3x+9\right)\cdot\left(42-6x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}27-x=0\\3x+9=0\\42-6x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=27\\x=-3\\x=7\end{matrix}\right.\)

Vậy \(x\in\left\{-3;7;27\right\}\)

\(\left(2x+1\right)^3=343\)

\(\left(2x+1\right)^3=7^3\)

\(2x+1=7\)

\(2x=7-1\)

\(2x=6\)

\(x=6:2\)

\(x=3\)

Vậy \(x=3\)

Ảnh lỗi kìa bạn.

\(\left(-32,25\right)+\left(\dfrac{29}{10}-67,75\right)-\left(-6,1\right)\)

\(=-32,25+\dfrac{29}{10}-67,75+6,1\)

\(=\left(-32,25-67,75\right)+\left(\dfrac{29}{10}+6,1\right)\)

\(=\left(-32,25-67,75\right)+\left(2,9+6,1\right)\)

\(=\left(-100\right)+9\)

\(=-91\)

Đề thiếu . Xem lại bài

Thiếu vế phải rồi

????. Mình chưa hiểu đề!

$\drafc {8}{10}$

\(\dfrac{2023}{2016}\) và \(\dfrac{2016}{2009}\)

Ta có:

\(\dfrac{2023}{2016}=1+\dfrac{7}{2016}\)

\(\dfrac{2016}{2009}=1+\dfrac{7}{2009}\)

Vì \(\dfrac{7}{2016}< \dfrac{7}{2009}\) nên

\(\Rightarrow1+\dfrac{7}{2016}< 1+\dfrac{7}{2009}\)

\(\Rightarrow\dfrac{2023}{2016}< \dfrac{2016}{2009}\)

Vậy \(\dfrac{2023}{2016}< \dfrac{2016}{2009}\)