`Answer:`

\(5.\left(x-2\frac{1}{5}\right)^2-\frac{1}{5}=9\frac{3}{5}\)

\(\Leftrightarrow5.\left(x-\frac{11}{5}\right)^2-\frac{1}{5}=\frac{48}{5}\)

\(\Leftrightarrow5.\left(x-\frac{11}{5}\right)^2=\frac{48}{5}+\frac{1}{5}\)

\(\Leftrightarrow5.\left(x-\frac{11}{5}\right)^2=\frac{49}{5}\)

\(\Leftrightarrow\left(x-\frac{11}{5}\right)^2=\frac{49}{5}:5\)

\(\Leftrightarrow\left(x-\frac{11}{5}\right)^2=\left(\frac{7}{5}\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{11}{5}=\frac{7}{5}\\x-\frac{11}{5}=-\frac{7}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{5}+\frac{11}{5}\\x=-\frac{7}{5}+\frac{11}{5}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{18}{5}\\x=\frac{4}{5}\end{cases}}\)

Không có học trò dốt

Mà chỉ có thầy chưa giỏi

Tô Thành Long đừng spam nữa.

`Answer:`

Ta đặt: \(A=-\frac{2018}{2019}.\frac{2}{7}-\frac{2018}{2019}.\frac{5}{7}+1\) và \(B=\frac{2018}{2019}\)

\(A=-\frac{2018}{2019}.\frac{2}{7}+-\frac{2018}{2019}.\frac{5}{7}+1\)

\(=-\frac{2018}{2019}.\left(\frac{2}{7}+\frac{5}{7}\right)+\frac{2019}{2019}\)

\(=-\frac{2018}{2019}.1+\frac{2019}{2019}\)

\(=\frac{\left(-2018\right)+2019}{2019}\)

\(=\frac{1}{2019}\)

Mà ta có: \(B=\frac{2018}{2019}\)

Mà \(2018>1\Rightarrow\frac{2018}{2019}>\frac{1}{2019}\) hay `B>A`

`Answer:`

\(\frac{4}{7}+\frac{3}{7}x-\left(\frac{11}{7}+x\right)=2\) (Mình sửa đề nhé.)

\(\Leftrightarrow\frac{3}{7}x-\frac{11}{7}-x=2-\frac{4}{7}\)

\(\Leftrightarrow\frac{3}{7}x-\frac{11}{7}-x.1=\frac{10}{7}\)

\(\Leftrightarrow x.\left(\frac{3}{7}-1\right)-\frac{11}{7}=\frac{10}{7}\)

\(\Leftrightarrow x.-\frac{4}{7}=\frac{10}{7}+\frac{11}{7}\)

\(\Leftrightarrow x.-\frac{4}{7}=3\)

\(\Leftrightarrow x=3:-\frac{4}{7}\)

\(\Leftrightarrow x=-\frac{21}{4}\)

Ta lấy:diện tíchx2:(đáy lớn+đáy bé)với điều kiện phải cùng đơn vị đo.

Chúc học tốt!