\(2\left|x-3\right|-3x=1\)

\(\Leftrightarrow2\left|x-3\right|=1+3x\)

\(\Leftrightarrow\left|x-3\right|=\frac{1+3x}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=\frac{1+3x}{2}\\x-3=-\frac{1+3x}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1+3x}{2}+3\\x=-\frac{1+3x}{2}+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1+3x+6}{2}\\x=-\frac{1+3x+6}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=1+3x+6\\2x=-1-3x-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3x=1+6\\2x+3x=-1-6\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=7\\5x=-7\end{cases}}\)

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

Vậy : \(x\in\left\{-7;-\frac{7}{5}\right\}\)

+) <=>\(\orbr{\begin{cases}x-1=\frac{3}{2}\\x-1=\frac{-3}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{1}{2}\end{cases}}}\)

+) <=>\(\orbr{\begin{cases}x-1=\frac{5}{3}\\x-1=-\frac{5}{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{2}{3}\end{cases}}}\)

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15 cặp

Đặt \(A=1^2+2^2+3^2+...+n^2\)

\(\Rightarrow A=1.1+2.2+3.3+...+n.n\)

\(\Rightarrow A=1.\left(2-1\right)+2.\left(3-1\right)+3.\left(4-1\right)+...+n\left[\left(n+1\right)-1\right]\)

 \(\Rightarrow A=1.2-1+2.3-2+3.4-3+...+n.\left(n+1\right)-n\)

\(\Rightarrow A=\left[1.2+2.3+3.4+...+n.\left(n+1\right)\right]-\left(1+2+3+...+n\right)\)

Đặt \(B=1.2+2.3+3.4+...+n\left(n+1\right)\)

\(\Rightarrow3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+n\left(n+1\right)\left(n+2\right)-\left(n-1\right).n.\left(n+1\right)\)

\(\Rightarrow3B=n\left(n+1\right)\left(n+2\right)\)

\(\Rightarrow B=\frac{n\left(n+1\right)\left(n+2\right)}{3}\)

Đặt \(C=1+2+3+...+n\)

Số số hạng: ( n - 1) : 1 + 1 = n - 1 + 1 = n

\(\Rightarrow C=\frac{n\left(n+1\right)}{2}\)

Thay B và C vào A

\(\Rightarrow A=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)

\(\Rightarrow A=n.\left(n+1\right).\left[\frac{n+2}{3}-\frac{1}{2}\right]\)

\(\Rightarrow A=n\left(n+1\right).\frac{2\left(n+2\right)-3}{6}=n\left(n+1\right).\frac{2n+4-3}{6}=\frac{n\left(n+1\right).\left(2n+1\right)}{6}\)

Ta có : x - y + x + y = 3x + 1 + 3y

           x + x - y + y = 3x + 3y + 1

           2x - 3x - 3y   = 1

           x - 3y           = 1 = 1 . 1 = ( - 1).( - 1)

\(\Rightarrow\)+ x = 1

            y = 1

         + x = - 1

            y = - 1

Vậy x = 1; y = 1

hoặc x = - 1; y = - 1