\(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)

\(\Rightarrow\dfrac{x^2}{2x}-\dfrac{2}{2x}=\dfrac{1}{12}\)

\(\Rightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)

\(\Rightarrow12\left(x^2-2\right)=2x\)

\(\Rightarrow12x^2-2x-24=0\)

\(\Rightarrow12x^2-18x+16x-24=0\)

\(\Rightarrow6x\left(2x-3\right)+8\left(2x-3\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(6x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\6x+8=0\end{matrix}\right.\)                        \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Có: \(f\left(x\right)=2ax^2-4\left(bx-1\right)+5x+c-11\)

\(=2ax^2-4bx+4+5x+c-11\)

\(=2ax^2+\left(-4b+5\right)x+\left(c-11\right)\)

\(\Rightarrow f\left(x\right)=x^2-5x+6\Leftrightarrow\left\{{}\begin{matrix}2a=1\\-4b+5=-5\\c-11=6\end{matrix}\right.\) (theo đồng nhất hệ số)

\(\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\\b=\dfrac{5}{2}\\c=17\end{matrix}\right.\)

2 thế kỉ 4 tháng = 200 năm 4 tháng

= 2400 tháng 4 tháng = 2404 tháng

Ta có: \(A=1+2+2^2+...+2^{2017}\)

\(2.A=2+2^2+2^3+...+2^{2018}\)

\(2A-A=2+2^2+2^3+...+2^{2018}-\left(1+2+2^2+...+2^{2017}\right)\)

\(A=2^{2018}-1\)

\(\Rightarrow A-B=2^{2018}-1-2^{2018}=-1\)

\(a,A=\left(2x+y\right)^2-\left(2x-y\right)^2\)

\(=\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)\)

\(=2y.4x=8xy\)

Vậy \(A=8xy\)

\(----------\)

\(b,B=\left(x-2y\right)^2-4\left(x-2y\right)y+4y^2\)

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

\(=\left(x-2y-2y\right)^2\)

\(=\left(x-4y\right)^2\)

Vậy \(B=\left(x-4y\right)^2\)