ĐK: ...
\(t=a\sqrt{\dfrac{x^2+1}{2x}}\)\(\Rightarrow t^2=a^2.\dfrac{x^2+1}{2x}\)
\(\Rightarrow\left\{{}\begin{matrix}t^2-a^2=a^2.\dfrac{x^2+1-2x}{2x}=a^2.\dfrac{\left(x-1\right)^2}{2x}\\t^2+a^2=a^2.\dfrac{x^2+1+2x}{2x}=a^2.\dfrac{\left(x+1\right)^2}{2x}\end{matrix}\right.\)
Thay vào M ta được:
\(M=\left(\dfrac{\sqrt{a^2.\dfrac{\left(x-1\right)^2}{2x}}+\sqrt{a^2.\dfrac{\left(x+1\right)^2}{2x}}}{\sqrt{a^2.\dfrac{\left(x-1\right)^2}{2x}}-\sqrt{a^2.\dfrac{\left(x+1\right)^2}{2x}}}\right)^4\)
\(M=\left(\dfrac{\dfrac{a.\left(x-1\right)}{\sqrt{2x}}+\dfrac{a.\left(x+1\right)}{\sqrt{2x}}}{\dfrac{a.\left(x-1\right)}{\sqrt{2x}}-\dfrac{a.\left(x+1\right)}{\sqrt{2x}}}\right)^4\)
\(M=\left(\dfrac{x-1+x+1}{x-1-\left(x+1\right)}\right)^4=\left(\dfrac{2x}{-2}\right)^4=\left(-x\right)^4=x^4=2012^4\)
