7 x 5 x 6 
3,4 x 2,5 x 1,2

\(\frac{5}{6}x\frac{4}{5}x\frac{3}{2}\)
 

=> m⁴ = 16m⁴

m⁴ - 16m⁴ = 0

m⁴.(1 - 16) = 0

m⁴.(-15) = 0

m⁴ = 0 : (-15)

m⁴ = 0

<=> m = 0

\(M\left(x\right)+N\left(x\right)=4x^4+5x^3-6x^2-3\)

\(M\left(x\right)-N\left(x\right)=-2x^4+5x^3+4x^2+2x+2\)

\(f\left(x\right)=\left(m-4\right)x^2+\left(m+1\right)x+2m-1\)

\(f\left(x\right)< 0,\forall x\in R\Leftrightarrow\left\{{}\begin{matrix}a< 0\\\Delta< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m-4< 0\\\left(m+1\right)^2-4\left(m-4\right)\left(2m-1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\m^2+2m+1-4\left(2m^2-m-8m+4\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow m^2+2m+1-8m^2+36m-16< 0\)

\(\Leftrightarrow-7m^2+38m-15< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\\left[{}\begin{matrix}m< \dfrac{3}{7}\\m>5\end{matrix}\right.\end{matrix}\right.\)

\(KL:m\in\left(5;+\infty\right)\)