a, f(-2)=(m^2-2).(-2)^2+2(m^2-1)=140
=(m^2-2).4+2m^2-2=140
=4m^2-8+2m^2-2=140
=6m^2-10=140
=6m^2=150
=m^2=25
=> m=+-5
b, h(3)-2h(1)=(n^3-2).3+2.(n^3-1)-3-2[(n^3-2).1+2(n^3-1)-3]=11
=3n^3-6+2n^3-2-3-2n^3+4-4n^3+4+6=11
=-n^3+3=11
=-n^3=8(loại)
vậy ko tìm đc n
a, Ta có : \(f\left(-2\right)=\left(m^2-2\right).\left(-2\right)^2+2.\left(m^2-1\right)=140\)
\(=4m^2-8+2m^2-2=140\)
\(=\left(4m^2+2m^2\right)+\left(-8-2\right)=140\)
\(=6m^2=140-\left(-10\right)\)
\(\Leftrightarrow6m^2=150\)
\(\Leftrightarrow m^2=25\)
\(\Leftrightarrow m=\pm5\)
Vậy : \(m=\pm5\) nếu \(f\left(-2\right)=140\)