a: 2k^2+kx-10=0
Khi x=2 thì ta sẽ có: 2k^2+2k-10=0
=>k^2+k-5=0
=>\(k=\dfrac{-1\pm\sqrt{21}}{2}\)
b: Khi x=-2 thì ta sẽ có:
\(\left(-2k-5\right)\cdot4-\left(k-2\right)\cdot\left(-2\right)+2k=0\)
=>-8k-20+2k-4+2k=0
=>-4k-24=0
=>k=-6
c: Theo đề, ta có:
9k-3k-72=0
=>6k=72
=>k=12