a; \(\Leftrightarrow\dfrac{x!}{\left(x-1\right)!\cdot1!}+\dfrac{x!}{\left(x-2\right)!\cdot2!}+\dfrac{x!}{\left(x-3\right)!\cdot3!}>\dfrac{7}{2}x\)
\(\Leftrightarrow x+\dfrac{1}{2}x\left(x-1\right)+\dfrac{1}{6}x\left(x-1\right)\left(x-2\right)>\dfrac{7}{2}x\)
=>\(1+\dfrac{1}{2}x-\dfrac{1}{2}+\dfrac{1}{6}\left(x^2-3x+2\right)>\dfrac{7}{2}\)
=>\(\dfrac{1}{2}x+\dfrac{1}{2}+\dfrac{1}{6}x^2-\dfrac{1}{2}x+\dfrac{1}{3}-\dfrac{7}{2}>0\)
=>\(\dfrac{1}{6}x^2-\dfrac{8}{3}>0\)
=>1/6x^2>8/3
=>x^2>16
=>x>4
b: \(\Leftrightarrow\dfrac{\left(x-2\right)!}{\left(x-2-2\right)!}+\dfrac{x!}{\left(x-x+2\right)!\cdot\left(x-2\right)!}=101\)
=>\(\left(x-3\right)\left(x-2\right)+\dfrac{1}{2}x\left(x-1\right)=101\)
=>\(x^2-5x+6+\dfrac{1}{2}x^2-\dfrac{1}{2}x=101\)
=>3/2x^2-11/2x-95=0
=>x=10
