tolong dibantu no 9 dan 10 mastah

[tex]\displaystyle f(x)=(x+2)(x-2)(x-3)h(x)+ax^2+bx+c\\\\f(2)=(2+2)(2-2)(2-3)h(2)+a(2)^2+b(2)+c\\0=4a+2b+c\\\\f(-2)=(-2+2)(-2-2)(-2-3)h(2)+a(-2)^2+b(-2)+c\\0=4a-2b+c\\\\f(3)=(3+2)(3-2)(3-3)h(2)+a(3)^2+b(3)+c\\f(3)=9a+3b+c\\\\4a+2b+c=0\\\underline{4a-2b+c=0}-\\~~~~~~~~4b~~~~~=0\\~~~~~~~~~b~~~~~=0\\\\4a+2b+c=0\\4a+2(0)+c=0\\-4a=c\\\\9a+3b+c=f(3)\\9a+3(0)-4a=f(3)\\5a=f(3)\\a=\frac15f(3)\\\\\therefore\\s(x)=ax^2+bx+c\\s(x)=\frac15f(3)x^2+(0)x-\frac{4}{5}f(3)\\\boxed{\boxed{s(x)=\frac15f(3)(x^2-4)}}[/tex]

[tex]\displaystyle 6^{4\log x}\cdot3^{-3\log x}=48\\2^{4\log x}\cdot3^{4\log x}\cdot3^{-3\log x}=48\\2^{4\log x}\cdot3^{4\log x-3\log x}=48\\2^{4\log x}\cdot3^{\log x}=16\cdot3\\\therefore\\\log x=1\\x=10\\\\\lim_{p\to10}\frac{2p^2-20p}{\sqrt{2p-6}-\sqrt{p+4}}=\lim_{p\to10}\frac{2p^2-20p}{\sqrt{2p-6}-\sqrt{p+4}}\cdot\frac{\sqrt{2p-6}+\sqrt{p+4}}{\sqrt{2p-6}+\sqrt{p+4}}\\\lim_{p\to10}\frac{2p^2-20p}{\sqrt{2p-6}-\sqrt{p+4}}=\lim_{p\to10}\frac{(2p^2-20p)(\sqrt{2p-6}+\sqrt{p+4})}{2p-6-p-4}[/tex]
[tex]\displaystyle \lim_{p\to10}\frac{2p^2-20p}{\sqrt{2p-6}-\sqrt{p+4}}=\lim_{p\to10}\frac{2p(p-10)(\sqrt{2p-6}+\sqrt{p+4})}{p-10}\\\lim_{p\to10}\frac{2p^2-20p}{\sqrt{2p-6}-\sqrt{p+4}}=\lim_{p\to10}2p(\sqrt{2p-6}+\sqrt{p+4})\\\lim_{p\to10}\frac{2p^2-20p}{\sqrt{2p-6}-\sqrt{p+4}}=2(10)(\sqrt{2(10)-6}+\sqrt{10+4})\\\boxed{\boxed{\lim_{p\to10}\frac{2p^2-20p}{\sqrt{2p-6}-\sqrt{p+4}}=40\sqrt{14}}}[/tex]