Помогите решить 6 и 8

Решите задачу:

$6)\frac{2-a}{a+2}+\frac{2a}{2-a}+\frac{4a^{2} }{a^{2}-4 }=\frac{2-a}{a+2}-\frac{2a}{a-2}+\frac{4a^{2} }{(a-2)(a+2)}=\frac{2a-4-a^{2}+2a-2a^{2}-4a+4a^{2}}{a^{2}-4 }=\frac{a^{2}-4 }{a^{2}-4 }=1$

$8)\frac{(x-3)^{2} }{12}-\frac{2x^{2}+5 }{4}+\frac{1,25x^{2}-1 }{3}=\frac{2}{3}\\\\\frac{(x-3)^{2} }{12}*12-\frac{2x^{2} +5}{4}*12+\frac{1,25x^{2}-1 }{3}*12=\frac{2}{3}*12\\\\x^{2}-6x+9-3(2x^{2}+5)+4(1,25x^{2}-1)=8\\\\x^{2}-6x+9-6x^{2}-15+5x^{2}-4=8\\\\-6x=8+10\\\\-6x=18\\\\x=-3$