Помогите решить ,выручайте

Решите задачу:

$\displaystyle A5. \sqrt{x^2+x+1} =x-4$
$\displaystyle x^2+x+1=x^2+16-8x$
$\displaystyle x^2+x+1-x^2-16+8x=0$
$\displaystyle 9x=15$
$\displaystyle x= \frac{5}{3}$

$\displaystyle A6. y=2e^x+3cosx$
$\displaystyle y'=2e^x-3sinx$
$y'(0)=2e^0-3sin0=2$

$\displaystyle A7. tg( \frac{ \pi }{4}+2x)=1$
$\displaystyle \frac{ \pi }{4}+2x= \frac{ \pi }{4}+ \pi n, n \in Z$
$\displaystyle 2x= \pi n, n \in Z$
$\displaystyle x= \frac{ \pi} {2}n, n \in Z$

$1. f(x)=lnxcosx$
$\displaystyle f'(x)= \frac{1}{cosx}*(-sinx)=-tgx$
$\displaystyle f'( \frac{ \pi }{4})=-tg \frac{ \pi }{4}=-1$