F`(x)=2e^2x-5e^x=e^x(2e^x-5)=0
2e^x-5=0
e^x=2,5
x=ln2,5≈0,9∈[-2;1]
f(-2)=1/e^4-5/e²-2≈0,02-0,68-2≈-2,66
f(ln2,5)=6,25-5*2,5-2=4,25-12,5=-8,25
f(1)=e²-5e-2≈4,4-5*2,7-2=2,4-13,5=-11,1 наим
-1 <= (x^2 - 5x + 4) / (x^2 - 4) <= 1;
(x^2 - 5x + 4) / (x^2 - 4<span>) - 1 <= 0;
</span>(-5x + 8) / ((x - 2) * (x + 2)) <= 0;
x ∈ (-2; -1.6] U (2; +00);
-1 <= (x^2 - 5x + 4) / (x^2 - 4<span>);
</span>(x^2 - 5x + 4) / (x^2 - 4<span>) + 1 >= 0;
</span>(2x^2 - 5x) / ((x - 2) * (x + 2)<span>) >= 0;
</span>x(2x - 5) / ((x - 2) * (x + 2)<span>) >= 0;
</span><span>x ∈ (-00; -2) U [0; 2) U [2.5; +00);
</span>Ответ: <span>x ∈ [2.5; +00).</span>
Решение в скане............