Y`=-12sinx+6√3=0
12sinx=6√3
sinx=√3/2
x=π/3∈[0;π/2]
y(0)=12cos0+6√3*0-2√3π+6=12-2√3π+6=18-2√3π≈7,3 -наим
y(π/3)=12cosπ/3+6√3*π/3-2√3π+6=6+2√3π-2√3π+6=12-наиб
y(π/2)=12cosπ/2+6√3*π/2-2√3π+6=3√3π-2√3π+6=√3π+6≈11,3
(a^n)^m=a^(nm)
((√(3)−1)^(2))^(0.5)-((√(3)+2)^(2))^(0.5)={√(√3 - 2)^2 = |√3 - 2| = 2 - √3}=(√3-1)^(2*0.5)+<span>2 - √3=
</span>√3-1+<span>2 - √3=1</span>
1=1
Ответ на 6 и 7
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