42х+ 20х=12710
62х=12710
х=12710:62
х=6355
BD=3√3см- медиана,АС=4⇒BD=CD=2см,<BDA=60⇒<BDC=120-смежные
AB²=AD²+BD²-2*AD*BD*cos60=4+27-2*2*3√3*1/2=31-6√3
AB=
![\sqrt{31-6 \sqrt{3} }](https://tex.z-dn.net/?f=+%5Csqrt%7B31-6+%5Csqrt%7B3%7D+%7D+)
BC²=BD²+CD²-2*BD*CD*cos120=4+27-2*2*3√3*(-1/2)=31+6√3
BC=
![\sqrt{31+6 \sqrt{3} }](https://tex.z-dn.net/?f=+%5Csqrt%7B31%2B6+%5Csqrt%7B3%7D+%7D+)
p=(AB+BC+AC)/2=(
![( \sqrt{31- 6\sqrt{3} } + \sqrt{31+6 \sqrt{3} } +4)/2](https://tex.z-dn.net/?f=%28+%5Csqrt%7B31-+6%5Csqrt%7B3%7D+%7D+%2B+%5Csqrt%7B31%2B6+%5Csqrt%7B3%7D+%7D+%2B4%29%2F2)
p-AB=(√(31-6√3)/2+√(31+6√3)/2+2-√(31-6√3)=√(31+6√3)/2+2-√(31-6√3)/2
p-BC=√(31-6√3)/2+2-√(31+6√3)/2
p-AC=√(31+6√3)/2+√(31-6√3)/2-2
S²=(√(31+6√3)/2+2+√(31-6√3)/2)*(√(31-6√3)/2+2-√(31+6√3)/2)*√(31+6√3)/2+2-√(31-6√3)/2)*(√(31+6√3)/2+√(31-6√3)/2-2)=81
S=9
2,4/7-1,2/3 = 18/7-5/3 = 54/21-35/21 = 19/21
ОДЗ:
12-х>0 -x > -12 x<12
log₂(12-x)>0 12-x > 1 x < 11
ОДЗ: х∈ (-∞; 11)
Решаем неравенство
<span> log1/3(log2(12-x))>-2·log1/3(1/3);
</span>
<span>log1/3(log2(12-x))>log1/3(1/3)⁻²;
</span>
<span>log1/3(log2(12-x))>log1/3(9) ⇒ log₂(12-x) < 9 ;
</span><span>log₂(12-x) < 9log₂2;
</span><span>log₂(12-x) < log₂2⁹;
12-x < 512;
</span>
x> -500
С учетом ОДЗ получаем ответ:
( - 500; 11)
Середина интервала - 255,5
О т в е т. - 255,5