Tq2x =tqπ/3 ;
2x =π/3 +πn ,n∈Z;
x =π/6 + πn/2 , n∈Z .
* * * * * * *
tq2x =tqπ/3 ;
tq2x =√3 ;
2x =π/3 +πn ,n∈Z ;
x =π/6 + πn/2 , n∈Z (n -целое число).
Y'=1-400/x^2
y'=0 x^2=400 x=-20 x=20
y(-20)=400-20=380
y(-28)=784-100/7>380
ответ 769 5/7
64-(4-3а)(16+12а+9а²)=64-(64-27а³)=27а³
3∛x=x-2
27x=x³-6x²+12x-8
27-x³+6x²-12x+8=0
15x-x³+6x²+8=0-x³+6x²+15x+8=0
-x³-x²+7x²+7x+8x+8=0
-x²×(x+1)+7x×(x+1)+8(x+1)=0
-(x+1)×(x²-7x-8)=0
-(x+1)×(x²+x-8x-8)=0
-(x+1)×(x×(x+1)-8(x+1))=0
-(x+1)×(x+1)×(x-8)=0
-(x+1)²×(x-8)=0
-(x+1)²=0
x-8=0
x₁=-1,x₂=8