81*(sqrt10+3)^5x-61=(3/sqrt10-3)^5x-61=0
<span>6х^2=х^3+8х
</span>х³-6х²<span>+8х=0
х*(х</span>²-6х+8<span>)=0
х=0
</span>х²-6х+8=0
D=36-32=4
x₁=(6+2)/2=4
x₂=(6-2)/2=2
итак получили три корня х=0 ;х=4 ; х=2
<span>sin^6(x)+cos^6(x)=(5/4)sin^2(2x)
</span>sin^6(x)+cos^6(x) = 5 sin^2(x)*cos^2(x)
(sin^2(x)+cos^2(x))(sin^4(x)+cos^4(x)-sin^2(x)cos^2(x)) = 5/4sin^2(2x)
1/8(3cos(4x)+5) = -5/8(cos(4x)-1)
<span>cos(4x) = 0
</span>x = πn-(7π)/8
x <span>= πn-(5π)/8
</span>x = πn-(3π)/8
<span>x = πn-π/8</span>
n ∈ Z
(x+1)(x²+x+1)-x(x-3)(x+3)=x^3+x^2+x^2+x+x+1-x(x^2-9)=
=x^3+2x^2+2x+1-x^3+9x=2x^2+11x+1