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Только 125 подходит
Начало координат (0,0), через нее проходит только y=-5x²+x
X^5*(x^2)^4=x^(5+2*4) = x^13
a^6*a/(a^8) = a^(6+1-8) = a^(-1) = 1/a
(-4a^4*b^3)^3=-64*a^12*b^9
8x^4y^3/(12x^8y)=2/3*x^(4-8)y^(3-1)=2/3*x^(-4)y^2=2/3*y^2/x^4
Решение
<span>1. <span>Sin2xcosx+cos2xsinx=<span><span> </span></span></span></span>√3/2
<span><span><span><span>sin(2x + x) = </span></span></span></span>√3/2
<span><span><span><span>sin3x = </span></span></span></span>√3/2
<span><span><span><span>3x = (-1)^x arcsin(</span></span></span></span>√3/2) + πn, n∈ Z
<span><span><span><span>3x = (-1)^n *(</span></span></span></span>π/3) + πn, n∈Z
<span><span><span><span>x = (-1)^n *(</span></span></span></span>π/9) + πn/3, n∈Z
<span>2. <span>Cosx+cos</span></span>²<span><span>x=<span> 1/2 </span>- sin</span></span>²<span><span>x</span></span>
cosx = 1/2 - 1
cosx = - 1/2
x = (+ -)arccos(-1/2) + 2πn, n ∈ Z
x = (+ -)*(π - arccos(1/2)) + 2πn, n ∈ Z
x = (+ -)*(π - π/3) + 2πn, n∈Z
x = (+ -)*(2π/3) + 2πn, n∈Z
<span>3. <span>Sin3xcosx-cos3xsinx = </span></span>√3/2
<span><span>sin(3x - x) = </span></span>√3/2
<span><span>sin2x = </span></span>√3/2
<span><span>2x = (-1)^n arcsin(</span></span>√3/2) + πk, k ∈ Z
<span><span>2x = (-1)^n * (</span></span>π/3) + πk, k ∈ Z
x = (-1)^n * (π/6) + πk/2, k ∈ Z
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