Sin4x-sin(π/2-6x)=2*sin(4x-(π/2-6x))/2
*cos(4x+π/2-6x)/2=
2*sin(4x-π/2+6x)/2*cos((π/2-2x)/2=
2*sin(10x/2-π/2:2)*cos(π/2:2-2x:2)=
2*sin(5x-π/4)*cos(π/4-x)
Cos x= 1/3<span>x=+- arccos 1/3 + 2 pn=n пренадлежит Z
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(5b²)³ = 5 * b² * 5 * b² * 5 * b² = 125b²⁺²⁺² = 125b⁶
1)0.125a⁹+0.15a⁶b⁴+0.06a³b^⁸+0.008b¹²=(0,5a³)³+3(0,5a³)²(0,2b⁴)+3(0,5a³)(0,2b⁴)²+(0,2b⁴)³=(<span><span>0,5a³+0,2b⁴)³
</span>2)0.216x¹²-0.54x⁸y⁵+0.45x⁴y¹⁰-0.125y¹⁵=
(0,6x⁴)³-3(0,6x⁴)²(0,5y⁵)+3(0,6x⁴)(0,5y)²-(0,5y⁵)³=(</span><span>0,6x⁴-</span><span>0,5y⁵)³</span>
Решение задания смотри на фотографии