Решение задания приложено
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Решение на картинке:
<span> (sin (3x/2)+ cos(3x\2))^2=1\2</span>
<span>sin^2(3x/2)+2sin3x/2cos3x/2+cos^2(3x/2)=1/2</span>
<span>1+sin3x=1/2</span>
<span>sin3x=-1/2</span>
<span>3x=(-1)^(n+1)*pi/6+pi n</span>
<span>x=(-1)^(n+1)*pi/18+1/3pi n</span>
<span>1)
2x</span>²<span>+ 24xy +72y</span>² = 2 ·(х²+12ху+36у²) = 2·(х+6у)² = 2·(х+6у)(х+6у)
<span>2)
-8a</span>⁵<span> +8а</span>³<span> - 2а = -2а</span>·(4а⁴-4a²+1) = -2a·(2a²-1)² = -2a·(2a²-1)(2a²-1)
<span>3)
5a</span>³<span> - 40b</span>⁶ = 5a³ ·(1-8b³) = 5a³ ·(1³ - (2b)³) = 5a³·(1-2b)(1+2b+4b²)
<span>4)
8а</span>³<span> -аb - a</span>²<span>b + а</span>² = a·(8a²-b-ab+a)