Question

Помогите пожалуйста 3,4,5,6,7

Answer 1

$3)\; \; log_4\, log_{16}\, log_2x=0\; \; ,\\\\ODZ:\; \left \{ {{x>0\; ,\; \; log_2x>0\; ,} \atop {log_{16}\, log_2x>0\; ,\; \; }} \right. \; \left \{ {{x>0\; ,\; \; x>1\; ,} \atop {log_2x>1\; ,\; \; \; \; }}\right. \; \left \{ {{x>1\; ,} \atop {x>2}} \right. \; \to \; \; x>2\\\\log_{16}log_2x=1\; ,\; \; log_2x=16\; ,\; \; \underline {x=2^{16}\; }.$

$4)\; \; cos^2(3\, arcctg\sqrt3+arctg1)=cos^2(3\cdot \frac{\pi}{6}+\frac{\pi}{4})=cos^2\, \frac{3\pi}{4}=\\\\=\Big (cos\frac{3\pi}{4}\Big )^2=\Big (-cos\frac{\pi}{4}\Big )^2=\Big (-\frac{\sqrt2}{2}\Big )^2=\frac{2}{4}=\frac{1}{2}$

$5)\; \; 2\, sin2x\cdot sin6x=cos4x\\\\\star \; \; sina\cdot sin\beta =\frac{1}{2}\, (cos(a-\beta )-cos(a+\beta ))\; \; \star \\\\cos4x-cos8x=cos4x\\\\cos8x=0\; ,\; \; 8x=\frac{\pi}{2}+\pi n\; ,\; \; \underline {x=\frac{\pi}{16}+\frac{\pi n}{8}\; ,\; n\in Z}$

$6)\; \; y=(x^2-1)(2x+5)\\\\y'=2x\cdot (2x+5)+2\cdot (x^2-1)=6x^2+10x-2\\\\y'(1)=6+10-2=14\\\\\\7)\; \; log_3(2x-7)>2\; \; ,\; \; \; ODZ:\; 2x-7>0\; \; \to \; \; x>3,5\\\\log_3(2x-7)>log_33^2\; \; \Rightarrow \; \; 2x-7>3^2\; \; (t.k.\; 3>1)\\\\2x-7>9\\\\2x>16\\\\x>8\\\\\underline {x\in (8,+\infty )}$