1. 4^(5/<span>(x</span>+1<span>)</span>)=(1/2)^(6-4<span>x</span>)
<span> </span>2^(2*5/x+1)=2^(-1*(6-4<span>x</span>))
<span> </span>10/(<span>x</span>+1)=-6+4<span>x</span>
<span> </span>10=-(6-4<span>x</span>)*(<span>x</span>+1)
<span><span> </span>10=(4x-6)*(x+1)</span>
<span><span> </span>10=4x^2-2x-6</span>
<span><span> </span>4x^2-2x-16=0</span>
<span><span> </span>2x^2-x-8=0</span>
<span><span> </span>D=b^2-4ac=1+ sqrt(65)</span>
<span><span> </span>x1=(-b±sqrt(65))/2a</span>
<span><span> </span>x1=(1+sqrt(65))/4</span>
<span><span> </span><span> </span>x2=(1-sqrt(65))/4</span>
<span> </span>
<span>2</span><span>. 27^(2/5)*2^(1/5)*2^(5/6) =3^(3*2/5)^(5/6)*2^(1/5)^(5/6)*2^(5/6) =3*2^((5/30)+(5/6))=3*2=6</span>
<span> </span>
<span>3. log(3*(2x-1))<4<span> </span>-?</span>
непонятно какое основание логарифма
<span> </span>
<span>4.<span> </span>sin^2(x)+cos^2(x)=1 => sin(x)=±sqrt(1-cos^2(x)) = ±sqrt(1-0,36)=±sqrt(0,64)=±0,8</span>
Учитывая<span>, </span>что<span> 0<</span><span>x</span><span><</span><span>pi</span><span>/2, </span>получим
<span><span> </span></span><span>sin(x)=+0,8</span>
<span> </span>
<span> </span>
<span>5. 2sin(pi/4)+3*tg(3*pi/4)-4*cos(pi/3) = 2*(1/sqrt(2) +3*(-1) – 4*(1/2) =</span>
<span><span> </span>(2/sqrt(2))-3-2=(2/sqrt(2)-5</span>
<span> </span>
<span>6. cos(pi/15)*cos(4pi/15)-sin(4pi/15)*sin(pi/15) =cos((pi/15)+(4pi/15))=cos(pi/3)=1/2</span>