Question

Помогите пожалуйста решить хоть что-то))

Answer 1

7. = ∛(m-2n)³ = m-2n = -1
        2n-m       -(m-2n)

8. = log₄ 4³ - log₄ 4⁻² - 3log₄ 4¹/² =3-(-2)-3*(1/2)=3+2-1.5=3.5

9.
tg(3π/4)=tg(4π/4 - π/4) = -tg(π/4)=-1
cos(5π/3)=cos(6π/3 - π/3) =cos(π/3)=1/2
sin(7π/6)=sin(6π/6 + π/6)=-sin(π/6)=-1/2
sin(2π/3)=sin(3π/3 - π/3) = sin (π/3)=√3/2

-1+(1/2)*(-1/2)-(√3/2)²=-1-(1/4)-(3/4)=-1-1=-2

10.
(1/4)^x + (1/4)^(x-1)>5
(1/4)^x (1+(1/4)⁻¹)>5
(1/4)^x (1+4)>5
(1/4)^x * 5>5
(1/4)^x>1
(1/4)^x >(1/4)⁰
x<0

Answer 2

$7) \frac{ \sqrt[3]{m ^{3} -6m ^{2} n+12mn ^{2}-8n ^{3} } }{2n-m}= \\=\frac{ \sqrt[3]{m ^{3} -3*2*m ^{2} *n+3*2 ^{2} *m*n ^{2} } -(2n) ^{3} }{2n-m} = \frac{ \sqrt[3]{(m-2n) ^{3} } }{2n-m} = \frac{m-2n}{2n-m} = \\ =- \frac{2n-m}{2n-m} =-1$

$8)log _{4} 64-log _{4} \frac{1}{16} -3log _{4} 2=log _{4} 64-log _{4} \frac{1}{16} -log _{4} 2 ^{3} = \\ =log _{4} 64-log _{4} \frac{1}{16} -log _{4} 8=log _{4} (64: \frac{1}{16} : 8)= \frac{64}{16*8} = \\ = \frac{8}{16*1} = \frac{1}{2}$

$9)tg \frac{3 \pi }{4} +cos \frac{5 \pi }{3} *sin \frac{7 \pi }{6} -(sin \frac{2 \pi }{3} ) ^{2} \\ \\ tg \frac{3 \pi }{4} =tg( \pi - \frac{ \pi }{4} )=-tg \frac{ \pi }{4} =-1\\ \\ cos \frac{5 \pi }{3} =cos(2 \pi - \frac{ \pi }{3} )=cos \frac{ \pi }{3} = \frac{1}{2} \\ \\ sin \frac{7 \pi }{6} =sin( \pi + \frac{ \pi }{6} )=-sin \frac{ \pi }{6} =- \frac{1}{2} \\ \\ sin ( \frac{2 \pi }{3} )^{2} =sin( \pi - \frac{ \pi }{3} ) ^{2} =-sin( \frac{ \pi }{3} )^{2} =(- \frac{ \sqrt{3} }{2} )^{2} = \frac{3}{4}$
$-1+ \frac{1}{2} *(- \frac{1}{2} )- \frac{3}{4} =-1- \frac{1}{4} - \frac{3}{4} =-(1+ \frac{1}{4} + \frac{3}{4} )=\\=-(1+ \frac{4}{4} )=-(1+1)=-2$


$10) ( \frac{1}{4} ) ^{x} +( \frac{1}{4} ) ^{x-1} \ \textgreater \ 5\\ \\ ( \frac{1}{4} ) ^{x} *(1+( \frac{1}{4} ) ^{-1} )\ \textgreater \ 5\\ ( \frac{1}{4} ) ^{x} *(1+4)\ \textgreater \ 5\\ ( \frac{1}{4} ) ^{x} \ \textgreater \ 1\\ ( \frac{1}{4} ) ^{x} \ \textgreater \ ( \frac{1}{4} ) ^{0} \\a= ( \frac{1}{4} )\ \textless \ 1\\x\ \textless \ 0$