Предмет: Математика,
автор: andrianava
напишіть будь ласка 13,14
Приложения:
Ответы
Автор ответа:
0
14
![4^{sin2x}+ frac{6}{4^{ frac{1}{2}(1-sin2x) }} -4=0\
2^{2sin2x}+ 3*2^{sin2x}-4=0\
2^{sin2x}=t\
t^2+3t-4=0\
D=9+16=25\
t= frac{-3+5}{2} =1\
2^{sin2x}=2^0\
sin2x=0\
2x= pi k\
x= frac{ pi k}{2} ,](https://tex.z-dn.net/?f=4%5E%7Bsin2x%7D%2B+frac%7B6%7D%7B4%5E%7B+frac%7B1%7D%7B2%7D%281-sin2x%29+%7D%7D+-4%3D0%5C%0A2%5E%7B2sin2x%7D%2B+3%2A2%5E%7Bsin2x%7D-4%3D0%5C%0A2%5E%7Bsin2x%7D%3Dt%5C%0At%5E2%2B3t-4%3D0%5C%0AD%3D9%2B16%3D25%5C%0At%3D+frac%7B-3%2B5%7D%7B2%7D+%3D1%5C%0A2%5E%7Bsin2x%7D%3D2%5E0%5C%0Asin2x%3D0%5C%0A2x%3D+pi+k%5C%0Ax%3D+frac%7B+pi+k%7D%7B2%7D+%2C "4^{sin2x}+ frac{6}{4^{ frac{1}{2}(1-sin2x) }} -4=0\
2^{2sin2x}+ 3*2^{sin2x}-4=0\
2^{sin2x}=t\
t^2+3t-4=0\
D=9+16=25\
t= frac{-3+5}{2} =1\
2^{sin2x}=2^0\
sin2x=0\
2x= pi k\
x= frac{ pi k}{2} ,")
k∈Z
13.
![(2+ sqrt{3} )^{x^2-2x+1}+(2- sqrt{3} )^{x^2-2x-1}= frac{4}{2- sqrt{3} } |*(2- sqrt{3})\
(2+ sqrt{3} )^{x^2-2x}+(2- sqrt{3} )^{x^2-2x}=4\
frac{1}{2- sqrt{3} } =2+ sqrt{3} \
(2+ sqrt{3} )^{x^2-2x}=t\
t + frac{1}{t} =4\
t^2-4t+1=0\
D=16-4=12\
t_1= frac{4+2 sqrt{3} }{2} = 2+ sqrt{3} =(2+ sqrt{3} )^{x^2-2x}\
x^2-2x-1=0\
D=2+4=6\
x_1=1-0,5 sqrt{6} \
x_2=1+0,5 sqrt{6} \
t_2=2- sqrt{3} =(2+ sqrt{3} )^{x^2-2x}\
x^2-2x+1=0\
x_3=1](https://tex.z-dn.net/?f=%282%2B+sqrt%7B3%7D+%29%5E%7Bx%5E2-2x%2B1%7D%2B%282-+sqrt%7B3%7D+%29%5E%7Bx%5E2-2x-1%7D%3D+frac%7B4%7D%7B2-+sqrt%7B3%7D+%7D+%7C%2A%282-+sqrt%7B3%7D%29%5C%0A%282%2B+sqrt%7B3%7D+%29%5E%7Bx%5E2-2x%7D%2B%282-+sqrt%7B3%7D+%29%5E%7Bx%5E2-2x%7D%3D4%5C%0A+frac%7B1%7D%7B2-+sqrt%7B3%7D+%7D+%3D2%2B+sqrt%7B3%7D+%5C%0A%282%2B+sqrt%7B3%7D+%29%5E%7Bx%5E2-2x%7D%3Dt%5C%0At+%2B++frac%7B1%7D%7Bt%7D+%3D4%5C%0At%5E2-4t%2B1%3D0%5C%0AD%3D16-4%3D12%5C%0At_1%3D+frac%7B4%2B2+sqrt%7B3%7D+%7D%7B2%7D++%3D+2%2B+sqrt%7B3%7D+%3D%282%2B+sqrt%7B3%7D+%29%5E%7Bx%5E2-2x%7D%5C%0Ax%5E2-2x-1%3D0%5C%0AD%3D2%2B4%3D6%5C%0Ax_1%3D1-0%2C5+sqrt%7B6%7D+%5C%0Ax_2%3D1%2B0%2C5+sqrt%7B6%7D+%5C%0At_2%3D2-+sqrt%7B3%7D+%3D%282%2B+sqrt%7B3%7D+%29%5E%7Bx%5E2-2x%7D%5C%0Ax%5E2-2x%2B1%3D0%5C%0Ax_3%3D1 "(2+ sqrt{3} )^{x^2-2x+1}+(2- sqrt{3} )^{x^2-2x-1}= frac{4}{2- sqrt{3} } |*(2- sqrt{3})\
(2+ sqrt{3} )^{x^2-2x}+(2- sqrt{3} )^{x^2-2x}=4\
frac{1}{2- sqrt{3} } =2+ sqrt{3} \
(2+ sqrt{3} )^{x^2-2x}=t\
t + frac{1}{t} =4\
t^2-4t+1=0\
D=16-4=12\
t_1= frac{4+2 sqrt{3} }{2} = 2+ sqrt{3} =(2+ sqrt{3} )^{x^2-2x}\
x^2-2x-1=0\
D=2+4=6\
x_1=1-0,5 sqrt{6} \
x_2=1+0,5 sqrt{6} \
t_2=2- sqrt{3} =(2+ sqrt{3} )^{x^2-2x}\
x^2-2x+1=0\
x_3=1")
k∈Z
13.
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