(три корня и 5)в квадрате деленное
на 15= (корень из 45)в квадрате, деленное на 15= 45:15=3
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Выражения под корнями взаимно обратные. Сделаем замену :
![(\sqrt{3-2\sqrt{2} })^{x}=m,m>0](https://tex.z-dn.net/?f=%28%5Csqrt%7B3-2%5Csqrt%7B2%7D%20%7D%29%5E%7Bx%7D%3Dm%2Cm%3E0)
Тогда :
![(\sqrt{3+2\sqrt{2} })^{x}=\frac{1}{m}](https://tex.z-dn.net/?f=%28%5Csqrt%7B3%2B2%5Csqrt%7B2%7D%20%7D%29%5E%7Bx%7D%3D%5Cfrac%7B1%7D%7Bm%7D)
![m+\frac{1}{m} \geq 6\\\\m^{2}-6m+1\geq0\\\\m^{2}-6m+1=0\\\\D=36-4=32=4\sqrt{2}\\\\m_{1}=\frac{6-4\sqrt{2} }{2}=3-2\sqrt{2}\\\\m_{2}=\frac{6+4\sqrt{2} }{2}=3+2\sqrt{2}](https://tex.z-dn.net/?f=m%2B%5Cfrac%7B1%7D%7Bm%7D%20%5Cgeq%206%5C%5C%5C%5Cm%5E%7B2%7D-6m%2B1%5Cgeq0%5C%5C%5C%5Cm%5E%7B2%7D-6m%2B1%3D0%5C%5C%5C%5CD%3D36-4%3D32%3D4%5Csqrt%7B2%7D%5C%5C%5C%5Cm_%7B1%7D%3D%5Cfrac%7B6-4%5Csqrt%7B2%7D%20%7D%7B2%7D%3D3-2%5Csqrt%7B2%7D%5C%5C%5C%5Cm_%7B2%7D%3D%5Cfrac%7B6%2B4%5Csqrt%7B2%7D%20%7D%7B2%7D%3D3%2B2%5Csqrt%7B2%7D)
![(m-(3-\sqrt{2}))(m-(3+2\sqrt{2}))\geq 0](https://tex.z-dn.net/?f=%28m-%283-%5Csqrt%7B2%7D%29%29%28m-%283%2B2%5Csqrt%7B2%7D%29%29%5Cgeq%200)
+ - +
0_________[3-2√2]__________[3 + 2√2]__________ m
1) 0 < m ≤ 3 - 2√2 2) m ≥ 3 + 2√2
![1)(\sqrt{3-2\sqrt{2} })^{x} \leq3-2\sqrt{2}\\\\(3-2\sqrt{2})^{\frac{x}{2} }\leq 3-2\sqrt{2}\\\\\frac{x}{2} \leq1\\\\x\leq2\\\\x\in(-\infty;2]\\\\2)(\sqrt{3-2\sqrt{2} })^{x}\geq3+2\sqrt{2}\\\\(3-2\sqrt{2})^{\frac{x}{2} }\geq (3-2\sqrt{2})^{-1}\\\\\frac{x}{2}\geq-1\\\\x\geq -2\\\\x\in[-2;+\infty)](https://tex.z-dn.net/?f=1%29%28%5Csqrt%7B3-2%5Csqrt%7B2%7D%20%7D%29%5E%7Bx%7D%20%5Cleq3-2%5Csqrt%7B2%7D%5C%5C%5C%5C%283-2%5Csqrt%7B2%7D%29%5E%7B%5Cfrac%7Bx%7D%7B2%7D%20%7D%5Cleq%203-2%5Csqrt%7B2%7D%5C%5C%5C%5C%5Cfrac%7Bx%7D%7B2%7D%20%5Cleq1%5C%5C%5C%5Cx%5Cleq2%5C%5C%5C%5Cx%5Cin%28-%5Cinfty%3B2%5D%5C%5C%5C%5C2%29%28%5Csqrt%7B3-2%5Csqrt%7B2%7D%20%7D%29%5E%7Bx%7D%5Cgeq3%2B2%5Csqrt%7B2%7D%5C%5C%5C%5C%283-2%5Csqrt%7B2%7D%29%5E%7B%5Cfrac%7Bx%7D%7B2%7D%20%7D%5Cgeq%20%283-2%5Csqrt%7B2%7D%29%5E%7B-1%7D%5C%5C%5C%5C%5Cfrac%7Bx%7D%7B2%7D%5Cgeq-1%5C%5C%5C%5Cx%5Cgeq%20-2%5C%5C%5C%5Cx%5Cin%5B-2%3B%2B%5Cinfty%29)
Ответ : x ∈ [- 2 ; 2]
По основному тригонометрическому тождеству:
![\sin {}^{2} (x ) = 1 - \cos {}^{2} (x) = 1 - \frac{16}{25} = \frac{9}{25} \\ \sin(x) = \frac{3}{5}](https://tex.z-dn.net/?f=%20%5Csin%20%7B%7D%5E%7B2%7D%20%28x%20%29%20%20%3D%201%20-%20%20%5Ccos%20%7B%7D%5E%7B2%7D%20%28x%29%20%20%20%3D%20%201%20-%20%20%5Cfrac%7B16%7D%7B25%7D%20%20%3D%20%20%5Cfrac%7B9%7D%7B25%7D%20%20%5C%5C%20%20%5Csin%28x%29%20%20%3D%20%20%5Cfrac%7B3%7D%7B5%7D%20)
т.к. х∈ III четверти, то синус в этой чатверти отрицательный, т.е.
![\sin(x) = - \frac{3}{5} = - 0.6](https://tex.z-dn.net/?f=%20%5Csin%28x%29%20%20%3D%20%20-%20%20%5Cfrac%7B3%7D%7B5%7D%20%20%3D%20%20-%200.6)
Ответ: -0,6.
A. <span>1-(sin⁶a+cos⁶a)=1-((sin²a)³+(cos²a)³)=1-(sin²a+cos²a)(sin⁴a-sin²acos²a+cos⁴a)=1-((sin²a+cos²a)²-2sin²acos²a-sin²acos²a)=1-1+3sin²acos²a=3sin²acos²a. </span>
<span>Тождество доказано.</span>
A^2 + a^2 = 2a^2
Потому что степени не складываются