![+ \left \{ {{ x^{2} +y=7} \atop {2 x^{2} -y=5}} \right.](https://tex.z-dn.net/?f=%2B+%5Cleft+%5C%7B+%7B%7B+x%5E%7B2%7D+%2By%3D7%7D+%5Catop+%7B2+x%5E%7B2%7D+-y%3D5%7D%7D+%5Cright.+)
____________
3x² = 12
x² = 4
x₁ = 2 x₂ = - 2
y₁ = 7 - x² = 7 - 2² = 7 - 4 = 3
y₂ = 7 - x² = 7 - (- 2)² = 7 - 4 = 3
Ответ : (2 ; 3) , (- 2 ; 3)
7 - 4√3 = 4 - 2 · 2 · √3 + 3 = 2² - 2 · 2 · √3 + (√3)² = (2 - √3)²
7 + 4√3 = 4 + 2 · 2 · √3 + 3 = 2² +<span> 2 · 2 · √3 +</span><span> (</span>√3)²= <span>(2 + √3)²
</span>√(2 - √3)² - √(2 + √3)² = |2 - √3| - |2 + √3| = 2 - √3 - 2 - √3 = -2√3
Вариант А правильный, т.к.
![\sqrt{ x^{2} } = \sqrt{4}](https://tex.z-dn.net/?f=+%5Csqrt%7B+x%5E%7B2%7D+%7D+%3D++%5Csqrt%7B4%7D)
IxI= +-2
x=2 или x=-2
![x^{2} -3x-10<0](https://tex.z-dn.net/?f=x%5E%7B2%7D+-3x-10%3C0)
![x^{2} -3x-10=0](https://tex.z-dn.net/?f=x%5E%7B2%7D+-3x-10%3D0)
![x_{1} =-2](https://tex.z-dn.net/?f=x_%7B1%7D+%3D-2)
![x_{2} =5](https://tex.z-dn.net/?f=x_%7B2%7D+%3D5)
![(x+2)(x-5)<0](https://tex.z-dn.net/?f=%28x%2B2%29%28x-5%29%3C0)
Точки
и ![-5](https://tex.z-dn.net/?f=-5)
∈![(-5;2)](https://tex.z-dn.net/?f=%28-5%3B2%29)
-----------------------------------------
![x^{2} -25>0](https://tex.z-dn.net/?f=x%5E%7B2%7D+-25%3E0)
![(x-5)(x+5)>0](https://tex.z-dn.net/?f=%28x-5%29%28x%2B5%29%3E0)
Точки
и ![5](https://tex.z-dn.net/?f=5)
x∈(-∞;-5) ∪ (5;+∞)
вроде так
Область допустимых значений уравнения : tgx≠0, х≠πn, n∈Z и х≠π/2+πk, k∈Z
![\frac{cos3x}{ \frac{sinx}{cosx} } =sin3x-2sinx, \\ cosx\cdot cos3x=sinx\cdot sin3x-2sin^{2}x, \\ cos3x\cdot cosx-sin3x\cdot sinx=-2sin^{2}x, \\ cos(3x+x)=-2sin^{2}x, \\ cos4x+2sin^{2}x=0, \\ (2cos ^{2} 2x-1)+(1-cos2x)=0, \\ 2cos ^{2} 2x-cos2x=0, \\ cos2x\cdot (2cos2x-1)=0 ](https://tex.z-dn.net/?f=+%5Cfrac%7Bcos3x%7D%7B+%5Cfrac%7Bsinx%7D%7Bcosx%7D+%7D+%3Dsin3x-2sinx%2C+%5C%5C++cosx%5Ccdot+cos3x%3Dsinx%5Ccdot+sin3x-2sin%5E%7B2%7Dx%2C+%5C%5C+cos3x%5Ccdot+cosx-sin3x%5Ccdot+sinx%3D-2sin%5E%7B2%7Dx%2C+%5C%5C+cos%283x%2Bx%29%3D-2sin%5E%7B2%7Dx%2C+%5C%5C++cos4x%2B2sin%5E%7B2%7Dx%3D0%2C+%5C%5C+%282cos+%5E%7B2%7D+2x-1%29%2B%281-cos2x%29%3D0%2C+%5C%5C+2cos+%5E%7B2%7D+2x-cos2x%3D0%2C+%5C%5C+cos2x%5Ccdot+%282cos2x-1%29%3D0%0A)
Произведение равно нулю, когда хотя один из множителей равен нулю, а другой при этом не теряет смысла.
![\left \{ {{cos2x=0} \atop {2cos2x-1=0}} \right. \Rightarrow \left \{ {{cos2x=0} \atop {cos2x= \frac{1}{2} }}\Rightarrow \left \{ {{2x= \frac{ \pi }{2} + \pi k,k\in Z} \atop {2x=\pm \frac{ \pi }{3}+2 \pi n,n\in Z }} \right. \right. \Rightarrow \left \{ {{x= \frac{ \pi }{4} + \frac{ \pi }{2}k,k\in Z } \atop {x=\pm \frac{ \pi }{6}= \pi n,n\in Z }} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bcos2x%3D0%7D+%5Catop+%7B2cos2x-1%3D0%7D%7D+%5Cright.+%5CRightarrow+%5Cleft+%5C%7B+%7B%7Bcos2x%3D0%7D+%5Catop+%7Bcos2x%3D+%5Cfrac%7B1%7D%7B2%7D+%7D%7D%5CRightarrow+%5Cleft+%5C%7B+%7B%7B2x%3D+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+%2B+%5Cpi+k%2Ck%5Cin+Z%7D+%5Catop+%7B2x%3D%5Cpm+%5Cfrac%7B+%5Cpi+%7D%7B3%7D%2B2+%5Cpi+n%2Cn%5Cin+Z+%7D%7D+%5Cright.++%5Cright.+%5CRightarrow+%5Cleft+%5C%7B+%7B%7Bx%3D+%5Cfrac%7B+%5Cpi+%7D%7B4%7D+%2B+%5Cfrac%7B+%5Cpi+%7D%7B2%7Dk%2Ck%5Cin+Z+%7D+%5Catop+%7Bx%3D%5Cpm++%5Cfrac%7B+%5Cpi+%7D%7B6%7D%3D+%5Cpi+n%2Cn%5Cin+Z+%7D%7D+%5Cright.+)
Ответ.
![\frac{ \pi }{4}+ \frac{ \pi }{2} k, \pm \frac{ \pi }{6} + \pi n,k,n\in Z](https://tex.z-dn.net/?f=+%5Cfrac%7B+%5Cpi+%7D%7B4%7D%2B+%5Cfrac%7B+%5Cpi+%7D%7B2%7D+k%2C++%5Cpm+%5Cfrac%7B+%5Cpi+%7D%7B6%7D+%2B+%5Cpi+n%2Ck%2Cn%5Cin+Z+)