Раскладываем по формуле разности квадратов:(а+b)(a-b)= a^2-b^2
(х^2-0,7у)(х^2+0,7у)
(х-4)/4 -8/4-2х/4 = 0
(х-4-8-2х)/4 = 0
-х-12 = 0
х=-12
2√6/(cos(π/6)cos(π/4))=2√6/(√3*√2/(2*2))=2√6/(√6/4)=2√6*4/√6=8
Ответ: {8}
1. ![\sqrt{x^2-24}=x+4\Leftrightarrow \left \{ {{x^2-24=x^2+8x+16} \atop {x+4\geq0}} \right. \left \{ {{8x=-40} \atop {x\geq-4}} \right. \left \{ {{x=-5} \atop {x\geq-4}} \right. \Rightarrow x\in\varnothing](https://tex.z-dn.net/?f=%5Csqrt%7Bx%5E2-24%7D%3Dx%2B4%5CLeftrightarrow+%5Cleft+%5C%7B+%7B%7Bx%5E2-24%3Dx%5E2%2B8x%2B16%7D+%5Catop+%7Bx%2B4%5Cgeq0%7D%7D+%5Cright.+%5Cleft+%5C%7B+%7B%7B8x%3D-40%7D+%5Catop+%7Bx%5Cgeq-4%7D%7D+%5Cright.+%5Cleft+%5C%7B+%7B%7Bx%3D-5%7D+%5Catop+%7Bx%5Cgeq-4%7D%7D+%5Cright.+%5CRightarrow+x%5Cin%5Cvarnothing)
Ответ: ∅
2. ![\sqrt{3x^2-x-6}=x\sqrt{2}\Leftrightarrow \left \{ {{3x^2-x-6=2x^2} \atop {x\sqrt{2}\geq0}} \right. \left \{ {{x^2-x-6=0} \atop {x\geq0} \right. \left \{ {{x=-2;3} \atop {x\geq0}} \right. \Rightarrow x=3](https://tex.z-dn.net/?f=%5Csqrt%7B3x%5E2-x-6%7D%3Dx%5Csqrt%7B2%7D%5CLeftrightarrow+%5Cleft+%5C%7B+%7B%7B3x%5E2-x-6%3D2x%5E2%7D+%5Catop+%7Bx%5Csqrt%7B2%7D%5Cgeq0%7D%7D+%5Cright.+%5Cleft+%5C%7B+%7B%7Bx%5E2-x-6%3D0%7D+%5Catop+%7Bx%5Cgeq0%7D+%5Cright.+%5Cleft+%5C%7B+%7B%7Bx%3D-2%3B3%7D+%5Catop+%7Bx%5Cgeq0%7D%7D+%5Cright.+%5CRightarrow+x%3D3)
x² - x - 6 = 0
По теореме Виета x₁ + x₂ = 1, x₁x₂ = -6 ⇒ x = -2; 3
Ответ: 3
3. ОДЗ: ![\left \{ {{2x-5\geq0} \atop {x-3\geq0}} \right. \left \{ {{x\geq2,5} \atop {x\geq3}} \right. \Rightarrow x\geq3](https://tex.z-dn.net/?f=%5Cleft+%5C%7B+%7B%7B2x-5%5Cgeq0%7D+%5Catop+%7Bx-3%5Cgeq0%7D%7D+%5Cright.+%5Cleft+%5C%7B+%7B%7Bx%5Cgeq2%2C5%7D+%5Catop+%7Bx%5Cgeq3%7D%7D+%5Cright.+%5CRightarrow+x%5Cgeq3)
![2x-5=1+2\sqrt{x-3}+x-3\\2\sqrt{x-3}=x-3\\t=\sqrt{x-3}\\t^2-2t=0\\t(t-2)=0\\t=0; t=2\\\sqrt{x-3}=0; \sqrt{x-3}=2\\x-3=0; x-3=4\\x=3;7](https://tex.z-dn.net/?f=2x-5%3D1%2B2%5Csqrt%7Bx-3%7D%2Bx-3%5C%5C2%5Csqrt%7Bx-3%7D%3Dx-3%5C%5Ct%3D%5Csqrt%7Bx-3%7D%5C%5Ct%5E2-2t%3D0%5C%5Ct%28t-2%29%3D0%5C%5Ct%3D0%3B+t%3D2%5C%5C%5Csqrt%7Bx-3%7D%3D0%3B+%5Csqrt%7Bx-3%7D%3D2%5C%5Cx-3%3D0%3B+x-3%3D4%5C%5Cx%3D3%3B7)
Ответ: 3; 7
<span>sinAsin(A+B) + cosAcos(A+B)=cos(A-A-B)=cos(-B)=cosB</span>
<span>cos(A+B) + cos(A-B)=cosAcosB - sinAsinB + cosAcosB + sinAsinB = 2cosAcosB </span>