1. а) 5х= п/2 + 2пk, k ∈ Z
х = п/10 + 2пk/5
б) 5х = +- п/4 + 2пk
х = +- п/20 + 2пk/5
![x^{2}+\frac{1}{x^{2} }+7(x-\frac{1}{x})+10=0\\\\x-\frac{1}{x}=m\\\\(x-\frac{1}{x})^{2}=x^{2}-2*x*\frac{1}{x}+\frac{1}{x^{2} }=x^{2}-2+\frac{1}{x^{2} }\\\\x^{2}+\frac{1}{x^{2} }=m^{2}+2\\\\m^{2}+2+7m+10=0\\\\m^{2}+7m+12=0\\\\m_{1}=-3;m_{2}=-4](https://tex.z-dn.net/?f=x%5E%7B2%7D%2B%5Cfrac%7B1%7D%7Bx%5E%7B2%7D+%7D%2B7%28x-%5Cfrac%7B1%7D%7Bx%7D%29%2B10%3D0%5C%5C%5C%5Cx-%5Cfrac%7B1%7D%7Bx%7D%3Dm%5C%5C%5C%5C%28x-%5Cfrac%7B1%7D%7Bx%7D%29%5E%7B2%7D%3Dx%5E%7B2%7D-2%2Ax%2A%5Cfrac%7B1%7D%7Bx%7D%2B%5Cfrac%7B1%7D%7Bx%5E%7B2%7D+%7D%3Dx%5E%7B2%7D-2%2B%5Cfrac%7B1%7D%7Bx%5E%7B2%7D+%7D%5C%5C%5C%5Cx%5E%7B2%7D%2B%5Cfrac%7B1%7D%7Bx%5E%7B2%7D+%7D%3Dm%5E%7B2%7D%2B2%5C%5C%5C%5Cm%5E%7B2%7D%2B2%2B7m%2B10%3D0%5C%5C%5C%5Cm%5E%7B2%7D%2B7m%2B12%3D0%5C%5C%5C%5Cm_%7B1%7D%3D-3%3Bm_%7B2%7D%3D-4)
![1)x-\frac{1}{x}=-3\\\\x^{2}-1=-3x\\\\x^{2}+3x-1=0\\\\D=3^{2}-4*(-1)=9+4=13\\\\x_{1} =\frac{-3+\sqrt{13} }{2}\\\\x_{2}=\frac{-3-\sqrt{13} }{2}](https://tex.z-dn.net/?f=1%29x-%5Cfrac%7B1%7D%7Bx%7D%3D-3%5C%5C%5C%5Cx%5E%7B2%7D-1%3D-3x%5C%5C%5C%5Cx%5E%7B2%7D%2B3x-1%3D0%5C%5C%5C%5CD%3D3%5E%7B2%7D-4%2A%28-1%29%3D9%2B4%3D13%5C%5C%5C%5Cx_%7B1%7D+%3D%5Cfrac%7B-3%2B%5Csqrt%7B13%7D+%7D%7B2%7D%5C%5C%5C%5Cx_%7B2%7D%3D%5Cfrac%7B-3-%5Csqrt%7B13%7D+%7D%7B2%7D)
![2)x-\frac{1}{x}=-4\\\\x^{2}-1=-4x\\\\x^{2}+4x-1=0\\\\D=4^{2}-4*(-1)=16+4=20=(2\sqrt{5})^{2}\\\\x_{3}=\frac{-4+2\sqrt{5} }{2}=\sqrt{5}-2\\\\x_{4}=\frac{-4-2\sqrt{5} }{2}=-\sqrt{5}-2](https://tex.z-dn.net/?f=2%29x-%5Cfrac%7B1%7D%7Bx%7D%3D-4%5C%5C%5C%5Cx%5E%7B2%7D-1%3D-4x%5C%5C%5C%5Cx%5E%7B2%7D%2B4x-1%3D0%5C%5C%5C%5CD%3D4%5E%7B2%7D-4%2A%28-1%29%3D16%2B4%3D20%3D%282%5Csqrt%7B5%7D%29%5E%7B2%7D%5C%5C%5C%5Cx_%7B3%7D%3D%5Cfrac%7B-4%2B2%5Csqrt%7B5%7D+%7D%7B2%7D%3D%5Csqrt%7B5%7D-2%5C%5C%5C%5Cx_%7B4%7D%3D%5Cfrac%7B-4-2%5Csqrt%7B5%7D+%7D%7B2%7D%3D-%5Csqrt%7B5%7D-2)
Сумма корней :
x₁ + x₂ + x₃ + x₄ = - 3 - 4 = - 7
Если в модуле число отрицательное , то мы выводя его из модуля делаем его положительным(модуль всегда положительный) , например |-4|=4,|-4,2|=4,2
|-1/6|=1\6
Если уровень жидкости в баке поднялся на 10 см, то объем увеличился на
20 × 20 ×10 (см³) = 4000 (см³)
Следовательно, объем детали 4000 см³