А)(у-4)(у+4)/10ху*5у/3(у+4)=у-4/6х
б)-а-в/а*3ав/(а-в)(а+в)=-3в/а+в
Ну как-то так
А:у=2х-1
б:у=0,5х+2
с:у=3
д:у=-х-2
№1 б)6х-(х+3)^2=4x-(x+2)^2-5
6x-x^2-9-6x=4x-x^2-4-4x-5
6x^2-x^2-9-6x-4x+x^2+4+4x+5=0
6x^2-6x=0
6x(x-1)=0
6x=0 или х-1=0
х=0 или х=1
![2*sin(x)-3=0\\\\ 2*sin(x)=3\\\\ sin(x)=\frac{3}{2}\\\\ x\not\in(-\infty;\ +\infty)](https://tex.z-dn.net/?f=2%2Asin%28x%29-3%3D0%5C%5C%5C%5C%0A2%2Asin%28x%29%3D3%5C%5C%5C%5C%0Asin%28x%29%3D%5Cfrac%7B3%7D%7B2%7D%5C%5C%5C%5C%0Ax%5Cnot%5Cin%28-%5Cinfty%3B%5C+%2B%5Cinfty%29)
решений нету, по скольку возможные значения синуса:
![-1 \leq sin(\alpha) \leq 1](https://tex.z-dn.net/?f=-1+%5Cleq+sin%28%5Calpha%29+%5Cleq+1)
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![3*cos^2(x)+1=0\\\\ 3*cos^2(x)=-1\\\\ cos^2(x)=-\frac{1}{3}\\\\ x\not\in(-\infty;\ +\infty)](https://tex.z-dn.net/?f=3%2Acos%5E2%28x%29%2B1%3D0%5C%5C%5C%5C%0A3%2Acos%5E2%28x%29%3D-1%5C%5C%5C%5C%0Acos%5E2%28x%29%3D-%5Cfrac%7B1%7D%7B3%7D%5C%5C%5C%5C%0Ax%5Cnot%5Cin%28-%5Cinfty%3B%5C+%2B%5Cinfty%29)
решений нету, по скольку возможные значения квадрата косинуса:
![0 \leq cos^2(\alpha) \leq 1](https://tex.z-dn.net/?f=0+%5Cleq+cos%5E2%28%5Calpha%29+%5Cleq+1)