-3/8x+15=5/6x+73
(-3/8-5/6)x=58
-29/24x=58
x=-48
y=5/6*(-48)+73=33
(-48;33) точка пересечения
подставим: 33-48p=0
p=33/48=11/16
Arccos(x)`=-1/(√(1-x²)
arccos²(x)`=-2*arccos(x)/√(1-x²)
1/arccos²(x)=(1`*arccos²(x)-1*arccos²(x)`)/arccos⁴(x)=
=(0-(-2*arccos(x)/√(1-x²))/arccos⁴(x)=2*arccos(x)/(arccos⁴(x)*√(1-x²)).
Воооооооооооооооооотттттттт
21.
ОДЗ: x-1>0 2x-4>0
x>1 2x>4
x>2
В итоге x∈(2; +∞)
log₂ (x-1)² - log₂ (2x-4) > log₂ 2
log₂ {(x-1)² / (2x-4)} > log₂ 2
![\frac{(x-1)^2}{2x-4} \ \textgreater \ 2 \\ \frac{(x-1)^2}{2x-4} -2\ \textgreater \ 0 \\ \frac{x^2-2x+1-2(2x-4)}{2x-4}\ \textgreater \ 0 \\ \frac{x^2-2x+1-4x+8}{2(x-2)}\ \textgreater \ 0 \\ \frac{x^2-6x+9}{2(x-2)}\ \textgreater \ 0 \\ \frac{(x-3)^2}{2(x-2)}\ \textgreater \ 0 \\ (x-3)(x-3)(x-2)\ \textgreater \ 0 \\ x=3 x=2](https://tex.z-dn.net/?f=+%5Cfrac%7B%28x-1%29%5E2%7D%7B2x-4%7D+%5C+%5Ctextgreater+%5C+2+%5C%5C+%0A+%5Cfrac%7B%28x-1%29%5E2%7D%7B2x-4%7D+-2%5C+%5Ctextgreater+%5C+0+%5C%5C+%0A+%5Cfrac%7Bx%5E2-2x%2B1-2%282x-4%29%7D%7B2x-4%7D%5C+%5Ctextgreater+%5C+0+%5C%5C+%0A+%5Cfrac%7Bx%5E2-2x%2B1-4x%2B8%7D%7B2%28x-2%29%7D%5C+%5Ctextgreater+%5C+0+%5C%5C+%0A+%5Cfrac%7Bx%5E2-6x%2B9%7D%7B2%28x-2%29%7D%5C+%5Ctextgreater+%5C+0+%5C%5C+%0A+%5Cfrac%7B%28x-3%29%5E2%7D%7B2%28x-2%29%7D%5C+%5Ctextgreater+%5C+0+%5C%5C+%0A%28x-3%29%28x-3%29%28x-2%29%5C+%5Ctextgreater+%5C+0+%5C%5C+%0Ax%3D3+++++++x%3D2++++++)
- + +
---------- 2
--------- 3
------------- \\\\\\\\\ \\\\\\\\\\\\\\\\
x=4 + + + | +
x=2.5 - - + | +
x= 0 - - - | -
x∈(2; 3)U(3; +∞)
Ответ: (2; 3)U(3; +∞)