......................................................
Х²+ 3х-4 = x²+ 4x - x -4 = x(x+4) + (-x-4) =
= x(x+4) - (x+4) = (x-1)(x+4)
x²-8x+15 = x²- 5x-3x - 5*(-3) =
= (x²-3x) + (-5x - 5(-3))=
= x(x-3) - 5(x-3)= (x-5)(x-3)
x²+8x+12= x²+6x+2x+6*2 =
= (x²+6x) + (2x+6*2) =
=x(x+6) + 2(x+6) =(x+2)(x+6)
7b(2b + 3) - (b + 6)(b - 5) = 14b² + 21b - (b² - 5b + 6b - 30) = 14b² + 21b - b2 -
- b + 30 = 13b²+ 20b
Если b = 0,1 , то
13 * 0,1² + 20 * 0,1 = 0,13 + 2 = 2,13
21-12√3=(2√3-3)²
√(2√3-3)²=|2√3-3|=2√3-3
1/(√13 -3 -√(21-12√3)=1/(√13 -3-2√3+3)=1/(√13-2√3)=
=(√13+2√3)/(√13-2√3)(√13+2√3)=(√13+2√3)/(13-12)=√13+2√3
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
- + +
---------- 2
--------- 3
------------- \\\\\\\\\ \\\\\\\\\\\\\\\\
x=4 + + + | +
x=2.5 - - + | +
x= 0 - - - | -
x∈(2; 3)U(3; +∞)
Ответ: (2; 3)U(3; +∞)