b)
log_{\frac{1}{2}x+1}4x=2
\frac{1}{2}x+1>0 \ |*2 \ => x+2>0 => x>-2
\frac{1}{2}x+1\ne 1\ |*2 \ => x+2\ne 2\ => x\ne 0
4x>0 \ => x>0
D: x\in (0;+\infty)
(\frac{1}{2}x+1)^2=4x wzór skróconego mnozenia (a+b)^2=a^2+2ab+b^2
\frac{1}{4}x^2+x+1-4x=0
\frac{1}{4}x^2+x-3x=0 \ |*4
x^2+4x-12x=0 , 4x=6x-2x
Zamieniam postac ogólną na iloczynową f(x) = a(x-x_1)(x-x_2)
x^2+6x-2x-12x=0
x(x+6)-2(x+6)=0
(x+6)(x-2)=0
x=-6 \not\in D, \ \vee x=2
x=2
d)
log_x(6x-9)=2
x>0, \ x\ne 1, \\ 6x-9>0 \\ 6x\ne 9 \\ x>1,5
D: x\in (1,5;+\infty)
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x^2=6x-9
x^2-6x+9=0
(x-3)^2=0
x=3