f/ $\sqrt{9x^4}+3x^2\\=|3x^2|+3x^2$
Với mọi $x$ ta luôn có $x^2\ge 0$
$→3x^2\ge 0\\→|3x^2|=3x^2\\→\sqrt{9x^4}+3x^2=3x^2+3x^2=6x^2$
Vậy $\sqrt{9x^4}+3x^2=6x^2$ với mọi $x$
g/ $x-4+\sqrt{16-8x+x^2}\\=x-4+\sqrt{x^2-8x+16}\\=x-4+\sqrt{(x-4)^2}\\=x-4+|x-4|$
Vì $x>4$
$↔x-4>0\\↔|x-4|=x-4\\→x-4+\sqrt{16-8x+x^2}=x-4+x-4=2x-8$
Vậy $x-4+\sqrt{16-8x+x^2}=2x-8$ với $x>4$
h/ $\sqrt{1-4a+4a^2}-2a\\=\sqrt{4a^2-4a+1}-2a\\=\sqrt{(2a-1)^2}-2a\\=|2a-1|-2a$
$|2a-1|=\begin{cases}2a-1\,\,nếu\,\,2a-1\ge 0\,\,hay\,\,a\ge \dfrac{1}{2}\\1-2a\,\,nếu\,\,2a-1<0\,\,hay\,\,a<\dfrac{1}{2}\end{cases}$
TH1: $a\ge \dfrac{1}{2}$
$→A=2a-1-2a=-1$
TH2: $a<\dfrac{1}{2}$
$→A=1-2a-2a=1-4a$
Vậy $A=-1$ với $a\ge \dfrac{1}{2}$, $A=1-4a$ với $a<\dfrac{1}{2}$
i/ $\sqrt{4x^2-12x+9}+2x-1\\=\sqrt{(2x-3)^2}+2x-1\\=|2x-3|+2x-1$
$|2x-3|=\begin{cases}2x-3\,\,nếu\,\,2x-3\ge 0\,\,hay\,\,x\ge \dfrac{3}{2}\\3-2x\,\,nếu\,\,2x-3<0\,\,hay\,\,x<\dfrac{3}{2}\end{cases}$
TH1: $x\ge \dfrac{3}{2}$
$→B=2x-3+2x-1=4x-4$
TH2: $x<\dfrac{3}{2}$
$→B=1-2x+2x-1=0$
Vậy $B=4x-4$ với $x\ge \dfrac{3}{2}$, $B=0$ với $x<\dfrac{3}{2}$
