Question

[Chử Bảo Nhi]

Rút gọn biểu thức

1) x + 3 + \(\sqrt{x^2-6x+9}\) (x \(\le\) 3)

2) \(\sqrt{x^2+4x+4}-\sqrt{x^2}\) (-2 \(\le\) x \(\le\) 0)

3) \(\sqrt{x^{2^{ }}+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}\)

4) \(\dfrac{\sqrt{x^2-2x+1}}{x-1}\) (x > 1)

5) |x - 2| + \(\dfrac{\sqrt{x^2-4x+4}}{x-2}\) (x < 2)

6) 2x - 1 - \(\dfrac{\sqrt{x^2-10x+25}}{x-5}\)

Answer 1

[Akai Haruma]

1.

$x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{(x-3)^2}=x+3+|x-3|$

$=x+3+(3-x)=6$

2.

$\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{(x+2)^2}-\sqrt{x^2}$

$=|x+2|-|x|=x+2-(-x)=2x+2$
3.

$\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}$

$=\sqrt{(\sqrt{x^2-1}+1)^2}-\sqrt{(\sqrt{x^2-1}-1)^2}$

$=|\sqrt{x^2-1}+1|+|\sqrt{x^2-1}-1|$

$=\sqrt{x^2-1}+1+|\sqrt{x^2-1}-1|$

Answer 2

[Akai Haruma]

4.

$\frac{\sqrt{x^2-2x+1}}{x-1}=\frac{\sqrt{(x-1)^2}}{x-1}$

$=\frac{|x-1|}{x-1}=\frac{x-1}{x-1}=1$

5.

$|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=2-x+\frac{\sqrt{(x-2)^2}}{x-2}$
$=2-x+\frac{|x-2|}{x-2}|=2-x+\frac{2-x}{x-2}=2-x+(-1)=1-x$

6.

$2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\frac{\sqrt{(x-5)^2}}{x-5}$

$=2x-1-\frac{|x-5|}{x-5}$