Question

Cho biểu thức

A= \(\dfrac{2}{\sqrt{x}+4}-\dfrac{3}{\sqrt{x}-4}-\dfrac{2\sqrt{x}+16}{16-x}\:\:\:\left(x\ge0,x\ne16\right)\) 

a) Rút gọn

b) Tìm giá trị A khi x = \(4-2\sqrt{3}\)

Answer 1

Lời giải:

a.

 \(A=\frac{2(\sqrt{x}-4)-3(\sqrt{x}+4)}{(\sqrt{x}-4)(\sqrt{x}+4)}+\frac{2\sqrt{x}+16}{(\sqrt{x}-4)(\sqrt{x}+4)}=\frac{-\sqrt{x}-20}{(\sqrt{x}-4)(\sqrt{x}+4)}+\frac{2\sqrt{x}+16}{(\sqrt{x}-4)(\sqrt{x}+4)}\\ =\frac{\sqrt{x}-4}{(\sqrt{x}-4)(\sqrt{x}+4)}=\frac{1}{\sqrt{x}+4}\)

b. Khi $x=4-2\sqrt{3}=(\sqrt{3}-1)^2\Rightarrow \sqrt{x}=\sqrt{3}-1$

$A=\frac{1}{\sqrt{3}-1+4}=\frac{1}{\sqrt{3}+3}$