√x√x−2−6√x−4x−4(x\(\ge\)0,x\(\ne\)4)
=\(\dfrac{\sqrt{x}.\left(\sqrt{x}+2\right)}{x-4}\)-\(\dfrac{6\sqrt{x}-4}{x-4}\)=\(\dfrac{x+2\sqrt{x}}{x-4}\)-\(\dfrac{6\sqrt{x}-4}{x-4}\)
=\(\dfrac{x+2\sqrt{x}-6\sqrt{x}+4}{x-4}\)=\(\dfrac{x-4\sqrt{x}+4}{x-4}\)=\(\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)(1)
b, với x=6-4\(\sqrt{2}\)=(2-\(\sqrt{2}\))^2 thay vào (1) ta được
\(\dfrac{\sqrt{\left(2-\sqrt{2}\right)}^2-2}{\sqrt{\left(2-\sqrt{2}\right)}^2+2}\)=\(\dfrac{2-\sqrt{2}-2}{2-\sqrt{2}+2}\)=\(\dfrac{-\sqrt{2}}{4-\sqrt{2}}\)=\(\dfrac{\sqrt{2}}{\sqrt{2}-4}\)
a)ĐKXĐ: x≠4;x≥0
=\(\dfrac{\sqrt{x}\cdot\left(\sqrt{x}+2\right)-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}\)
=\(\dfrac{x+2\sqrt{x}-6\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
b) thế x=\(6-4\sqrt{2}\) (thỏa mãn) vào bt ta đc:
\(\dfrac{\sqrt{6-4\sqrt{2}}-2}{\sqrt{6-4\sqrt{2}}+2}\)=\(\dfrac{\sqrt{\left(2-\sqrt{2}\right)^2}-2}{\sqrt{\left(2-\sqrt{2}\right)^2}+2}\)=\(\dfrac{-\sqrt{2}}{4-\sqrt{2}}\)=\(\dfrac{-1}{\sqrt{2}-1}\)=\(-\sqrt{2}-1\)
