`ĐKXĐ:x\geq0;x\ne0`
`(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a})(\frac{1+\sqrt{a}}{1-a})^{2}`
`=[\frac{(1+\sqrt{a})(1-\sqrt{a}+a)}{1+\sqrt{a}}-\sqrt{a}][\frac{1+\sqrt{a}}{(1+\sqrt{a})(1-\sqrt{a})}]^{2}`
`=(1-\sqrt{a}+a-\sqrt{a})(\frac{1}{1-\sqrt{a}})^{2}`
`=(1-2\sqrt{a}+a).\frac{1}{(1-\sqrt{a})^{2}}`
`=(1-\sqrt{a})^{2}.\frac{1}{(1-\sqrt{a})^{2}}`
`=1`
Vậy với `x\geq0;x\ne1` thì bth`=1`