Chủ đề này có 1 trả lời

[ShadowNguyen]

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

[DOTOANNANG]

Sau khi xét mẫu thức$:$ $\it{1}- \sqrt{\it{2}\,\it{(}\,\,\it{x}^{\,\it{2}}- \it{x}+ \it{1}\,\,\it{)}}\neq \it{0}$$,$ ta được$:$

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

$$\Leftrightarrow \frac{\it{x}^{\,\it{2}}- \it{3}\,\it{x}+ \it{1}}{\it{1}- \it{x}+ \sqrt{\it{x}}}= \frac{\it{2}\it{(}\,\,\it{x}^{\,\it{2}}- \it{3}\,\it{x}+ \it{1}\,\,\it{)}}{\it{2}- \it{2}\,\it{x}+ \sqrt{\it{2}\,\it{(}\,\,\it{x}^{\,\it{2}}- \it{x}+ \it{1}\,\,\it{)}}}$$

Ta sẽ chứng minh nữa$:$

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

$$\Leftrightarrow \it{2}\,\sqrt{\it{x}}= \sqrt{\it{2}\it{(}\,\,\it{x}^{\,\it{2}}- \it{x}+ \it{1}\,\,\it{)}}\Leftrightarrow \it{x}^{\,\it{2}}- \it{3}\,\it{x}+ \it{1}= \it{0}$$

Kết hợp $\it{x}> \it{0}$ nên $\it{x}= \frac{\it{3}}{\it{2}}- \frac{\sqrt{\,\it{5}}}{\it{2}}$$.$