Giải phương trình: x + x / (căn x^2 - 1) = 35/12

Đề bài: Giải phương trình: x + x / (căn x^2 - 1) = 35/12

Trả lời

Hướng dẫn giải:

ĐKXĐ: x > 1 hoặc x < −1

Từ pt $\Rightarrow$ x > 0. Kết hợp vs ĐKXĐ, ta có: x > 1

Pt $\iff {\text{x}}^{\text{2}}\text{ + }\frac{{\text{x}}^{\text{2}}}{{\text{x}}^{\text{2}}\text{ - 1}}\text{ + }\frac{{\text{2x}}^{\text{2}}}{\sqrt{{\text{x}}^{\text{2}}\text{ - 1}}}\text{ = }\frac{\text{1225}}{\text{144}}$

$\iff {\text{x}}^{\text{2}}\text{ + }\frac{{\text{x}}^{\text{2}}}{{\text{x}}^{\text{2}}\text{ - 1}}\text{ + }\frac{{\text{2x}}^{\text{2}}}{\sqrt{{\text{x}}^{\text{2}}\text{ - 1}}}\text{ = }\frac{\text{1225}}{\text{144}}$

Đặt $\frac{{\text{x}}^{\text{2}}}{\sqrt{{\text{x}}^{\text{2}}\text{ - 1}}}\text{ = y }(y>0)$

Pt trở thành: ${\text{y}}^{\text{2}}\text{ + 2y = }\frac{\text{1225}}{\text{144}}\iff \text{y = }\frac{\text{25}}{\text{12}}$ (nhận) hoặc $\text{y = }\frac{\text{- 49}}{\text{12}}$ (loại)

$\text{y = }\frac{\text{25}}{\text{12}}$ hay $\frac{{\text{x}}^{\text{2}}}{\sqrt{{\text{x}}^{\text{2}}\text{ - 1}}}\text{ = }\frac{\text{25}}{\text{12}}$

$$\begin{gathered} \iff \frac{{\text{x}}^{\text{4}}}{{\text{x}}^{\text{2}}\text{ - 1}}\text{ = }\frac{\text{1225}}{\text{144}} \\ \iff 1{\text{44x}}^{\text{4}}{\text{ - 625x}}^{\text{2}}\text{ + 625 = 0} \end{gathered}$$

$\iff \text{x = }\frac{\text{5}}{\text{3}}$ hoặc $\text{x = }\frac{\text{5}}{\text{4}}$ (vì x>1x>1)