Giải phương trình: x+ x/ căn bậc hai x^2-1= 35/ 12 c

Trả lời

ĐKXĐ: x > 1 hoặc x < −1

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

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

$\iff \frac{{\text{x}}^{\text{4}}}{{\text{x}}^{\text{2}}\text{ - 1}}\text{ + }\frac{{\text{2x}}^{\text{2}}}{\sqrt{{\text{x}}^{\text{2}}\text{ - 1}}}\text{ = }\frac{\text{1225}}{\text{144}}$

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

$\iff \frac{{\text{x}}^{\text{4}}}{{\text{x}}^{\text{2}}\text{ - 1}}\text{ = }\frac{\text{1225}}{\text{144}}$

$\iff 1{\text{44x}}^{\text{4}}{\text{ - 625x}}^{\text{2}}\text{ + 625 = 0}$

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