Ta có:
\(\frac{3(x^2+5x-1)}{x^2+x-1}+23=\frac{24(x^2+3x-1)}{x^2+2x-1};(0< x<2)\\\Leftrightarrow \frac{3\left(x+5-\frac{1}{x}\right)}{x+1-\frac{1}{x}}-\frac{24\left(x+3-\frac{1}{x}\right)}{x+2-\frac{1}{x}}=-23\text{ (1)}\)
Đặt \(x+1-\frac{1}{x}=t\), khi đó phươn trình (1) trở thành:
\(\frac{3(t+4)}{t}-\frac{24(t+2)}{t+1}=-23\) (ĐKXĐ: \(t\ne0;t\ne-1\))
\( \Leftrightarrow \frac{3(t+4)(t+1)-24t(t+2)}{t(t+1)}=-23\\\Rightarrow 3(t^2+5t+4)-24t^2-48t=-23t^2-23t\\\Leftrightarrow -21t^2-33t+12=-23t^2-23t\\\Leftrightarrow 2t^2-10t+12=0\\\Leftrightarrow t^2-5t+6=0\Leftrightarrow \left[\begin{array}{} t=3(tm)\\ t=2(tm) \end{array} \right. \\\Rightarrow \left[\begin{array}{} x+1-\frac{1}{x}=3\\ x+1-\frac{1}{x}=2 \end{array} \right. \Leftrightarrow \left[\begin{array}{} x^2+x-1=3x\\ x^2+x-1=2x \end{array} \right. (\text{vì } x\ne0)\\\Leftrightarrow \left[\begin{array}{} x^2-2x-1=0\\ x^2-x-1=0 \end{array} \right.\)
\(\Leftrightarrow \left[\begin{array}{} x=1\pm\sqrt2\\ x=\frac{1\pm\sqrt5}{2} \end{array} \right.\)
Kết hợp với ĐK 0 < x < 2, ta được $x=\frac{1+\sqrt5}{2}$
Khi đó:
\(\begin{cases} x^2-x-2=\left(\frac{1+\sqrt5}{2}\right)^2-\frac{1+\sqrt5}{2}-2=\frac{1+2\sqrt5+5-2-2\sqrt5-8}{4}=-1\\ x^2-x=-1+2=1 \end{cases} \) (2)
Thay (2) vào $T$, ta được:
$T=(-1)^{2024}+\frac{1}{1^{2023}}=1+1=2$
Vậy $T=2$
$\text{#}Toru$
