Đáp án + Giải thích các bước giải:

$2x^2 - 10x + 3 = 0(a = 2; b = -10; c = 3)$

$\Delta = b^2 - 4ac = (-10)^2 - 4 \cdot 2 \cdot 3 = 76 > 0$

$\Rightarrow$ Phương trình có $2$ nghiệm $x_1$, $x_2$ phân biệt

Theo hệ thức Viet, ta có:

$\begin {cases} x_1 + x_2 = \dfrac{-b}{a} =\dfrac{10}{2} = 5 \\ x_1x_2 = \dfrac{c}{a} = \dfrac{3}{2} \end {cases}$

Ta có: $2x_1^2 - 10x_1 + 3 = 0$

$\Leftrightarrow 4x_1^2 - 20x_1 + 6 = 0$

$T = \dfrac{\sqrt{24x_1 - 5} + 2x_2 + 2026}{25 - 2x_1 - 8x_2}$

$= \dfrac{\sqrt{4x_1^2 - 20x_1 + 6 + 24x_1 - 5} + 2x_2 + 2026}{25 - 2x_1 - 2x_2 - 6x_2}$

$= \dfrac{\sqrt{4x_1^2 + 4x_1 + 1} +2x_2 + 2026}{25 - 10 - 6x_2}$

$= \dfrac{\sqrt{(2x_1 + 1)^2} + 2x_2 + 2026}{15 - 6x_2}$

$= \dfrac{|2x_1 + 1| + 2x_2 + 2026}{3(x_1 + x_2) - 6x_2}$

$= \dfrac{|2x_1 + 1| + 2x_2 + 2026}{3(x_1 - x_2)}$

Ta có: $\begin {cases} x_1 + x_2 = 5 > 0 \\ x_1x_2 = \dfrac{3}{2} > 0 \end {cases}$

$\Rightarrow x_1 > x_2 > 0$

$\Rightarrow T = \dfrac{2x_1 + 1 + 2x_2 + 2026}{3\sqrt{(x_1 - x_2)^2}}$

$= \dfrac{10 + 2027}{3\sqrt{x_1^2 - 2x_1x_2 + x_2^2}}$

$= \dfrac{2037}{3\sqrt{x_1^2 + 2x_1x_2 + x_2^2 - 4x_1x_2}}$

$= \dfrac{2037}{3\sqrt{(x_1 + x_2)^2 - 4x_1x_2}}$

$= \dfrac{2037}{3\sqrt{5^2 - 6}}$

$= \dfrac{2037}{3\sqrt{19}}$

$= \dfrac{679\sqrt{19}}{19}$