Ta có \(a . c = - 1 < 0\) nên phương trình đã cho luôn có hai nghiệm \(x_{1} , x_{2}\) phân biệt.

Theo định lí Viète ta có: \(x_{1} + x_{2} = 1\) và \(x_{1} . x_{2} = - 1\)

Ta có:

\(P \left(\right. x_{1} \left.\right) = P \left(\right. x_{2} \left.\right)\)

\(3 x_{1} - \sqrt{33 x_{1} + 25} = 3 x_{2} - \sqrt{33 x_{2} + 25}\)

\(3 \left(\right. x_{1} - x_{2} \left.\right) - \left(\right. \sqrt{33 x_{1} + 25} - \sqrt{33 x_{2} + 25} \left.\right) = 0\)

\(3 \left(\right. x_{1} - x_{2} \left.\right) - \frac{33 \left(\right. x_{1} - x_{2} \left.\right)}{\sqrt{33 x_{1} + 25} + \sqrt{33 x_{2} + 25}} = 0\)

\(1 - \frac{11}{\sqrt{33 x_{1} + 25} + \sqrt{33 x_{2} + 25}} = 0\)

\(\sqrt{33 x_{1} + 25} + \sqrt{33 x_{2} + 25} = 11\)

\(\left(\right. \sqrt{33 x_{1} + 25} + \sqrt{33 x_{2} + 25} \left.\right)^{2} = 121\)

\(33 \left(\right. x_{1} + x_{2} \left.\right) + 50 + 2 \sqrt{\left(\right. 33 x_{1} + 25 \left.\right) \left(\right. 33 x_{2} + 25 \left.\right)} = 121\) (*)

Ta có VT(*) \(= 33.1 + 50 + 2 \sqrt{3 3^{2} x_{1} x_{2} + 33.25 \left(\right. x_{1} + x_{2} \left.\right) + 2 5^{2}}\)

\(= 83 + 2 \sqrt{- 3 3^{2} + 2 533 + 2 5^{2}}\)

\(= 83 + 2 \sqrt{361} = 83 + 83 = 121 =\) VP.

Theo Vi-et, ta có:

\(\left{\right. x_{1} + x_{2} = - \frac{b}{a} = - 2024 \\ x_{1} x_{2} = \frac{c}{a} = 2\)

Theo Vi-et, ta có:

\(\left{\right. x_{3} + x_{4} = - \frac{b}{a} = - 2025 \\ x_{3} x_{4} = \frac{c}{a} = 2\)

\(A = \left(\right. x_{1} + x_{3} \left.\right) \left(\right. x_{2} - x_{3} \left.\right) \left(\right. x_{1} + x_{4} \left.\right) \left(\right. x_{2} - x_{4} \left.\right)\)

\(= \left(\right. x_{1} + x_{3} \left.\right) \left(\right. x_{2} - x_{4} \left.\right) \left(\right. x_{2} - x_{3} \left.\right) \left(\right. x_{1} + x_{4} \left.\right)\)

\(= \left(\right. x_{1} x_{2} - x_{1} x_{4} + x_{2} x_{3} - x_{3} x_{4} \left.\right) \left(\right. x_{1} x_{2} + x_{2} x_{4} - x_{3} x_{1} - x_{3} x_{4} \left.\right)\)

\(= \left(\right. 2 - x_{1} x_{4} + x_{2} x_{3} - 2 \left.\right) \left(\right. 2 + x_{2} x_{4} - x_{3} x_{1} - 2 \left.\right)\)

\(= \left(\right. - x_{1} x_{4} + x_{2} x_{3} \left.\right) \left(\right. x_{2} x_{4} - x_{3} x_{1} \left.\right)\)

\(= - x_{1} \cdot x_{2} \cdot x_{4}^{2} + x_{1}^{2} \cdot x_{3} \cdot x_{4} + x_{2}^{2} \cdot x_{3} \cdot x_{4} - x_{2} \cdot x_{1} \cdot x_{3}^{2}\)

\(= - 2 \cdot x_{4}^{2} + 2 x_{1}^{2} + 2 x_{2}^{2} - 2 x_{3}^{2}\)

\(= 2 \left[\right. \left(\left(\right. x_{1} + x_{2} \left.\right)\right)^{2} - 2 x_{1} x_{2} \left]\right. - 2 \left[\right. \left(\left(\right. x_{3} + x_{4} \left.\right)\right)^{2} - 2 x_{3} x_{4} \left]\right.\)

\(= 2 \left[\right. \left(\left(\right. - 2024 \left.\right)\right)^{2} - 2 \cdot 2 \left]\right. - 2 \left[\right. \left(\left(\right. - 2025 \left.\right)\right)^{2} - 2 \cdot 2 \left]\right.\)

\(= 2 \cdot 202 4^{2} - 8 - 2 \cdot 202 5^{2} + 8 = 2 \left(\right. 202 4^{2} - 202 5^{2} \left.\right)\)

\(= 2 \left(\right. 2024 - 2025 \left.\right) \left(\right. 2024 + 2025 \left.\right) = - 8098\)

a) \(\Delta^{'} = m^{2} + 3 > 0\) với mọi \(m\) nên phương trình (1) luôn có hai nghiệm phân biệt.

b) Theo định lí Viète ta có: \(x_{1} + x_{2} = 2 \left(\right. m + 1 \left.\right)\).

Vì \(x_{1}\) là nghiệm của phương trình nên ta có:

\(x_{1}^{2} - 2 \left(\right. m + 1 \left.\right) x_{1} + 2 m - 2 = 0\) hay \(x_{1}^{2} + 2 m - 2 = 2 \left(\right. m + 1 \left.\right) x_{1}\).

Suy ra \(B = 2 \left(\right. m + 1 \left.\right) x_{1} + 2 \left(\right. m + 1 \left.\right) x_{2} = 2 \left(\right. m + 1 \left.\right) \left(\right. x_{1} + x_{2} \left.\right) = 4 \left(\right. m + 1 \left.\right)^{2}\).

a, Phương trình (1) luôn có hai nghiệm phân biệt và trái dấu vì  Δ = \(m^2\) + 4 > 0 và tích hai nghiệm \(x1x2=-1<0\)

b,A=0