Giải phương trình: x³ +x³/(x-1)³ + 3x²/(x - 1) - 2 = 0 Tham khảo câu trả lời từ Trợ lý ảo Lazi To solve the equation \( x^3 + \frac{x^3}{(x-1)^3} + \frac{3x^2}{x-1} - 2 = 0 \), let's start by simplifying and analyzing the terms. First, let's rewrite the equation for clarity: \[ x^3 + \frac{x^3}{(x-1)^3} + \frac{3x^2}{x-1} - 2 = 0 \] To simplify the terms involving fractions, let's find a common denominator for the fractional terms. The common denominator is \((x-1)^3\). Rewrite each term with this common denominator: \[ x^3 + \frac{x^3}{(x-1)^3} + \frac{3x^2 \cdot (x-1)^2}{(x-1)^3} - 2 = 0 \] Now, simplify the numerator of the second fractional term: \[ \frac{3x^2 \cdot (x-1)^2}{(x-1)^3} = \frac{3x^2 \cdot (x^2 - 2x + 1)}{(x-1)^3} = \frac{3x^4 - 6x^3 + 3x^2}{(x-1)^3} \] So the equation becomes: \[ x^3 + \frac{x^3 + 3x^4 - 6x^3 + 3x^2}{(x-1)^3} - 2 = 0 \] Combine like terms in the numerator: \[ x^3 + \frac{3x^4 - 5x^3 + 3x^2}{(x-1)^3} - 2 = 0 \] Now, let's multiply through by \((x-1)^3\) to clear the fraction: \[ x^3 (x-1)^3 + 3x^4 - 5x^3 + 3x^2 - 2(x-1)^3 = 0 \] Expand \( x^3 (x-1)^3 \): \[ x^3 (x^3 - 3x^2 + 3x - 1) = x^6 - 3x^5 + 3x^4 - x^3 \] Expand \(-2(x-1)^3\): \[ -2(x^3 - 3x^2 + 3x - 1) = -2x^3 + 6x^2 - 6x + 2 \] So the equation becomes: \[ x^6 - 3x^5 + 3x^4 - x^3 + 3x^4 - 5x^3 + 3x^2 - 2x^3 + 6x^2 - 6x + 2 = 0 \] Combine like terms: \[ x^6 - 3x^5 + 6x^4 - 8x^3 + 9x^2 - 6x + 2 = 0 \] This is a polynomial equation of degree 6. Solving such high-degree polynomials analytically can be very challenging and often requires numerical methods or special techniques. However, we can check for possible rational roots using the Rational Root Theorem, which states that any rational root, \( \frac{p}{q} \), of the polynomial equation \( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 = 0 \) must be a factor of the constant term \( a_0 \) divided by a factor of the leading coefficient \( a_n \). For the polynomial \( x^6 - 3x^5 + 6x^4 - 8x^3 + 9x^2 - 6x + 2 \), the constant term \( a_0 \) is 2, and the leading coefficient \( a_n \) is 1. The possible rational roots are \( \pm 1, \pm 2 \). Let's test these possible roots: 1. \( x = 1 \): \[ 1^6 - 3(1)^5 + 6(1)^4 - 8(1)^3 + 9(1)^2 - 6(1) + 2 = 1 - 3 + 6 - 8 + 9 - 6 + 2 = 1 \neq 0 \] 2. \( x = -1 \): \[ (-1)^6 - 3(-1)^5 + 6(-1)^4 - 8(-1)^3 + 9(-1)^2 - 6(-1) + 2 = 1 + 3 + 6 + 8 + 9 + 6 + 2 = 35 \neq 0 \] 3. \( x = 2 \): \[ 2^6 - 3(2)^5 + 6(2)^4 - 8(2)^3 + 9(2)^2 - 6(2) + 2 = 64 - 96 + 96 - 64 + 36 - 12 + 2 = 26 \neq 0 \] 4. \( x = -2 \): \[ (-2)^6 - 3(-2)^5 + 6(-2)^4 - 8(-2)^3 + 9(-2)^2 - 6(-2) + 2 = 64 + 96 + 96 + 64 + 36 + 12 + 2 = 370 \neq 0 \] None of these possible rational roots satisfy the equation. Therefore, the polynomial does not have any rational roots, and we may need to resort to numerical methods or other techniques to find the roots. Given the complexity of the polynomial, it is often practical to use numerical solvers or graphing tools to approximate the roots.