Question

[Truong Cong Vinh]

giải phương trình tìm x

(x² + x + 1)² = 5( x⁴ + x² + 1)

ai nhanh và chính xác nhất cho 1 like

Answer 1

[trần quang nhật]

(x2 + x + 1)2 = 5( x4 + x2 + 1)

<=> (x² + x + 1)² = 5 [(x⁴ + 2x² + 1) - x²]

<=> (x² + x + 1)² = 5 [(x² + 1)² - x²]

<=> (x² + x + 1)² = 5 (x² - x + 1)(x² + x + 1)

<=> (x² + x + 1)² = (5x² - 5x + 5)(x² + x + 1)

<=> (x² + x + 1)² - (5x² - 5x + 5)(x² + x + 1) = 0

<=> (x² + x + 1)(x² + x + 1 - 5x² + 5x - 5) = 0

<=> (x² + x + 1)(-4x² + 6x - 4) = 0

<=> (x² + x + 1)(x² - \(\dfrac{3}{2}\)x + 1) = 0 (chia cả hai vế cho -4)

<=> (\(x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\))(x² - 2. x. \(\dfrac{3}{4}\)+ \(\dfrac{9}{16}\)+\(\dfrac{7}{16}\)) = 0

<=> [\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)][\(\left(x-\dfrac{3}{4}\right)^2+\dfrac{7}{16}\)] = 0

Vì \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0,\forall x\); \(\left(x-\dfrac{3}{4}\right)^2+\dfrac{7}{16}>0,\forall x\)

=> [\(\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)][\(\left(x-\dfrac{3}{4}\right)^2+\dfrac{7}{16}\)] > 0, \(\forall x\)

Vậy phuong trình vô nghiệm.