Huong dan
1) (x² - 5x + 1)(x² - 4) = 6(x - 1)²
<=> [(x² - 4) - 5(x - 1)](x² - 4) - 6(x - 1)² = 0
<=> (x² - 4)² - 5(x - 1)(x² - 4) - 6(x - 1)² = 0
Nhan thay x = 1 khong phai la nghiem => x - 1 ≠ 0 nen co the chia 2 ve cua pt cho (x - 1)² ≠ 0 va dat y = (x² - 4)/(x - 1) ta co pt bac 2 theo y
y² - 5y - 6 = 0 => y = - 1; y = 6
Ban tu giai tip
2) 3√(x³ + 8) = 2x² - 6x + 4 (x ≥ - 2 )
<=> 3√[(x + 2)(x² - 2x + 4)] = 2(x² - 2x + 4) - 2(x + 2)
<=> 2(x + 2) + 3√[(x + 2)(x² - 2x + 4)] - 2(x² - 2x + 4) = 0
Chia 2 ve pt cho √(x² - 2x + 4) = √[(x - 1)² + 3]> 0 va dat y = √[(x + 2)/(x² - 2x + 4)] ta co pt bac 2 theo y:
2y² + 3y - 2 = 0 => y = 1/2 ( loai nghiem y = - 2)
Ban tu giai tiep
