Question

Bài 2:
a) Giải phương trình \(\dfrac{x\left(3-x\right)\left(x^2+3\right)}{\left(x+1\right)^2}=2\)
b) Giải bất phương trình 2x + \(\dfrac{49}{2x+1}\le13\)

Answer 1

\(\dfrac{x\left(3-x\right)\left(x^2+3\right)}{\left(x+1\right)^2}=2\)

<=> \(\dfrac{\left(3x-x^2\right)\left(x^2+3\right)}{x^2+2x+1}-2=0\)

<=> \(\dfrac{3x^3+9x-x^4-3x^2}{x^2+2x+1}-\dfrac{2\left(x^2+2x+1\right)}{x^2+2x+1}=0\)

=> 3x³ + 9x - x⁴ - 3x² - 2x² - 4x - 2 = 0

<=> - x⁴ + 3x³ - 5x² + 5x - 2 = 0

<=> - x⁴ + x³ - 2x² + 2x³ - 2x² + 4x - x^2 + x - 2 = 0

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

<=> (x² - x + 2)(- x² + 2x - 1) = 0

<=> - (x - 1)²(x² - x + 2) = 0

<=> x = 1 (vì x² - x + 2 \(\ge\) 1,75 > 0)

Vậy S = {1}

Answer 2

b/ \(2x+\dfrac{49}{2x+1}\le13\)

\(\Leftrightarrow\dfrac{x^2-6x+9}{2x+1}\le0\)

Vì \(x^2-6x+9=\left(x-3\right)^2\ge0\) nên

\(\Leftrightarrow2x+1\le0\)

\(\Leftrightarrow x\le-\dfrac{1}{2}\)