Toán Giải bpt: 2x^2 + 3x >= 3.căn(2x^2 + 3x + 9) + 9 19/09/2021 By Hadley Giải bpt: 2x^2 + 3x >= 3.căn(2x^2 + 3x + 9) + 9
Đặt $2x^2+3x=a$ Bất pt ⇔ $a\geq3√(a+9)+9\\⇔\frac{a-9}{3}\geq√(a+9)$ (điều kiện $a\geq9$)$\\$ $⇔\frac{a^2-18a+81}{9}\geq a+9\\⇔\frac{a^2-27a}{9}\geq 0 $ mà $a\geq9$$\\$ $⇔a\geq 27 \\ ⇔2x^2+3x \geq 27 \\ ⇔ (x+\frac{9}{2})(x-3) \geq 0 $ mà $x+\frac{9}{2} \geq x-3 $ $\\$ $⇔$ \(\left[ \begin{array}{l}x+\frac{9}{2} \leq 0 \\x-3 \geq 0 \end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x \leq -\frac{9}{2} \\x\geq 3 \end{array} \right.\)√ Trả lời
Đặt $2x^2+3x=a$
Bất pt ⇔ $a\geq3√(a+9)+9\\⇔\frac{a-9}{3}\geq√(a+9)$ (điều kiện $a\geq9$)$\\$ $⇔\frac{a^2-18a+81}{9}\geq a+9\\⇔\frac{a^2-27a}{9}\geq 0 $ mà $a\geq9$$\\$ $⇔a\geq 27 \\ ⇔2x^2+3x \geq 27 \\ ⇔ (x+\frac{9}{2})(x-3) \geq 0 $ mà $x+\frac{9}{2} \geq x-3 $ $\\$ $⇔$ \(\left[ \begin{array}{l}x+\frac{9}{2} \leq 0 \\x-3 \geq 0 \end{array} \right.\) $⇔$ \(\left[ \begin{array}{l}x \leq -\frac{9}{2} \\x\geq 3 \end{array} \right.\)√