Bài 4 (toán nâng cao)

H

harrypham

a) $$\begin{aligned} |5(2x+3)|+|2(2x+3)|+|2x+3|=16 & \implies 8|2x+3|=16 \\ & \implies |2x+3|=2 \\ & \implies \left[ \begin{array} 2x+3=2 \\ 2x+3=-2 \end{array} \right. \\ & \implies \left[ \begin{array} 2x=-1 \\ 2x=-5 \end{array} \right. \\ & \implies \left[ \begin{array} x= \dfrac{-1}{2} \\ x= \dfrac{-5}{2} \end{array} \right. \end{aligned}$$

b) $|x^2+|6x-2||=x^2+4 \qquad (1)$.
Do $x^2 \ge 0$ nên $$(1) \implies x^2+|6x-2|=x^2+4 \implies |6x-2|=4 \implies 2|3x-1|=4 \implies |3x-1|=2$$
$$\implies \left[ \begin{array} 3x-1=2 \\ 3x-1=-2 \end{array} \right. \implies \left[ \begin{array} 3x=3 \\ 3x=-1 \end{array} \right. \implies \left[ \begin{array} x=1 \\ x= \dfrac{-1}{3} \end{array} \right.$$
 
You must log in or register to reply here.
Top Bottom