Toán [toán 9] giải phương trình

L

leminhnghia1

Giải:

ĐK: $x \ge \dfrac{-3}{2}$

$\iff (x^3+3x^2+3x+1)+3(x^2+2x+1)+2(x+1)=(2x+3)\sqrt{2x+3}+3(2x+3)+2\sqrt{2x+3}$

$\iff (x+1)^3+3(x+1)^2+2(x+1)=\sqrt{2x+3}^3+3\sqrt{2x+3}^2+2\sqrt{2x+3}$

$\rightarrow x+1=\sqrt{2x+3}$

$\rightarrow x^2+2x+1=2x+3$ (ĐK: $x \ge -1$)

$\iff x^2=2$

$\rightarrow x=\sqrt{2}$
 
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