DK:.....+) x=−1 k là no of pt+) x≠−1pt ⇔x+4=(√x+2+1)(1+4(3√2x+3−1)2x+3−1 ⇔(x+4)(x+1)=(√x+2+1)(23√2x+3+x−1) ⇔(x+4)(√x+2+1)(√x+2−1)=(√x+2+1)(23√2x+3+x−1) ⇔(x+4)√x+2−x−4=23√2x+3+x−1 ⇔(√x+2)3+2√x+2=2x+3+23√2x+3 ⇔(√x+2−3√2x+3)(....)=0⇔√x+2=3√2x+3 Đến đây bạn giải tip nhé!!! :D
DK:.....+)
x=−1 k là no of pt+)
x≠−1pt $\Leftrightarrow x+4=(\sqrt{x+2}+1)(1+\frac{4(\sqrt[3]{2x+3}-1)}{2x+3-1}
)\Leftrightarrow (x+4)(x+1)=(\sqrt{x+2}+1)(2\sqrt[3]{2x+3}+x-1)
\Leftrightarrow (x+4)(\sqrt{x+2}+1)(\sqrt{x+2}-1)=(\sqrt{x+2}+1)(2\sqrt[3]{2x+3}+x-1)
\Leftrightarrow (x+4)\sqrt{x+2}-x-4=2\sqrt[3]{2x+3}+x-1
\Leftrightarrow (\sqrt{x+2})^{3}+2\sqrt{x+2}=2x+3+2\sqrt[3]{2x+3}
\Leftrightarrow (\sqrt{x+2}-\sqrt[3]{2x+3})(....)=0$$\Leftrightarrow \sqrt{x+2}=\sqrt[3]{2x+3}$ Đến đây bạn giải tip nhé!!! :D