Question

[Thái Thị Bích Thủy]

Giúp em cậu C ở dưới với ạ :(((

Answer 1

[Nguyễn Việt Lâm]

c.2

ĐKXĐ: \(x\ge0\)

\(2x^2+5x+4=6\sqrt{2x^3+4x}\)

\(\Leftrightarrow2\left(x^2+2\right)+5x=6\sqrt{2x\left(x^2+2\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{2x}=a\ge0\\\sqrt{x^2+2}=b>0\end{matrix}\right.\)

\(\Rightarrow2b^2+\dfrac{5a^2}{2}=6ab\)

\(\Leftrightarrow5a^2-12ab+4b^2=0\)

\(\Leftrightarrow\left(5a-2b\right)\left(a-2b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5a=2b\\a=2b\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}5\sqrt{2x}=2\sqrt{x^2+2}\\\sqrt{2x}=2\sqrt{x^2+2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}50x=4\left(x^2+2\right)\\2x=4\left(x^2+2\right)\end{matrix}\right.\) (chuyển vế bấm máy)

Answer 2

[Nguyễn Việt Lâm]

ĐKXĐ: \(x\ge-3\)

\(\sqrt{x^2-2x+3}=\sqrt{x+3}\)

\(\Leftrightarrow x^2-2x+3=x+3\)

\(\Leftrightarrow x^2-3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (thỏa mãn)