\(PT\Leftrightarrow\left(x^2+4\right)\sqrt{2x+4}+\left(x^2+4\right)=4x^2+6x\\ \Leftrightarrow\dfrac{\left(x^2+4\right)\left(2x+3\right)}{\sqrt{2x+4}-1}-2x\left(2x+3\right)=0\\ \Leftrightarrow\left(2x+3\right)\left(\dfrac{x^2+4}{\sqrt{2x+4}-1}-2x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\\dfrac{x^2+4}{\sqrt{2x+4}-1}=2x\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2x\sqrt{2x+4}-2x=x^2+4\\ \Leftrightarrow2x\sqrt{2x+4}=x^2+2x+4\\ \Leftrightarrow8x^3+16x^2=x^4+4x^3+12x^2+16x+16\\ \Leftrightarrow x^4-4x^3-4x^2+16x+16=0\\ \Leftrightarrow\left(x^2-2x-4\right)^2=0\\ \Leftrightarrow x^2-2x-4=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{matrix}\right.\)

Thử lại ta thấy \(x=-\dfrac{3}{2}\text{ không thỏa mãn; }x=1-\sqrt{5}\text{ không thỏa mãn}\)

Vậy PT có nghiệm \(x=1+\sqrt{5}\)

Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé

pt $2x^3+6x^2=x^2+3x$

giải

<=>\(2x^3+6x^2-x^2-3x=0\)

<=>\(2x^3+5x^2-3x=0\)

<=>\(x\left(2x^2+5x-3\right)\)=0

<=>\(x\left(2x^2-x+6x-3\right)=0\)

<=>\(x\left(2x-1\right)\left(x+3\right)=0\)

<=>x=0

x=\(\dfrac{1}{2}\)

x=-3

tham khảo nhé

\(2x^3+6x^2=x^2+3x\)

\(\Leftrightarrow2x^3+6x^2-x^2-3x=0\)

\(\Leftrightarrow2x^3+5x^2-3x=0\)

\(\Leftrightarrow x\left(2x^2+5x-3\right)=0\)

\(\Leftrightarrow x\left(2x^2+6x-x-3\right)=0\)

\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)

=.= hok tốt!!

a)        \(9x^2-1=\left(3x+1\right)\left(4x+1\right)\)

\(\Leftrightarrow\)\(\left(3x-1\right)\left(3x+1\right)-\left(3x+1\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left(3x-1-4x-1\right)=0\)

\(\Leftrightarrow\)\(\left(3x+1\right)\left(-x-2\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+1=0\\-x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)

Vậy...