Question

[Lê Anh Ngọc]

Giải PT sau:\(2\left(x^2+2\right)=3\left(\sqrt{x^3+8}+2x\right)\)

Answer 1

[Trần Thanh Phương]

fix lai chut...

...

Ta có : \(a=2b\Leftrightarrow\sqrt{x^2-2x+4}=2\sqrt{x+2}\)

\(\Leftrightarrow x^2-2x+4=4x+8\)

\(\Leftrightarrow x^2-6x-4=0\)

\(\Delta=6^2-4\cdot\left(-4\right)=52\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{6+\sqrt{52}}{2}=3+\sqrt{13}\\x=\frac{6-\sqrt{52}}{2}=3-\sqrt{13}\end{matrix}\right.\)

Vậy....

Answer 2

[Trần Thanh Phương]

ĐK: \(x\ge-2\)

\(2\left(x^2+2\right)=3\left(\sqrt{x^3+8}+2x\right)\)

\(\Leftrightarrow2x^2+4=3\sqrt{x^3+8}+6x\)

\(\Leftrightarrow2x^2-6x+4=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

\(\Leftrightarrow2\left(x^2-3x+2\right)=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-2x+4}=a\\\sqrt{x+2}=b\end{matrix}\right.\)( \(a,b\ge0\) )

Ta có : \(a^2-b^2=x^2-2x+4-x-2=x^2-3x+2\)

\(pt\Leftrightarrow2\left(a^2-b^2\right)=3ab\)

\(\Leftrightarrow2a^2-3ab-2b^2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}a=2b\left(chon\right)\\2a=-b\left(loai\right)\end{matrix}\right.\)

Ta có \(a=2b\Leftrightarrow\sqrt{x^2-2x+4}=2\sqrt{x+2}\)

\(\Leftrightarrow x^2-4x+4=4x+8\)

\(\Leftrightarrow x^2-8x-4=0\)

\(\Delta=8^2-4\cdot\left(-4\right)=80\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{8+\sqrt{80}}{2}\\x=\frac{8-\sqrt{80}}{2}\end{matrix}\right.\)( thỏa )

Vậy...