Question

[Nguyễn Xuân An]

2(x²-3x+2)=3\(\sqrt{x^3+8}\)

Answer 1

[Edogawa Conan]

Đk: x \(\ge\)-2

Ta có: \(2\left(x^2-3x+2\right)=3\sqrt{x^3+8}\)

<=> \(2\left[x^2-2x+4-\left(x+2\right)\right]=3\sqrt{\left(x+2\right)\left(x^2-2x+4\right)}\)

Đặt \(\sqrt{x^2-2x+4}=a\)(a > 0) ; \(\sqrt{x+2}=b\)(b \(\ge\)0)

Do đó: \(2\left(a^2-b^2\right)=3ab\)

<=> \(2a^2-2b^2-3ab=0\)

<=> \(2a^2-4ab+ab-2b^2=0\)

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

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

<=> \(\orbr{\begin{cases}2a+b=0\\a-2b=0\end{cases}}\)(a + 2b = 0 (loại) vì a > 0 và b  >  = 0)

<=> a = 2b

<=> \(\sqrt{x^2-2x+4}=2\sqrt{x+2}\)

<=> \(x^2-2x+4=4x+8\)

<=> \(x^2-6x-4=0\)

\(\Delta'=\left(-3\right)^2+4=13>0\) 

=> pt có 2 nghiệm pb

\(x_1=3+\sqrt{13}\);(tm) \(x_2=3-\sqrt{13}\)(tm)

Answer 2

[Đoàn Đức Hà]

ĐK: \(x\ge-2\).

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

\(\Leftrightarrow2\left(x^2-3x+2\right)-2\left(3x+6\right)=3\sqrt{x^3+8}-3\left(2x+4\right)\)

\(\Leftrightarrow2\left(x^2-6x-4\right)=3\frac{x^3+8-\left(2x+4\right)^2}{\sqrt{x^3+8}+2x+4}\)

\(\Leftrightarrow2\left(x^2-6x-4\right)=3\frac{x^3-4x^2-16x-8}{\sqrt{x^3+8}+2x+4}\)

\(\Leftrightarrow2\left(x^2-6x-4\right)-3\frac{\left(x+2\right)\left(x^2-6x-4\right)}{\sqrt{x^3+8}+2x+4}=0\)

\(\Leftrightarrow\left(x^2-6x-4\right)\left[2-\frac{3\left(x+2\right)}{\sqrt{x^3+8}+2x+4}\right]=0\)

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

\(\Leftrightarrow x=3\pm\sqrt{13}\)(thỏa mãn)