Question

[Đinh Doãn Nam]

\(\sqrt{3x^2-6x-6}=3\sqrt{\left(2-x\right)^5}+\left(7x-19\right)\sqrt{2-x}\)

Answer 1

[Nguyen]

ĐK:\(3x^2-6x-6\ge0;\left(2-x\right)^5\ge0;x\le2\)

\(\Leftrightarrow\sqrt{3x^2-6x-6}-\sqrt{3}=3\sqrt{\left(2-x\right)^5}-27\sqrt{3}+\left(7x-19\right)\sqrt{2-x}+26\sqrt{3}\)

\(\Leftrightarrow\frac{3x^2-6x-9}{\sqrt{3x^2-6x-6}+\sqrt{3}}=3\left(\frac{\left(2-x\right)^5-243}{\sqrt{\left(2-x\right)^5}+9\sqrt{3}}\right)+\frac{\left(7x-19\right)^2\left(2-x\right)-2028}{\left(7x-19\right)\sqrt{2-x}-26\sqrt{3}}\)

\(\Leftrightarrow\left(x+1\right)\left[...\right]=0\)

Ta c/m đc [...] khác 0.

Vậy x=-1(TM)

Answer 2

[Nguyễn Việt Lâm]

ĐKXĐ: \(x\le1-\sqrt{3}\)

\(\Leftrightarrow\sqrt{3x^2-6x-6}=3\left(2-x\right)^2\sqrt{2-x}+\left(7x-19\right)\sqrt{2-x}\)

\(\Leftrightarrow\sqrt{3x^2-6x-6}=\left(3x^2-5x-7\right)\sqrt{2-x}\)

\(\Leftrightarrow\sqrt{3x^2-6x-6}=\left(3x^2-6x-6-\left(2-x\right)+1\right)\sqrt{2-x}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{3x^2-6x-6}=a\ge0\\\sqrt{2-x}=b\ge0\end{matrix}\right.\) ta được:

\(a=\left(a^2-b^2+1\right)b\Leftrightarrow b\left(a^2-b^2\right)+b-a=0\)

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

\(\Leftrightarrow\left(ab+b^2-1\right)\left(a-b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\ab=1-b^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x^2-6x-6}=\sqrt{2-x}\\\sqrt{\left(3x^2-6x-6\right)\left(2-x\right)}=1-\left(2-x\right)\end{matrix}\right.\)

TH1: \(3x^2-5x-8=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=\frac{8}{3}>1-\sqrt{3}\left(l\right)\end{matrix}\right.\)

TH2: \(\sqrt{\left(3x^2-6x-6\right)\left(2-x\right)}=x-1\)

Do \(x\le1-\sqrt{3}\Rightarrow x-1\le-\sqrt{3}\Rightarrow VP< 0\) và \(VT\ge0\Rightarrow ptvn\)

Vậy pt có nghiệm duy nhất \(x=-1\)