Question

[:vvv]

Giải pt:

\(2x^2+5x-1=7\sqrt{x^3-1}\)

Em cảm ơn ạ.

Answer 1

[An Thy]

ĐKXĐ: \(x^3-1\ge0\Rightarrow\left(x-1\right)\left(x^2+x+1\right)\ge0\) 

mà \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

\(\Rightarrow x-1\ge0\Rightarrow x\ge1\)

\(2x^2+5x-1=7\sqrt{x^3-1}\Leftrightarrow2x^2+2x+2+3x-3=7\sqrt{x-1}\sqrt{x^2+x+1}\)

\(\Leftrightarrow2\left(x^2+x+1\right)+3\left(x-1\right)=7\sqrt{x-1}\sqrt{x^2+x+1}\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-1}\\b=\sqrt{x^2+x+1}\end{matrix}\right.\left(a,b\ge0\right)\)

\(\Rightarrow\) pt trở thành \(2b^2+3a^2=7ab\Rightarrow2b^2-7ab+3a^2=0\)

\(\Rightarrow2b^2-6ab-ab+3a^2=0\Rightarrow2b\left(b-3a\right)-a\left(b-3a\right)=0\)

\(\Rightarrow\left(b-3a\right)\left(2b-a\right)=0\Rightarrow\left[{}\begin{matrix}b=3a\\2b=a\end{matrix}\right.\)

\(TH_1:b=3a\Rightarrow\sqrt{x^2+x+1}=3\sqrt{x-1}\)

\(\Rightarrow x^2+x+1=9\left(x-1\right)\Rightarrow x^2-8x+10=0\)

\(\Delta=\left(-8\right)^2-4.10=24\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{8-\sqrt{24}}{2}=4-\sqrt{6}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{8+\sqrt{24}}{2}=4+\sqrt{6}\end{matrix}\right.\)

\(TH_2:2b=a\Rightarrow2\sqrt{x^2+x+1}=\sqrt{x-1}\)

\(\Rightarrow4\left(x^2+x+1\right)=x-1\Rightarrow4x^2+3x+5=0\)

mà \(4x^2+3x+5=\left(2x\right)^2+2.2x.\dfrac{3}{4}+\left(\dfrac{3}{4}\right)^2+\dfrac{71}{16}=\left(2x+\dfrac{3}{4}\right)^2+\dfrac{71}{16}>0\)

\(\Rightarrow\) loại 

Vậy pt có tập nghiệm \(S=\left\{4+\sqrt{6};4-\sqrt{6}\right\}\)