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NT
30 tháng 11 2023 lúc 18:20

ĐKXĐ: \(5x^2+2x-3>=0\)

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

=>\(\left(x+1\right)\left(5x-3\right)>=0\)

TH1: \(\left\{{}\begin{matrix}x+1>=0\\5x-3>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=-1\\x>=\dfrac{3}{5}\end{matrix}\right.\)

=>\(x>=\dfrac{3}{5}\)

TH2: \(\left\{{}\begin{matrix}x+1< =0\\5x-3< =0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =-1\\x< =\dfrac{3}{5}\end{matrix}\right.\)

=>\(x< =-1\)

\(\left(x+1\right)\cdot\sqrt{5x^2+2x-3}=5x^2+4x-5\)

=>\(\left(x+1\right)\sqrt{5x^2+2x-3}=5x^2+2x-3+2x-2\)

=>\(\left(x+1\right)\sqrt{5x^2+2x-3}-\left(5x^2+2x-3\right)-\left(2x-2\right)=0\)

=>\(\sqrt{5x^2+2x-3}\left(x+1-\sqrt{5x^2+2x-3}\right)-2\left(x-1\right)=0\)

=>\(\sqrt{5x^2+2x-3}\cdot\dfrac{\left(x+1\right)^2-\left(5x^2+2x-3\right)}{x+1+\sqrt{5x^2+2x-3}}-2\left(x-1\right)=0\)

=>\(\sqrt{5x^2+2x-3}\cdot\dfrac{x^2+2x+1-5x^2-2x+3}{x+1+\sqrt{5x^2+2x-3}}-2\left(x-1\right)=0\)

=>\(\dfrac{\sqrt{5x^2+2x-3}}{x+1+\sqrt{5x^2-2x+3}}\cdot\left(-4x^2+4\right)-2\left(x-1\right)=0\)

=>\(\dfrac{2\sqrt{5x^2+2x-3}}{x+1+\sqrt{5x^2-2x+3}}\cdot\left(x^2-1\right)+\left(x-1\right)=0\)

=>\(\dfrac{2\sqrt{5x^2+2x-3}\cdot\left(x+1\right)\left(x-1\right)}{x+1+\sqrt{5x^2-2x+3}}+\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(\dfrac{2\sqrt{5x^2+2x-3}\cdot\left(x+1\right)}{x+1+\sqrt{5x^2-2x+3}}+1\right)=0\)

=>x-1=0

=>x=1(nhận)

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