Question

tìm x, biết:

\(\sqrt{x^2+x+\dfrac{5}{4}}\) + \(\sqrt{x^2+x+\dfrac{17}{4}}=3\)

Answer 1

<=>

\(\sqrt{x^2+x+\dfrac{5}{4}}=3+\sqrt{x^2+x+\dfrac{17}{4}}\)

<=>

\(x^2+x+\dfrac{5}{4}=3^2+x^2+x+\dfrac{17}{4}\)

<=>

\(x^2+x+\dfrac{5}{4}-9-x^2-x-\dfrac{17}{4}=0\)

<=>

\(\dfrac{5}{4}-9-\dfrac{17}{4}=-12.khác.0\)

=> pt vô nghiệm

Answer 2

Ta có : \(\sqrt{x^2+x+\dfrac{5}{4}}=\sqrt{\left(x+\dfrac{1}{2}\right)^2+1}\ge1\)(1) 

\(\sqrt{x^2+x+\dfrac{17}{4}}=\sqrt{\left(x+\dfrac{1}{2}\right)^2+4}\ge2\) (2) 

Từ (1) ; (2) => VT \(\ge3=VP\)

Dấu "=" xảy ra <=> \(x+\dfrac{1}{2}=0\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy tập nghiệm phương trình \(S=\left\{-\dfrac{1}{2}\right\}\)