Question

giải pt

\(\sqrt{\frac{1}{4}x^2+x+1}-\sqrt{6-2\sqrt{5}=0}\)

Answer 1

\(\sqrt{\frac{1}{4}x^2+x+1}=\sqrt{\left(\frac{1}{2}x\right)^2+2.\frac{1}{2}x.1+1^2}=\sqrt{\left(\frac{1}{2}x+1\right)^2}=\left|\frac{1}{2}x+1\right|\)

\(\sqrt{6-2\sqrt{5}}=\sqrt{5-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)

phương trình <=> \(\left|\frac{1}{2}x+1\right|=\sqrt{5}-1\)

<=> \(\frac{1}{2}x+1=\sqrt{5}-1\) hoặc \(\frac{1}{2}x+1=-\sqrt{5}+1\)

+)  \(\frac{1}{2}x+1=\sqrt{5}-1\)<=> \(x=2\sqrt{5}+4\)

+) \(\frac{1}{2}x+1=-\sqrt{5}+1\) <=> \(x=-2\sqrt{5}\)

Vậy pt có 2 nghiệm \(x=2\sqrt{5}+4\); \(x=-2\sqrt{5}\)