Question

Giải Pt :

\(\sqrt{x^2+2x}+\sqrt{2x-1}=\sqrt{3x^2+4x+1}\)

Answer 1

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

\(x^2+4x-1+2\sqrt{2x^3+3x^2-2x}=3x^2+4x+1\)

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

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

\(2x^3+3x^2-2x=x^4+2x^2+1\)

\(x^4-2x^3-x^2+2x+1=0\)

pt đối xứng bậc 4 tự làm được chưa?

Answer 2

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

\(x^2+4x-1+2\sqrt{2x^3+3x^2-2x}=3x^2+4x+1\)

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

\(2x^3+3x^2+2x=x^4+2x^2+1\)

\(x^4-2x^3-x^2+2x+1=0\)

\(\left(x^2-x-1\right)^2=0\)

\(x^2-x-1=0\)

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

\(\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)

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

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

lên thánh nhé