Question

[Đinh Thuận]

\(6x^2+2x+\sqrt[3]{3x^2+x+4}-10=0\)

\(x+1+\sqrt{x^24x+1}=3\sqrt{x}\)

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

\(\sqrt{x+8}+\dfrac{9x}{\sqrt{x+8}}-6\sqrt{x}=0\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\Leftrightarrow6x^2+2x+8+\sqrt{3x^2+x+4}-18=0\)

\(\Leftrightarrow2\left(\sqrt[3]{3x^2+x+4}\right)^3+\sqrt[3]{3x^2+x+4}-18=0\)

=>\(3x^2+x+4=8\)

=>3x^2+x-4=0

=>x=1 hoặc x=-4/3

b: ĐKXĐ: x>0

Pt sẽ là \(x+8+9x-6\sqrt{x\left(x+8\right)}=0\)

=>\(10x+8=\sqrt{36x\left(x+8\right)}\)

=>36x^2+288x=100x^2+160x+64

=>x=1