Question

[Hoàng Linh Chi]

Giải các phương trình sau:

a) \(\sqrt{2\left(x^4+4\right)}=3x^2-10x+6\)

b) \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)

d) \(x+\frac{3x}{\sqrt{x^2-9}}=6\sqrt{2}\)

Answer 1

[Nguyễn Việt Lâm]

a/ ĐK: \(3x^2-10x+6\ge0\)

Nhận thấy \(x=0\) không phải nghiệm

\(\Leftrightarrow2\left(x^2+4\right)=\left(3x^2-10x+6\right)^2\)

\(\Leftrightarrow2\left(x^2+\frac{4}{x^2}\right)=\left(3x-10+\frac{6}{x}\right)^2=\left(3\left(x+\frac{2}{x}\right)-10\right)^2\)

Đặt \(x+\frac{2}{x}=a\Rightarrow x^2+\frac{4}{x^2}=a^2-4\)

\(\Leftrightarrow2\left(a^2-4\right)=\left(3a-10\right)^2\)

\(\Leftrightarrow7a^2-60a+108=0\Rightarrow\left[{}\begin{matrix}a=6\\a=\frac{18}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+\frac{2}{x}=6\\x+\frac{2}{x}=\frac{18}{7}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+2=0\\7x^2-18x+14=0\left(vn\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3+\sqrt{7}\\x=3-\sqrt{7}\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

b/ \(x\ge-\frac{1}{4}\)

Đặt \(\sqrt{x+\frac{1}{4}}=a\ge0\Rightarrow x=a^2-\frac{1}{4}\)

\(\Leftrightarrow a^2-\frac{1}{4}+\sqrt{a^2-\frac{1}{4}+\frac{1}{2}+a}=2\)

\(\Leftrightarrow a^2-\frac{1}{4}+\sqrt{\frac{1}{4}\left(4a^2+4a+1\right)}=2\)

\(\Leftrightarrow a^2-\frac{1}{4}+\frac{1}{2}\left(2a+1\right)=2\)

\(\Leftrightarrow4a^2+4a-7=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{-1+2\sqrt{2}}{2}\\a=\frac{-1-2\sqrt{2}}{2}< 0\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x+\frac{1}{4}}=\frac{-1+2\sqrt{2}}{2}\Rightarrow x=2-\sqrt{2}\)