Question

[Mai Thị Thúy]

giải phương trình :

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

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

Answer 1

[Nguyễn Việt Lâm]

a.

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow x^2+4-3\sqrt{x\left(x^2+4\right)}+2x=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+4}=a\\\sqrt{x}=b\end{matrix}\right.\)

\(\Rightarrow a^2-3ab+2b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=2b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+4}=\sqrt{x}\\\sqrt{x^2+4}=2\sqrt{x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+4=x^2\left(vn\right)\\x^2+4=4x\end{matrix}\right.\)

\(\Leftrightarrow x=2\)

Answer 2

[Nguyễn Việt Lâm]

b,

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow x^2+1-\sqrt{\dfrac{x\left(x^2+1\right)}{2}}-x=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+1}=a>0\\\sqrt{\dfrac{x}{2}}=b\ge0\end{matrix}\right.\) ta được:

\(a^2-ab-2b^2=0\)

\(\Leftrightarrow\left(a+b\right)\left(a-2b\right)=0\)

\(\Leftrightarrow a-2b=0\) (do \(a+b>0\))

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

\(\Leftrightarrow x^2+1=2x\)

\(\Leftrightarrow x=1\)