Question

[Nguyễn Minh Tuấn]

\(^{x^4+\sqrt{x^2+2002}=2002}\)

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

Answer 1

[alibaba nguyễn]

\(x^4+\sqrt{x^2+2002}=2002\)

Đặt \(\sqrt{x^2+2002}=a^2>0\)

\(\Rightarrow\hept{\begin{cases}x^4+a^2=2002\left(1\right)\\a^4-x^2=2002\left(2\right)\end{cases}}\)

Lấy (1) - (2) ta được

\(x^4-a^4+x^2+a^2=0\)

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

\(\Leftrightarrow x^2+1=a^2=\sqrt{x^2+2002}\)

\(\Leftrightarrow x^4+2x^2+1=x^2+2002\)

\(\Leftrightarrow x^4+x^2-2001=0\)

Tới đây thì đơn giản rồi

Answer 2

[alibaba nguyễn]

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

\(\Leftrightarrow\left(x^2+3x+1\right)^2=\left(x+3\right)^2\left(x^2+1\right)\)

\(\Leftrightarrow x^2=8\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{8}\\x=-\sqrt{8}\end{cases}}\)