Question

GPT

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

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

\(x^{4n}+\sqrt{x^{2n}+2022}=2022\)

Answer 1

a.

ĐKXĐ: \(x\ge-\dfrac{1}{3}\)

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

Đặt \(\left\{{}\begin{matrix}x+1=a\\\sqrt{3x+1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a^3+a=b^3+b\)

\(\Leftrightarrow a^3-b^3+a-b=0\)

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

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{3x+1}=x+1\)

\(\Leftrightarrow3x+1=x^2+2x+1\)

\(\Leftrightarrow x^2-x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Answer 2

b.

ĐKXĐ: \(\left[{}\begin{matrix}x\le-1\\0< x\le1\end{matrix}\right.\)

- Với \(x\le-1\Rightarrow\left\{{}\begin{matrix}VT=\left(x^2+x\right)^2+\left(x-1\right)^2>0\\VP=x\left(x^2+1\right)\sqrt{\dfrac{1-x^2}{x}}< 0\end{matrix}\right.\)

Phương trình vô nghiệm

- Với \(0< x\le1\) pt tương đương:

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

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

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

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

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

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

\(\Leftrightarrow a=2b\)

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

\(\Leftrightarrow x^4+2x^2+1=4x\left(1-x^2\right)\)

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

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

\(\Rightarrow\left[{}\begin{matrix}x=-1-\sqrt{2}\left(loại\right)\\x=-1+\sqrt{2}\end{matrix}\right.\)

Answer 3

c.

Đặt \(\sqrt{x^2+2022}=a>0\)

\(\Rightarrow2022=a^2-x^{2n}\)

Phương trình trở thành:

\(x^{4n}+a=a^2-x^{2n}\)

\(\Leftrightarrow x^{4n}-a^2+x^{2n}+a=0\)

\(\Leftrightarrow\left(x^{2n}+a\right)\left(x^{2n}-a\right)+x^{2n}+a=0\)

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

\(\Leftrightarrow x^{2n}-a+1=0\)

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

\(\Leftrightarrow x^{2n}+1=\sqrt{x^{2n}+2022}\)

\(\Leftrightarrow x^{4n}+2x^{2n}+1=x^{2n}+2022\)

\(\Leftrightarrow x^{4n}+x^{2n}-2021=0\)

\(\Rightarrow\left[{}\begin{matrix}x^{2n}=\dfrac{-1-7\sqrt{165}}{2}\left(loại\right)\\x^{2n}=\dfrac{-1+7\sqrt{165}}{2}\end{matrix}\right.\)

\(\Rightarrow x=\pm\sqrt[2n]{\dfrac{-1+7\sqrt{165}}{2}}\)