Question

[Nguyễn Thị Thùy Dung]

Giải pt: a)\(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)

b)\(\sqrt[3]{2-x}+\sqrt{x-1}=1\)

c)\(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{\left(x-1\right)^2}+\sqrt[3]{x^2-1}=1\)

Answer 1

[Nguyễn Việt Lâm]

a/ ĐKXĐ: \(\left|x\right|\ge1\)

- Với \(x\le-1\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2+6}>0\\x-2\sqrt{x^2-1}< 0\end{matrix}\right.\) \(\Rightarrow\) pt vô nghiệm

- Với \(x>1\) ta luôn có \(\sqrt{x^2+6}>x\) (dễ dàng chứng minh bằng cách bình phương 2 vế)

Mà \(x>x-2\sqrt{x^2-1}\Rightarrow\sqrt{x^2+6}>x-2\sqrt{x^2-1}\)

Phương trình vô nghiệm

Bạn có nhầm đề ko?

b/ ĐKXĐ: \(x\ge1\)

\(\sqrt[3]{2-x}+\sqrt{x-1}=1\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2-x}=a\\\sqrt{x-1}=b\end{matrix}\right.\) ta có hệ:

\(\left\{{}\begin{matrix}a+b=1\\a^3+b^2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=1-a\\a^3+b^2=1\end{matrix}\right.\) \(\Rightarrow a^3+\left(1-a\right)^2=1\)

\(\Leftrightarrow a^3+a^2-2a=0\) \(\Leftrightarrow a\left(a-1\right)\left(a+2\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt[3]{2-x}=0\\\sqrt[3]{2-x}=1\\\sqrt[3]{2-x}=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=10\end{matrix}\right.\)

Answer 2

[Nguyễn Việt Lâm]

c/

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{x+1}=a\\\sqrt[3]{x-1}=b\end{matrix}\right.\) ta có hệ:

\(\left\{{}\begin{matrix}a^3-b^3=2\\a^2+b^2+ab=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)\left(a^2+ab+b^2\right)=2\\a^2+b^2+ab=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a-b=2\Rightarrow a=b+2\\a^2+b^2+ab=1\end{matrix}\right.\) \(\Rightarrow\left(b+2\right)^2+b^2+\left(b+2\right)b-1=0\)

\(\Leftrightarrow3b^2+6b+3=0\Rightarrow3\left(b+1\right)^2=0\Rightarrow b=-1\)

\(\Rightarrow\sqrt[3]{x-1}=-1\Rightarrow x=0\)