\(a,ĐK:-9\le x\le16\\ PT\Leftrightarrow\left(\sqrt{16-x}-3\right)+\left(\sqrt{x+9}-4\right)=0\\ \Leftrightarrow\dfrac{7-x}{\sqrt{16-x}+3}+\dfrac{x-7}{\sqrt{x+9}+4}=0\\ \Leftrightarrow\left(x-7\right)\left(\dfrac{1}{\sqrt{x+9}+4}-\dfrac{1}{\sqrt{16-x}+3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=7\left(tm\right)\\\dfrac{1}{\sqrt{x+9}+4}-\dfrac{1}{\sqrt{16-x}+3}=0\end{matrix}\right.\)
Với \(x\ge-9\) thì \(\dfrac{1}{\sqrt{x+9}+4}-\dfrac{1}{\sqrt{16-x}+3}>0\)
Do đó PT có nghiệm duy nhất \(x=7\)
\(b,ĐK:-\sqrt{2}\le x\le\sqrt{2}\\ PT\Leftrightarrow\left(\sqrt{2-x^2}-1\right)+\left(\sqrt{x^2+8}-3\right)=0\\ \Leftrightarrow\dfrac{1-x^2}{\sqrt{2-x^2}+1}+\dfrac{x^2-1}{\sqrt{x^2+8}+3}=0\\ \Leftrightarrow\left(x^2-1\right)\left(\dfrac{1}{\sqrt{x^2+8}+3}-\dfrac{1}{\sqrt{2-x^2}+1}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\\dfrac{1}{\sqrt{x^2+8}+3}-\dfrac{1}{\sqrt{2-x^2}+1}=0\end{matrix}\right.\)
Với \(x\ge-\sqrt{2}\) thì \(\dfrac{1}{\sqrt{x^2+8}+3}-\dfrac{1}{\sqrt{2-x^2}+1}>0\)
Vậy pt có tập nghiệm \(x=\pm1\)
a) Đk: \(\left\{{}\begin{matrix}16-x\ge0\\x+9\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le16\\x\ge-9\end{matrix}\right.\) \(\Rightarrow x\in\left[-9;16\right]\)
Pt: \(\Rightarrow\left(\sqrt{16-x}+\sqrt{x+9}\right)^2=7^2\)
\(\Rightarrow16-x+x+9+2\sqrt{144+7x-x^2}=49\)
\(\Rightarrow\sqrt{144+7x-x^2}=12\)
\(\Rightarrow144+7x-x^2=144\)
Bạn tự tìm x nhé rồi đối chiếu đk ta đc \(x=0\) hoặc \(x=7\)
