b)\(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\)
Đk:\(-\sqrt{10}\le x\le\sqrt{10}\)
\(pt\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}=\left(x-4\right)\left(x+3\right)\)
\(\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-\left(x-4\right)\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x+3=0\\10-x^2=x^2-8x+16\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\-2x^2+8x-6=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-3\\-\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=-3\) (thỏa)
c)\(\sqrt{\dfrac{x^3+1}{x+3}}+\sqrt{x+3}=\sqrt{x^2-x+1}+\sqrt{x+1}\)
\(\Leftrightarrow\sqrt{\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x+3}}+\sqrt{x+3}-\sqrt{x^2-x+1}-\sqrt{x+1}=0\)
Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x+1}=b;\sqrt{x+3}=c\left(a,b,c>0\right)\)
\(\Leftrightarrow\dfrac{ab}{c}+c-a-b=0\)
\(\Leftrightarrow\dfrac{\left(a-c\right)\left(b-c\right)}{c}=0\)
\(\Leftrightarrow\left(a-c\right)\left(b-c\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-c=0\\b-c=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}a=c\\b=c\end{matrix}\right.\)
*)Xét \(a=c\)\(\Rightarrow\sqrt{x^2-x+1}=\sqrt{x+3}\)
\(\Rightarrow x^2-x+1=x+3\Rightarrow x=\dfrac{2\pm\sqrt{12}}{2}\) (thỏa)
*)Xét \(b=c\)\(\Rightarrow\sqrt{x+1}=\sqrt{x+3}\)
\(\Rightarrow x+1=x+3\Rightarrow-2=0\) (loại)
a)Xem câu hỏi
Xem thêm về liên hợp ở đây