Câu a : \(4\sqrt{x+1}=x^2-5x+14\)
\(\Leftrightarrow x^2-5x+14-4\sqrt{x+1}=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left(x+1-4\sqrt{x+1}+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(\sqrt{x+1}-2\right)^2=0\end{matrix}\right.\Leftrightarrow x=3\)
Câu b : \(\left\{{}\begin{matrix}y=x^2\\z=xy\\\dfrac{1}{x}=\dfrac{1}{y}+\dfrac{6}{z}\end{matrix}\right.\) ( ĐK : \(x,y,z\ne0\)
\(\Leftrightarrow\left\{{}\begin{matrix}y=x^2\left(1\right)\\z=x^3\left(2\right)\\\dfrac{1}{x}=\dfrac{1}{x^2}+\dfrac{6}{x^3}\left(3\right)\end{matrix}\right.\)
Xét phương trình (3) :
\(\left(3\right)\Leftrightarrow x^2=x+6\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Thay từng giá trị của x vào pt (1) và (2) . Ta được những cặp nghiệm :
\(\left\{{}\begin{matrix}\left(x;y;z\right)=\left(-2;4;-8\right)\\\left(x;y;z\right)=\left(3;9;27\right)\end{matrix}\right.\)
