Ta có : \(\Delta=\left[5\left(m+1\right)\right]^2-4.25m=25m^2+50m+25-100m=25m^2-50m+25=25\left(m-1\right)^2\ge0\)
Nên phương trình luôn có 2 nghiệm phân biệt theo \(x_1\) và \(x_2\)
Mặt khác :
\(\sqrt{x_1+4}+\sqrt{x_2+4}=5\)
\(\Leftrightarrow\left(\sqrt{x_1+4}+\sqrt{x_2+4}\right)^2=25\)
\(\Leftrightarrow x_1+4+2\sqrt{\left(x_1+4\right)\left(x_2+4\right)}+x_2+4=25\)
\(\Leftrightarrow\left(x_1+x_2\right)+2\sqrt{x_1x_2+4\left(x_1+x_2\right)+16}=17\) (1)
Theo định lý vi - et ta lại có :
\(\left\{{}\begin{matrix}x_1+x_2=5m+5\\x_1x_2=25m\end{matrix}\right.\)
Thay vào phương trình (1) ta được :
\(5m+5+2\sqrt{25m+4\left(5m+5\right)+16}=17\)
\(\Leftrightarrow2\sqrt{25m+20m+20+16}=12-5m\)
\(\Leftrightarrow\sqrt{45m+36}=\dfrac{12-5m}{2}\)
\(\Leftrightarrow45m+36=\dfrac{25m^2-120m+144}{4}\)
\(\Leftrightarrow180m+144=25m^2-120m+144\)
\(\Leftrightarrow180m+144-25m^2+120m-144=0\)
\(\Leftrightarrow-25m^2+300=0\)
\(\Leftrightarrow-25\left(m^2-12\right)=0\)
\(\Leftrightarrow m^2-12=0\)
\(\Leftrightarrow\left(m+\sqrt{12}\right)\left(m-\sqrt{12}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}m+\sqrt{12}=0\\m-\sqrt{12}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=-\sqrt{12}\\m=\sqrt{12}\end{matrix}\right.\)
