Question

giải phương trình

\(x+\sqrt{17-x^2}+x\cdot\sqrt{17-x^2}=9\)

Answer 1

Đk:\(-\sqrt{17}\le x\le\sqrt{17}\)

Đặt \(t=x+\sqrt{17-x^2}\left(t>0\right)\)

\(\Rightarrow t^2=17+2x\sqrt{17-x^2}\)

\(\Rightarrow x\sqrt{17-x^2}=\frac{t^2-17}{2}\)

thay vào pt 

\(t+\frac{t^2-17}{2}=9\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}t=-7\left(loai\right)\\t=5\left(tm\right)\end{array}\right.\)

\(\Rightarrow x+\sqrt{17-x^2}=5\)

\(\Leftrightarrow\sqrt{17-x^2}=5-x\)

Với \(x< \sqrt{17}\) bình 2 vế ta có:

\(17-x^2=x^2-10x+25\)

\(\Leftrightarrow2x^2-10x+8=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=4\end{cases}\left(tm\right)}\)

 

Answer 2

dòng cuối là \(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=4\end{array}\right.\)(thỏa mãn)