Question

Giải phương trình

\(x=\sqrt{2-x}\sqrt{3-x}+\sqrt{3-x}\sqrt{5-x}+\sqrt{5-x}\sqrt{2-x}\)

Answer 1

điệu kiện \(\hept{\begin{cases}x\ge0\\2-x\ge0;3-x\ge0;5-x\ge0\end{cases}< =>0\le x\le2;}\)

ta có 2x = \(2\sqrt{2-x}\sqrt{3-x}+2\sqrt{3-x}\sqrt{5-x}+2\sqrt{5-x}\sqrt{2-x}\)

<=> 2x = \(\sqrt{2-x}\left(\sqrt{3-x}+\sqrt{5-x}\right)+\sqrt{3-x}\left(\sqrt{5-x}+\sqrt{2-x}\right)\)+\(\sqrt{5-x}\left(\sqrt{2-x}+\sqrt{3-x}\right)\)

<=> 2x = \(\sqrt{2-x}\left(x-\sqrt{2-x}\right)+\sqrt{3-x}\left(x-\sqrt{3-x}\right)+\sqrt{5-x}\left(x-\sqrt{5-x}\right)\)

<=> 2x = x (\(\sqrt{2-x}+\sqrt{3-x}+\sqrt{5-x}\)) - (2-x +3-x + 5-x) 

<=> 2x= x.x - 10 +3x <=> x²+x-10 = 0 <=> \(\orbr{\begin{cases}x=\frac{-1+\sqrt{41}}{2}\left(loai\right)\\x=\frac{-1-\sqrt{41}}{2}\left(loai\right)\end{cases}}\) cả 2 nghiệm đều không thỏa mãn \(0\le x\le2\)

=> phương trình vô nghiệm

Answer 2

ò khó quá vì mk mới hc lp 5 à