and it is provided that x2−3x+2&≠0 x≠1,2 Now solve the above equation. yx2−(3y+1)y+(2y+P)‌‌=0‌‌‌‌‌∵x∈R,‌ so ‌D≥0 (3y+1)2−4y(2y+P)≥0 y2+(6−4P)y+1≥0‌‌‌‌‌‌‌‌∵y∈R,soD≤0 (6−4P)2−4≤0 (4P−6+2)(4P−6−2)≤0 (P−1)(P−2)≤0 P∈[1,2]