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Giải pt
x − 3 + 5 − x = x 2 − 8 x + 18 \sqrt{x-3}+\sqrt{5-x}=x^2-8x+18 x − 3 + 5 − x = x 2 − 8 x + 1 8 3 ≤ x ≤ 5 3\le x\le 5 3 ≤ x ≤ 5 u = x − 3 ≥ 0 u=\sqrt{x-3}\ge 0 u = x − 3 ≥ 0 v = 5 − x ≥ 0 → u 2 + v 2 = 2 → ( u + v ) 2 = 2 u v + 2 v=\sqrt{5-x}\ge 0\to u^2+v^2=2\to (u+v)^2=2uv+2 v = 5 − x ≥ 0 → u 2 + v 2 = 2 → ( u + v ) 2 = 2 u v + 2 u + v = − u 2 v 2 + 3 u+v=-u^2v^2+3 u + v = − u 2 v 2 + 3 ( u + v ) = t ≥ 0 (u+v)=t\ge 0 ( u + v ) = t ≥ 0 t 4 − 4 t 2 + 4 t − 8 = 0 t^4-4t^2+4t-8=0 t 4 − 4 t 2 + 4 t − 8 = 0 → t 2 ( t 2 − 4 ) + 4 ( t − 2 ) = 0 → ( t − 2 ) ( t 3 + 2 t 2 + 4 ) = 0 → t = 2 ; t = − 2 , 59... ( l ) \to t^2(t^2-4)+4(t-2)=0\to (t-2)(t^3+2t^2+4)=0\to t=2 ; t=-2,59...(l) → t 2 ( t 2 − 4 ) + 4 ( t − 2 ) = 0 → ( t − 2 ) ( t 3 + 2 t 2 + 4 ) = 0 → t = 2 ; t = − 2 , 5 9 . . . ( l ) → x 2 − 8 x + 18 = 2 → ( x − 4 ) 2 = 0 → x = 4 \to x^2-8x+18=2\to (x-4)^2=0\to x=4 → x 2 − 8 x + 1 8 = 2 → ( x − 4 ) 2 = 0 → x = 4
Theo mình nên đặt x - 3 = u và 5 - x = vu + v = u v \sqrt{u} + \sqrt{v} = uv u + v = u v
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
