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对于形如y=(x2+b1x+c1)~(1/2)±(x2+b2+c2)~(1/2)x的函数,可以联想直角坐标系内两点间距离公式,利用三角形三边长的关系来求最小(大)值.例如,为求函数y=(x2-2x+2)~(1/2)+(x2-12x+40)~(1/2)的最小值,先配方成y=((x-1)2+(0-1)2)~(1/2)+((x-6)2+(0-2)2)~(1/2),再设定点A(1,1),B(6,2),A′(1,-1)及x轴上动点P(x,0),那么y=︱PA︱+︱PB︱;因为︱PA︱+︱PB︱=︱PA′︱+︱PB︱≥︱A′B︱,所以当点P恰和A′B与x轴交点Q重合时,︱PA︱+︱PB︱最小等于︱A′B︱,即
For functions of the form y = (x2 + b1x + c1) ~ (1/2) ± (x2 + b2 + c2) ~ (1/2) x, one can think of the distance between two points in a Cartesian coordinate system, For example, to find the minimum value of the function y = (x2-2x + 2) ~ (1/2) + (x2-12x + 40) ~ (1/2) , The first formulation is y = ((x-1) 2+ (0-1) 2) ~ (1/2) + ((x-6) 2+ (0-2) 2) Then set the points A (1,1), B (6,2), A ’(1, -1) and the x-axis upward motion point P (x, 0) Because ︱ PA ︱ + ︱ PB ︱ = ︱ PA ’︱ + ︱ PB ︱ ︱ A’B ︱, so when the point P and A’B and the intersection of the x-axis Q coincidence, ︱ PA ︱ + ︱ PB ︱ minimum equal to ︱ A’B︱, that is