用换元法解下列方程X平方+3X+根号X平方+3X=6 3X平方+6X—2根号X平方+2X=1
人气:148 ℃ 时间:2020-01-26 22:05:32
解答
设根号x^2+3x=y,则y>0
y^2+y=6
解得:y=2,
y=-3(舍去)
把y=2代入根号x^2+3x=y
得x^2+3x=y^2=4
解得:x1=4,x2=-1。。。谢谢还有一道呢设根号x^2+2x=y,则y>0,x^2+2x=y^2原方程可化为3(x^2+2x)-2根号x^2+2x=1即3y^2-2y=1解得:y1=-1/3(舍去),y2=1x^2+2x=y^2=1即:x^2+2x-1=0X=-1+√2或X=-1-√2 请采纳,谢谢
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