已知方程组mx+y=3,x2+2y=5有一组实数解,求m的值和这个方程组的解
人气:258 ℃ 时间:2019-10-14 02:13:27
解答
mx+y=3
y=3-mx
代入x^2+2y=5
x^2+2(3-mx)-5=0
x^2-2mx+1=0
方程组有一组实数解,所以△=0
(-2m)^2-4*1*1=0
m^2=1
m=±1
m=1时,x=1
m=-1时,x=-1
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