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当k为何值时,方程组①kx+y+z=1②x+ky+z=k③x+y+kz=k^2(1)有无穷多组解(2)无解
人气:287 ℃ 时间:2020-05-08 20:43:45
解答
kx+y+z=1 ①x+ky+z=k ②x+y+kz=k² ③①+②+③得;(k+2)(x+y+z)=k²+k+1x+y+z=(k²+k+1)/(k+2)当k+2=0时;k=-2:(k²+k+1)/(k+2)无意义∴该方程组无解当该方程组左边系数相等,右边值也相等时该方程...
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