解三元一次方程组:①0=4a-2b+c
②0=16a+4b+c
③3=c
人气:220 ℃ 时间:2020-03-31 01:49:04
解答
2a+b+c=0①(两边同时×2)得(4)4a+2b+c=5②4a-2b+c=-1③(2)(3)相加8a+2c=4(4)4a+2b+2c=0与(2)相减(2)4a+2b+c=5c=-58a+2*(-5)=48a=14a=14/8=7/4把a=7/4;c=-5代入到(1)b=5-2*7/4=5-7/2=3/2是否...
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