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解三元一次方程组{x/7=y/10=z/5,2x+3y-2z=34
人气:108 ℃ 时间:2020-03-23 16:23:36
解答
用换元法最简单.
令x/7=y/10=z/5=t,则x=7t,y=10t,z=5t
代入2x+3y-2z=34
2×7t+3×10t-2×5t=34
34t=34
t=1
x=7t=7 y=10t=10 z=5t=5
x=7 y=10 z=5
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