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2x+1/8=11/8求解方程

简便运算: 0.5×3/11+1/2x8/11 解: 0.5×3/11+1/2×8/11 =0.5×(3/11)+0.5×(8/11) =0.5×(3/11+8/11) =0.5×1 =0.5

888x8.7+11.2x9.9-11.2x1.2 =888x8.7+11.2x(9.9-1.2) =888x8.7+11.2x8.7 =(888+11.2)x8.7 =899.2x8.7 =7823.04 如果是考简便元算的话,个人觉得,前面不应该是888,而应该是88.8最合适!如果是88.8的话: 88.8x8.7+11.2x9.9-11.2x1.2 =88.8x...

1/2*5=1/3(1/2-1/5) 下面以此类推 原式=1/3(1/2-1/5+1/5-1/8+......+1/98-1/101) =1/3(1/2-1/101) =33/202

=10*1/15 =2/3

=412×0.81+11×(8+1.25)+(412+125)×0.19 =412×0.81+11×8+11×1.25+412×0.19+125×0.19 =412×(0.81+0.19)+88+125×0.11+125×0.19 =412+88+125×(0.11+0.19) =500+125×0.3 =537.5

=1/3×(1/2-1/5+1/5-1/8+1/8-1/11+...+1/2009-1/2012) =1/3×(1/2-1/2012) =1/3×1005/2012 =335/2012

增广矩阵 = 4 2 -1 2 3 -1 2 10 11 3 0 8 r2+2r1 4 2 -1 2 11 3 0 14 11 3 0 8 r3-r2 4 2 -1 2 11 3 0 14 0 0 0 -6 所以 r(A)=2≠3=r(A,b) 故方程组无解.

(1)∵(2x-1)(x+3),∴2x-1=0或x+3=0,∴x1=12,x2=-3;(2)x?32+3≥x+1①1?3(x?1)<8?x②,解①得x≤1,解②得,x>-2,∴原不等式组的解为-2<x≤1.

z=3x1+12x2 约束条件2x1+2x2=8 x1,x2>=0.没有最大值,只有最小值。解答如下: 解:2x1+2x2≤11 -x1+x2≥8 它们的交点解得(-1.25,6.75) 而x1,x2都≧0 所以在约束条件下Z 有最小值。没有最大值。 Z的 其最小值数是x2=8 x1=0时 Z最小=3x1+12x2=12×8=96

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