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计算(1)(-12)-(-15)+(-8)-(-10)(2)(16...

(1)原式=12+(-14)+56+(-27),=12+56+(-14)+(-27),=68+(-41),=27;(2)原式=(-3.2)+8.1+(-16.9)+(-1 2 15 ),=(-3.2)+(-16.9)+8.1+(-1 2 15 ),=-12+(-1 2 15 ),=-13 2 15 ;(3)原式=3 2 5 +(-2 7 8 )+(-3 5...

①(-8)-(-1),=-8+1,=-7;②8+(-10)+(-2)-(-5),=8-10-2+5,=8+5-10-2,=1;③392324×(-12),=(40-124)×(-12),=-40×12+12,=-47912;④-24×(-12+34-13),=24×12-24×34+24×13,=12-18+8,=2;⑤8×(-45)÷|-16|,=-8×45×116,=-25;⑥-...

(1)原式=12-8+11-2-12=4+11-2-12=15-2-12=13-12=1;(2)原式=118÷13=118×3=16;(3)原式=(-1)3×(-5)÷[-9+4]=(-1)×(-5)×(-15)=5×(-15)=-1.

(1)解:原式=16×(-48)+136×(-48)-14×(-48)+112×(-48)=-8+(-43)-(-12)+(-4)=-8-43+12-4=-43;(2)解:原式=4+(-2)×(-32)+116×(-16)=4+3-1=6.

(1) 3.41÷2 15 16 ×5.875-(21 5 37 -19.18) = 341 100 × 16 47 × 47 8 -(21 5 37 -19 9 50 ),=6 41 50 +19 9 50 -21 5 37 ,=26-21 5 37 ,=4 32 37 ;(2) [(13.75-7 11 12 )×2 3 13 ]÷[(1 1 12 +12.5%)÷(2 4 7 ÷9 3 13 )] =[(13 3 4 -...

#include#includemain(){int n=0;double sum=1,k=-0.5,t=1,m;while(fabs(t)>0.0001){t=t*k;m=t;sum=sum+m;n++;}printf("n=%d ,sum=%lf\n",n,sum);}

(1)-16+25+16-15+4-10 =-16+16+25-15+4-10 =0+10-10+4 =4 (2)-5.4+0.2-0.6+0.8 =-5.4-0.6+(0.2+0.8) =-6+1 =-5

(1)原式=(-3-6-11)+(2+9)=-20+11=-9;(2)原式=16×(-24)?34×(-24)+112×(-24)=-4+18-2=14-2=12;(3)原式=-10-(-5)×12×(-20)=-10-50=-60;(4)原式=?1?12×13×(2-9)=-1+76=16;(5)原式=(?12)×(?12)×(?1100)=144×(?1100)=-1...

(1)-32-(-17)-|-23|+(-15)=-32+17-23-15=-70+17=-53; (2)-22-(-2)3×14-16×(12?34+58)=-4-(-8)×14-8+12-10=-4+2-8+12-10=-8.

(1)12-(-18)+(-7)-15=12+18-7-15=30-22=8;(2)?4÷23?(?23)×(?30)=-6-20=-6;(3)(-16+34-112)×(-48)=16×48-34×48+112×48=8-36+4=-24;(4)(-1)10×2+(-2)3÷4+(-22)=(-1)×2+(-8)÷4+(-4)=-2-2-4=-8;(5)3(x2-y2)+(y...

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