Question

初一数学题、求解。

求代数式3b-2c-[-4a+(c+3b)]+c的值,其中a=1/4,b=1/5,c=1/2.

Answer 1

3b-2c-[-4a+(c+3b)]+c
=3b-2c-(-4a+c+3b)+c
=3b-2c+4a-c-3b+c
=4a-2c
将a=1/4. c=1/2代入、得、
4a-2c=4×1/4-2×1/2
=1-1
=0

Answer 2

3b-2c-[-4a+(c+3b)]+c
=3b-2c+4a-(c+3b)+c
=3b-2c+4a-c-3b+c
=4a-2c
=4*1/4-2*1/2
=1-1
=0

Answer 3

3b-2c-[-4a+(c+3b)]+c
=3b-2c+4a-(c+3b)+c
=3b-c+4a-c-3b
=4a-2c
=1-1
=0

Answer 4

3b-2c-[-4a+(c+3b)]+c
=3b-2c+4a-c-3b+c
=4a-2c
=4*(1/4)-2*(1/2)
=0

Answer 5

3b-2c-[-4a+(c+3b)]+c=3b-2c+4a-(c+3b)+c=3b-2c+4a-c-3b+c=4a-2c
=4*(1/4)-2*(1/2)=0

Answer 6

3b-2c-[-4a+(c+3b)]+c=3b-2c+4a-c-3b+c=4a-2c=1-1=0