初一的題目

解：
合并同類項(xiàng)：
(1) x-f+5x-4f=5x-x-f-4f=4x-4f.[=4(x-f)]
(2) 30a^2b+2b^2c-15a^2b-4b^2c=30a^2b-15a^2b+2b^2c-4b^2c
=15a^2b-2b^2c. [=(15ab-2bc)b]

求代數(shù)值：
(1)m/2-(3/2)n-(5/6)n-m/6=(m/2-m/6)-[(3/2)n+(5/6)n].
=(3m-m)/6-(9n+5n)/6.
=2m/6-14n/6.
=m/3-7n/3
=6/3-7*2/3 (m=6,n=2)
=-8/3
∴原式=-2又2/3.

(2) 3pq-(4/5)m-pq=3pq-pq-(4/5)m
=2pq-(4/5)m
=2*(1/3)*(-3/2)-(4/5)*5 (p=1/3,q=-3/2,m=5)
=-1-4=-5
∴原式=-5.

去括號，并合并同類項(xiàng)：
(1) -2n-(3n-1)=-2n-3n+1
∴原式=-5n+1.

(2) a-(5a-3b)+(2b-a)=a-5a+3b+2b-a
∴原式=-5a+5b [=5(b-a)].

(3) -3(2s-5)+6s=－6s+15
∴原式=-6s+15.

(4) 1-(2a-1)-(3a+3)=1-2a+1-3a-3=-5a-1 .
∴原式=-5a-1 .

(5) 3(-2b+2a)-(3a-b)=-6b+6a-3a+b
=3a-5b .
∴原式=3a-5b.

(6) 14(abc-2a)+3(6a-2abc)=14abc-28a+18a-6abc
∴原式=8abc-28a [=4a(2bc-7)].

(7) 3(xy-2z)+(-xy+3z)=3xy-6z-xy+3z
∴原式=2xy-3z .

(8) -4(pq+pr)+(4pq+pr=-4pq-4pr+4pq+pr
∴原式=-3pr .