加法
$(i)$. $3mn,\ -5mn,\ 8mn,\ -4mn$
$(ii)$. $t-8tz,\ 3tz-z,\ z-t$
$(iii)$. $-7mn+5,\ 12mn+2,\ 9mn-8,\ -2mn-3$
$(iv)$. $a+b-3,\ b-a+3,\ a-b+3$
$(v)$. $14x+10y-12xy-13,\ 18-7x-10y+8xy,\ 4xy$
$(vi)$. $5m-7n,\ 3n-4m+2,\ 2m-3mn-5$
$(vii)$. $4x^2y,\ -3xy^2,\ -5xy^2,\ 5x^2y$
$(viii)$. $3p^2q^2-4pq+5,\ -10p^2q^2,\ 15+9pq+7p^2q^2$
$(ix)$. $ab-4a,\ 4b-ab,\ 4a-4b$
$(x)$. $x^2-y^2-1,\ y^2-1-x^2,\ 1-x^2-y^2$

需要完成的任务：求和

$(i)$. $3mn,\ -5mn,\ 8mn,\ -4mn$

$(ii)$. $t-8tz,\ 3tz-z,\ z-t$

$(iii)$. $-7mn+5,\ 12mn+2,\ 9mn-8,\ -2mn-3$

$(iv)$. $a+b-3,\ b-a+3,\ a-b+3$

$(v)$. $14x+10y-12xy-13,\ 18-7x-10y+8xy,\ 4xy$

$(vi)$. $5m-7n,\ 3n-4m+2,\ 2m-3mn-5$

$(vii)$. $4x^2y,\ -3xy^2,\ -5xy^2,\ 5x^2y$

$(viii)$. $3p^2q^2-4pq+5,\ -10p^2q^2,\ 15+9pq+7p^2q^2$

$(ix)$. $ab-4a,\ 4b-ab,\ 4a-4b$

$(x)$. $x^2-y^2-1,\ y^2-1-x^2,\ 1-x^2-y^2$

解答： 

i) $3mn,\ -5mn,\ 8mn,\ -4mn$

$=(3mn)+(-5mn)+(8mn)+(-4mn)$

$=(3-5+8-4)mn$

$=2mn$

ii) $t-8tz,\ 3tz-z,\ z-t$

$=(t-8tz)+(3tz-z)+(z-t)$

$=t-8tz+3tz-z+z-t$

$=t-t-8tz+3tz-z+z$

$=-5tz$

iii) $– 7mn + 5 + 12mn + 2 + (9mn – 8) + (- 2mn – 3) = – 7mn + 5 + 12mn + 2 + 9mn – 8 – 2mn – 3$

$= – 7mn + 12mn + 9mn – 2mn + 5 + 2 – 8 – 3$

$= mn (-7 + 12 + 9 – 2) + (5 + 2 – 8 – 3)$

$= mn (- 9 + 21) + (7 – 11)$

$= mn (12) – 4$

$= 12mn – 4$

iv) $a+b-3,\ b-a+3,\ a-b+3$

$=(a+b-3)+(b-a+3)+(a-b+3)$

$=a-a+a+b+b-b-3+3+3$

$=a(1-1+1)+b\ (1+1-1)+3\ (-1+1+1)$

$=a+b+3$

v) $14x+10y-12xy-13,\ 18-7x-10y+8xy,\ 4xy$

$=(14x+10y-12xy-13)+(18-7x-10y+8yx)+4xy$

$=14x-7x+10y-10y-12xy+8yx+4xy-13+18$

$=x(14-7)+y(10-10)+xy\ (-12+8+4)-13+18$

$=7x+5$

vi) $5m-7n,\ 3n-4m+2,\ 2m-3mn-5$

$=(5m-7n)+(3n-4m+2)+(2m-3mn-5)$

$=5m-4m+2m-7n+3n-3mn+2-5$

$=m(5-4+2)+n(-7+3)-3mn+2-5$

$=3m-4n-3mn-3$

vii) $4x^2y,-3xy^2,-5xy^2,\ 5x^2y$

$=x^2y\ (4+5)+xy^2\ (-3-5)$

$=9x^2y-8xy^2$

viii) $3p^2q^2-4pq+5,\ -10p^2q^2,\ 15+9pq+7p^2q^2$

$=(3p^2q^2-4pq+5)+(-10p^2q^2)+(15+9pq+7p^2q^2)$

$=3p^2q^2-10p^2q^2+7p^2q^2-4pq+9pq+5+15$

$=p^2q^2(3-10+7)+pq(-4+9)+5+15$

$=5pq+20$

ix) $ab-4a,\ 4b-ab,\ 4a-4b$

$=(ab-4a)+(4b-ab)+(4a-4b)$

$=ab-ab-4a+4a+4b-4b$

$=ab(1-1)+a(-4+4)+b(4-4)$

$=0$

x) $x^2-y^2-1,\ y^2-1-x^2,\ 1-x^2-y^2$

$=(x^2-y^2-1)+(y^2-1-x^2)+(1-x^2-y^2)$

$=x^2-x^2-x^2-y^2+y^2-y^2-1-1+1$

$=x^2\ (1-1-1)+y^2(-1+1-1)+(-1-1+1)$

$=-x^2-y^2-1$

教程点

更新于： 2022年10月10日

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