Question

已知m,x,y满足

:①。3/5（x-5)+丨m丨=0:②-3a^2b^y+1与a^2b^3是同类项，求代数式（2x^2-3xy+6y^2)-m(3x^2-xy+9y^2)的值

Answer 1

3/5（x-5)^2+丨m丨=0
x=5，m=0

-3a^2b^y+1与a^2b^3是同类项
y=3

(2x^2-3xy+6y^2)-m(3x^2-xy+9y^2)
= (2x^2-3xy+6y^2)-0*(3x^2-xy+9y^2)
= 2x^2-3xy+6y^2
= 2*5^2-3*5*3+6*3^2
= 50-30+54
= 74

Answer 2

楼上的，y+1要等于指数3，所以y=2，不是等于3.

3/5（x-5)^2+丨m丨=0
x=5，m=0

-3a^2b^y+1与a^2b^3是同类项
y=2

(2x^2-3xy+6y^2)-m(3x^2-xy+9y^2)
= (2x^2-3xy+6y^2)-0*(3x^2-xy+9y^2)
= 2x^2-3xy+6y^2
= 2*5^2-3*5*2+6*2^2
= 50-30+24
= 44

Answer 3

3/5（x-5)^2+丨m丨=0
x=5，m=0

-3a^2b^y+1与a^2b^3是同类项
y=2

(2x^2-3xy+6y^2)-m(3x^2-xy+9y^2)
= (2x^2-3xy+6y^2)-0*(3x^2-xy+9y^2)
= 2x^2-3xy+6y^2
= 2*5^2-3*5*2+6*2^2
= 50-30+24
= 44
是这样的