Question

初一数学的几道题

(4k2+7k)+(-k2+3k-1)

(5y+3x-15z2)-(12y+7x+z2).

(3k2+7k)+(4k2-3k+1)

(3x2+2xy-1/2x)-(2x2-xy+x)

(1/2xy+y2+1)+(x2-1/2xy-2y2-1)

(m2-3m+2)-(4m-2n-1)

-5(x2-3)-2(3x2+5)
帮帮忙 有60分

Answer 1

(4k2+7k)+(-k2+3k-1)=3k²+10k-1

(5y+3x-15z2)-(12y+7x+z2)=-7y-4x-16z²
.
(3k2+7k)+(4k2-3k+1)=7k²+4k+1

(3x2+2xy-1/2x)-(2x2-xy+x)=x²+3xy-3x/2

(1/2xy+y2+1)+(x2-1/2xy-2y2-1)=x²-y²

(m2-3m+2)-(4m-2n-1)=m²-7m+2n+3

-5(x2-3)-2(3x2+5)=-11x²+5

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Answer 2

(4k2+7k)+(-k2+3k-1)=3k2+10k-1
(5y+3x-15z2)-(12y+7x+z2).=-7y+10x-16z2
(3k2+7k)+(4k2-3k+1)=7k2+4k+1
(3x2+2xy-1/2x)-(2x2-xy+x)=x2+3xy+x/2
(1/2xy+y2+1)+(x2-1/2xy-2y2-1)=x2-y2
(m2-3m+2)-(4m-2n-1)=m2-7m+3
-5(x2-3)-2(3x2+5)=-11x2+5

Answer 3

4k2+7k)+(-k2+3k-1)=3k²+10k-1

(5y+3x-15z2)-(12y+7x+z2)=-7y-4x-16z²
(3k2+7k)+(4k2-3k+1)=7k²+4k+1

(3x2+2xy-1/2x)-(2x2-xy+x)=x²+3xy-3x/2

(1/2xy+y2+1)+(x2-1/2xy-2y2-1)=x²-y²

(m2-3m+2)-(4m-2n-1)=m²-7m+2n+3

-5(x2-3)-2(3x2+5)=-11x²+5

Answer 4

1.解：原式=8k+7k-2k+3k-1
=16k-1
2.解：原式=5y+3x-30z-12y-7x-2z
=-4x-7y-32z
3.解：原式=6k+7k+8k-3K+1
=18K+1
4.解：原式=6x+2xy-1/2x-4x+xy-x
=-1/2x+3xy
5.解：原式=1/2xy+2y+1+2x-1/2Xy-4y-1
=2x-2y
6.解：原式=2m-3m+2-4m+2n+1
=-5m+2n+3
7.解：原式=-10x+15-12x-10
=-22x+5

Answer 5

3k^2+10k-1 -7y+10x-14z^2 7k^2+4k+1 2+3xy-3/2x x^2-y^2 m^2-7n+1+2n 5x^2-7 用手机好累

Answer 6

做一条直线垂直于AB、DE，则围成一个五边形，又有公式多边形的内角和为（n-2）*180°，所以得出∠C=540°-90°-90°-145°-135°=90°