1.已知實數x,y滿足(x-2y+8)² +(2x-y+1)² =0 試求x,y之值 2.展開下列各式 (1) (2a-3)² (2) (a-3b+2c)² (3) (1+3x)(1-3x+9x² )?

Sol
1
(x－2y+8)^2+(2x－y+1)^2=0
x－2y+8=0
2x－y+1=0
(x－2y+8)－2*(2x－y+1)=0
－3x+6=0
x=2
y=2x+1=5
2
(1)
(2a-3)^2=4a^2-12a+9
(2)
(a－3b+2c)^2
=a^2+(－3b)^2+(2c)^2+2*a*(－3b)+2*a*(2c)+2*(－3b)*(2c)
=a^2+9b^2+4c^2－6ab+4ac－12bc
(3)
(1+3x)(1－3x+9x^2)
=(1+3x)*[1－(3x)+(3x)^2]
=1+(3x)^3
=1+27x^3

(x-2y+8)² +(2x-y+1)² =0
[(x-2y)+8]²+[(2x-y)+1]² =0
(x-2y)²+16(x-2y)+64+(2x-y)²+(2x-y)+1=0
x²-4xy+4y²+16x-32y+x²-4xy+y²+2x-y+65=0
5x²-8xy+5y²+18x-33y+65=0

(2)(1)
(2a-3)²
=4a²-12a+9

(2)(2)
(a-3b+2c)²
=[(a-3b)+3c]²
=(a-3b)²+9(a-3b)c+9c²
=a²-6ab+9b²+9ac-27bc+9c²
=a²+9b²+9c²-6ab+9ac-27bc

(2)(3)
(1+3x)(1－3x+9x^2)
=1+27x^3