把下列二次函数配方f（x）=3+5x-2x²

f(x)=-2x²+5x+3
=-2(x²-5x/2)+3
=-2(x²-5x/2+25/16-25/16)+3
=-2(x²-5x/2+25/16)+2×25/16+3
=-2(x-5/4)²+49/8

f(x)=3/4*x²-2x
=(3/4)(x²-8x/3)
=(3/4)(x²-8x/3+16/9-16/9)
=(3/4)(x²-8x/3+16/9)-(3/4)(16/9)
=(3/4)(x-4/3)²-4/3

f(x)=-2(x²+2x+1)+9x+5
=-2(x+1)^2+9x+9-4
=-2(x+1)^2+9(x+1)-4
=(-2x-1)(x-3) (十字交叉相乘法)

f(x)=1/4(3x^2-8x)
=1/4(3x^2-6x+3-2x+2-5)
=1/4{3(x^2-2x+1)-2(x-1)-5}
=1/4{3(x+1)^2-2(x-1)-5}
=1/4{3(x-1)-5}{(x-1)+1} (十字交叉相乘法)

F(X)=3/4X~2-2X=(3/4X-5/4)(X-1)-5/4
F(X)=3+5X-2X~2=(-2X+3)(X-1)+6