x2-4x+3分之x2-5x+6+s2-2x+1分之x2-4x+3+1-x2分之x2-3x-4的过程
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x2-4x+3分之x2-5x+6+s2-2x+1分之x2-4x+3+1-x2分之x2-3x-4的过程x2-4x+3分之x2-5x+6+s2-2x+1分之x2-4x+3+1-x2分之x2-3x-4的过程
x2-4x+3分之x2-5x+6+s2-2x+1分之x2-4x+3+1-x2分之x2-3x-4的过程
x2-4x+3分之x2-5x+6+s2-2x+1分之x2-4x+3+1-x2分之x2-3x-4的过程
x2-4x+3分之x2-5x+6+s2-2x+1分之x2-4x+3+1-x2分之x2-3x-4的过程
x2-4x+3分之x2-5x+6+s2-2x+1分之x2-4x+3+1-x2分之x2-3x-4
=(x²-5x+6)/(x²-4x+3)+(x²-4x+3)/(x²-2x+1)-(x²-3x-4)/(1-x²)
=(x-2)(x-3)/(x-1)(x-3)+(x-1)(x-3)/(x-1)²-(x-4)(x+1)/(1+x)(1-x)
=(x-2)/(x-1)+(x-3)/(x-1)+(x-4)/(x-1)
=[x-2+x-3+x-4]/(x-1)
=(3x-9)/(x-1)
x2-4x+3分之x2-5x+6+x2-2x+1分之x2-4x+3+1-x2分之x2-3x-4
=(x-3)(x-2)/[(x-3)(x-1)]+(x-3)(x-1)/[(x-1)²]+(x-4)(x+1)/[(1+x)(1-x)]
=(x-2)/(x-1)+(x-3)/(x-1)-(x+4)/(x-1)
=x-2+x-3-x-4
=x-9
则ax1+b,ax2+b,,axn+b的平均数’=(ax1+b+ax2+b++axn+b把3X1-2看成a又因为方差为三分之一则3X2-2=a+1 3x3-3=a+2 这样和
x2-4x+3分之x2-5x+6+s2-2x+1分之x2-4x+3+1-x2分之x2-3x-4的过程
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