x的平方减去5x减去2007等于0,求x-2分之(x-2)的三次方-(x-1)的平方加1的值

如题所述

第1个回答  2019-12-19
2011

x的平方减去5x减去2007等于0,求x-2分之(x-2)的三次方-(x-1)的平方加...
化简:(x - 2)^3\/(x - 2) - (x - 1)^2 + 1 = - 2x + 4。设:- 2x + 4=a,则:x=2-(1\/2)a,代入:x^2-5x-2007=0,再化简得:a1=-1+√8053, a2=-1-√8053;检验知:a2=-1-√8053不合题意,舍去。即:x-2分之(x-2)的三次方-(x-1)的平方加1的值=a1=-...

已知x的平方减5x减2000=0,求x减2分之(x-2)的三次方减(x-1)的平方+1...
解:x²-5x-2000=0 x²-5x=2000 于是 [(x-2)³-(x-1)²+1]\/(x-2)=[(x-2)³-(x²-2x+1)+1]\/(x-2)=[(x-2)³-x²+2x]\/(x-2)=[(x-2)³-x(x-2)]\/(x-2)=(x-2)[(x-2)²-x]\/(x-2)=(x-2)²...

已知x²-5x-2007=0,求(x-2)分之(x-2)³-(x-1)²+1的值
(x-2)分之(x-2)³-(x-1)²+1 解:原式=(x-2)²-[(x-1)²-1]/(x-2)=(x-2)²-(x²-2x)/(x-2)=(x-2)²-[x(x-2)]\/(x-2)=(x-2)²-x =x²-4x+4-x =x²-5x+4 ∵x²-5x-2007=0 ∴x²...

已知x平方-5x-2007=0,求(x-2)的立方-(x-1)的平方+1 x-2
因为X^2-5X-2007=0,即X^2-5X=2007 所以 [(X-2)^3-(X-1)^2+1]\/(X-2)=(x-2)^2-(x^2-2x+1)\/(x-2)+1\/(x-2)=(x-2)^2-(x^2-2x)\/(x-2)-1\/(x-2)+1\/(x-2)=x^2-4x+4-x- 1\/(x-2)+1\/(x-2)=x^2-5x+4 =2007+4 =2011 ...

已知x方-5x-2007=0,求代数式{x-2}的三次方-{x-1}的二次方+1...
x方-5x-2007=0 可变式为 x^2-5x+4=2007+4 先留着这个式子 (x-2)^3-(x-1)^2+1中将减号后面的拆开合并变为(x-2)^3-x(x-2)其中一个(x-2)与分子约掉,式子变为 (x-2)^2-x=x^2-5x+4 把最上面的带进去 答案为2011 ...

已知x方减5X减2003等于0,求x-2\/(x-2)的三次方减(x-1)的平方加1的值
x^2-5x-2003=0 x^2-5x=2003 x-2\/(x-2)的三次方减(x-1)的平方加1的值 =(x-2)\/((x-2)^3-((x-1)^2-1))=(x-2)\/((x-2)^3-((x-1-1)(x-1+1))=(x-2)\/((x-2)^3-x(x-2))=(x-2)\/(x-2)((x-2)^2-x)=1\/(x^2-4x+4-x)=1\/(x^2-5x+4)=1...

x的平方减5x减1997等于0,则代数式x-2分之x-2的三次方减x-1的平方+1的...
首先代数式化简:[(x-2)�0�6-(x-1)�0�5+1]\/(x-2)=(x�0�6-7x�0�5+14x-8)\/(x-2)=(x�0�6-2x�0�5-5x�0�5+10x+4x-8)\/(x-2)=[(x-2...

x的平方-5x-2004=0,求(x-2)的3次方-(x-1)的2次方+1除以x-2的值
解答:[(x-2)^3-(x-1)^2+1]\/(x-2)=[x^3-6x^2+12x-8-x^2+2x-1+1]\/(x-2)=[x^3-7x^2+14x-8]\/(x-2)=[(x-2)(x^2+2x+4)-7x(x-2)]\/(x-2)=x^2-5x+4=2004+4=2008

...=0,求代数式(x-2)的三次方-(x-1)的平方+1\/x-2的值
由:已知x的平方-5x-2006=0,考虑(x-2)=y,(x-1)=z的值,再整体代入代数式(x-2)的三次方-(x-1)的平方+1\/x-2

已知x^2-5x-2010=0,,则代数式[(x-2)^3-(x-1)^2+1]\/(x-2)的值??
x^2-5x-2010=0 x^2-5x = 2010 [(x-2)^3-(x-1)^2+1]\/(x-2)= (x-2)^2 - [(x-1)^2-1]\/(x-2)= (x-2)^2 - [x²-2x]\/(x-2)= (x-2)^2 - x = x^2-4x+4-x = x^2-5x+4 = 2010+4 = 2014 ...

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