(x三次方-27分之x²+3x+9)+(9-x²分之6)-(6+2x分之x-1)=?
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![(x三次方-27分之x²+3x+9)+(9-x²分之6)-(6+2x分之x-1)=?](/uploads/image/z/1325264-32-4.jpg?t=%EF%BC%88x%E4%B8%89%E6%AC%A1%E6%96%B9-27%E5%88%86%E4%B9%8Bx%26%23178%3B%2B3x%2B9%EF%BC%89%2B%EF%BC%889-x%26%23178%3B%E5%88%86%E4%B9%8B6%EF%BC%89-%EF%BC%886%2B2x%E5%88%86%E4%B9%8Bx-1%EF%BC%89%3D%3F)
(x三次方-27分之x²+3x+9)+(9-x²分之6)-(6+2x分之x-1)=?
(x三次方-27分之x²+3x+9)+(9-x²分之6)-(6+2x分之x-1)=?
(x三次方-27分之x²+3x+9)+(9-x²分之6)-(6+2x分之x-1)=?
根据立方差公式a³-b³=(a-b)(a²+ab+b²)可得
(x²+3x+9)/(x³-27)+6/(9-x²)-(x-1)/(6+2x)=1/(x-3)-6/(x²-9)-(x-1)/(6+2x)=(x+3)/(x²-9)-6/(x²-9)-(x-1)/(6+2x)=1/(x+3)-(x-1)/(6+2x)=(3-x)/(2x+6)
x三次方-27分之x²+3x+9)+(9-x²分之6)-(6+2x分之x-1)
=(x²+3x+9/x三次方-27)+(6/9-x²)-(x-1/6+2x)
=[x²+3x+9/(x-3)x(x²+3x+9)]-[6/(x+3)(x-3)]-[x-1/2(x+3)]
=(1/x+3)-[x-1/2(x+3)]
=(3-x)/2(x+3)
=3-x/6+2x
(x三次方-27分之x²+3x+9)+(9-x²分之6)-(6+2x分之x-1)
=(x-3)(x²+3x+9)分之(x²+3x+9)+(3-x)(3+x)分之6 - 2(3+x)分之(x-1)
=(x-3)分之1-(x-3)(3+x)分之6 - 2(3+x)分之(x-1)
=2(x-3)(3+x)分之(6+2x-12-x²...
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(x三次方-27分之x²+3x+9)+(9-x²分之6)-(6+2x分之x-1)
=(x-3)(x²+3x+9)分之(x²+3x+9)+(3-x)(3+x)分之6 - 2(3+x)分之(x-1)
=(x-3)分之1-(x-3)(3+x)分之6 - 2(3+x)分之(x-1)
=2(x-3)(3+x)分之(6+2x-12-x²+4x-3)
=2(x-3)(3+x)分之[ -(x-3)²]
= (6+2x)分之(3-x)
收起
-1 6/(x∧2 9)
兄弟,这道题化出来结果比较长,看起来比较复杂。结果直接写成:X三次方 12-27分之X平方 3x-2的x平方分之12 x的平方-x。大胆写吧!