2012年到2013年成都发现了 3^2013-5*3^2012+6*3^2011= 请问大神怎么做啊
3^2013-5*3^2012+6*3^2011= 请问大神怎么做啊
3^2013-5*3^2012+6*3^2011= 请问大神怎么做啊
3^2013-5*3^2012+6*3^2011= 3*3^2012-5*3^2012+2*3^2012=0
分解因式x^3(y-z)+y^3(z-x)+z^3(x-y)
x^3(y-z)+y^3(z-x)+z^3(x-y)
=x^3y-x^3z+y^3z-xy^3+z^3(x-y)
=(x^3y-xy^3)-(x^3z-y^3z)+z^3(x-y)
=xy(x^2-y^2)-z(x^3-y^3)+z^3(x-y)
=xy(x+y)(x-y)-z(x-y)(x^2+xy+y^2)+z^3(x-y)
=(x-y)[xy(x+y)-z(x^2+xy+y^2)+z^3]
=(x-y)(x^2y+xy^2-x^2z--y^2z+z^3)
=(x-y)[(x^2y-x^2z)+(xy^2-)-(y^2z-z^3)]
=(x-y)[x^2(y-z)+xy(y-z)-z(y^2-z^2)]
=(x-y)[(y-z)(x^2+xy-yz-z^2)]
=(x-y)(y-z)[(x^2-z^2)+(xy-yz)]
=(x-y)(y-z)(x-z)(x+y+z).
①计算:(?3)2?3?64?(3)2?|?4|②求x的值:4x2=9
(1)原式=3-(-4)-3-4
=3+4-3-4
=0;
(2)x2=
9 4,
∴x=±
,
∴x=±
.
15÷3=5 150÷3=50 1500÷3=500 请问有什么规律
当除数相同时,被除数以相同倍数(十倍)增长时,商也以相同倍数增长(十倍)。

2(x一3)(x十2)一(3十a)(3一a),其中a二一2,x二1
2(x-3)(x+2)-(3+a)(3-a)
=2*(1-3)*(1+2)-(3-2)(3+2)
=2*(-2)*3-1*5
= - 12 - 5
= - 17
25×3又1/3×0.4×3/10怎么简便计算
25×3又1/3×0.4×3/10
=25×0.4×3又1/3×3/10
=10×1
=10
计算:(1)-|-5|+(-3)3÷(-22);(2)52+94+178+3316+6532+12964?13;(3)(180°-91°32′24″)
(1)原式=-5+(-27)÷(-4),
=?5+
,
=
;
(2)原式=(
5 2?2)+(
9 4?2)+(
17 8?2)+(
33 16?2)+(
65 32?2)+(
129 64?2)?1,
=
+
1 4+
1 8+
1 16+
1 32+
1 64?1,
=1?
+
1 2?
1 4+
1 4?
1 8+
1 8?
1 16+
1 16?
1 32+
1 32?
1 64?1,
=?
;
(3)(180°-91°32′24″)÷2,
=88°27′36″÷2,
=44°13′98″;
(4)原式=5abc-2a2b-[3abc-4ab2+a2b],
=5abc-a2b-3abc+4ab2,
=8abc-a2b-4ab2.
已知4x-3y+z=0,x+3y-3z=0,则x:y:z=( )
4x-3y+z=0 (1) x+3y-3z=0 (2)解:方程(1)+(2)得到5x-2z=0x=2/5z把x=2/5z带入方程(1)中得到8/5z+z=3yy=13/15z所以x:y:z=2/5z:13/15z:z=6:13:15补充用心作答,希望采纳,谢谢!
(a+b-2x)^3-(a-x)^3-(b-x)^3因式分解
可以用;立方和公式a3+b3=(a+b)(a2-ab+b2)(a+b-2x)^3-(a-x)^3-(b-x)^3
=(a+b-2x)^3-[(a-x)^3+(b-x)^3]
=(a+b-2x)^3-[(a-x)+(b-x)][(a-x)^2-(a-x)(b-x)+(b-x)^2]
=(a+b-2x)*[(a+b-2x)^2-(a-x)^2+(a-x)(b-x)-(b-x)^2]
=(a+b-2x)*{[(a-x)+(b-x)]^2-(a-x)^2+(a-x)(b-x)-(b-x)^2}
=(a+b-2x)*{(a-x)^2+(b-x)^2+2*(a-x)*(b-x)-(a-x)^2+(a-x)(b-x)-(b-x)^2}
=(a+b-2x)*3*(a-x)*(b-x)
=3*(a-x)*(b-x)*(a+b-2x)
因式分解(x+a)3+(x+b)3-(2x+a+b)3
(x+a)^3+(x+b)^3-(2x+a+b)^3
=(x+a+x+b)[(x+a)^2-(x+a)(x+b)+(x+b)^2]-(2x+a+b)^3
=(2x+a+b)[(x+a+x+b)^2+(x+a)^2+(x+b)^2-(x+a)(x+b)]
=(2x+a+b)[2(x+a)^2+2(x+b)^2+(x+a)(x+b)]
=(2x+a+b)[5x^2+5ax+5bx+2a^2+2b^2+ab]