化简比 化简4(x-2y)^2-3[(x-2y)^2+2(x+y)]+2x+2y
化简4(x-2y)^2-3[(x-2y)^2+2(x+y)]+2x+2y
化简4(x-2y)^2-3[(x-2y)^2+2(x+y)]+2x+2y
解:
4(x-2y)²-3[(x-2y)²+2(x+y)]+2x+2y
=4(x-2y)²-3(x-2y)²-6(x+y)+2(x+y)
=(x-2y)²-4(x+y)
=x²+4y²-4xy-4x-4y
1+2+……+n=n(n+1)/2 1∧2+2∧2+……n∧2=n(n+1)(2n+1)/6 1
你现在高中吗 这些都是定理 已经证明好了 你记住会用就行了 没必要在这上面花太多时间的
用递回方法求1+2^2+3^2+4^2+5^2+6^2```````+i^2
var n:longint;
function f(t:Longint):longint;
begin
if t=1 then f:=1
else f:=f(t-1)+t*t;
end;
begin
readln(n);
writeln(f(n));
end.
![化简比 化简4(x-2y)^2-3[(x-2y)^2+2(x+y)]+2x+2y](http://img.zhputi.com/uploads/57a7/57a7412ae24720f5221542bb65ae8b6524697.jpg)
sn=2^2/(1*3)+4^2/(3*5)+....+(2n)^2/(2n-1)(2n+1)=
(2n)^2/(2n-1)(2n+1)=4n^2/(4n^2-1)=1+1/(4n^2-1)=1+1/(2n-1)(2n+1)
=1+[1/(2n-1)-1/(2n+1)]/2
将该通项式代入Sn,可以消解符号相反的各项,既得解.具体自己做.
已知a^2+ab=3,ab+b^2=1,试求a^2+2ab+b^2,a^2-b^2的值
a^2+2ab+b^2
=a²+ab+ab+b²
=3+1
=4
a^2-b^2
=a²+ab-ab-b²
=3-1
=2
2(x-2y)^2-7(x+2y)^3-3(2y-x)^2+5(x+2y)^3
2(x-2y)^2-7(x+2y)^3-3(2y-x)^2+5(x+2y)^3
=[2(x-2y)^2-3(x-2y)^2]+[5(x+2y)^3-7(x+2y)^3]
=-(x-2y)^2-2(x+2y)^3
(-2) 2004 +(-2) 2005 的结果是( ) A.(-2) 2004 B.-2 2004 C.(-2) 2005 D.2 2
原式=(-2) 2004 +(-2) 2004 ?(-2),=(-2) 2004 (1-2),
=-(-2) 2004 ,
=-2 2004 .
故选B.
化简复数z=(1-i)2+4i得( )A.2-2iB.2+2iC.-2iD.2
∵z=(1-i)2+4i=-2i+4i=2i.
故选D.
1-2^2+3^2-4^2+5^2-6^2+……+24^2 =? 是24/300/-300?
1-2^2+3^2-4^2+5^2-6^2+……+99^2-100^2
=(1+2)(1-2)+(3+4)(3-4)+……+(99+100)(99-100)
=-(1+2+3+4+……+99+100)
=-(1+100)X100/2
=-5050
已知a^2+ab=6,b^2+ab=4,求a^2-b^2和a^2+2ab+b^2的值
a^2+ab=6,b^2+ab=4
相减
a^2-b^2=6-4=2
a^2+ab=6,b^2+ab=4
相加
a^2+2ab+b^2=6+4=10