Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
làm a) thui nhé,b) bn tu lam
a) P = (x+y +x-y)(x+y -x+y) = 4xy = 1
xong rui do,toan là vay, noi it,hiu nhiu
D= \(\frac{x^3+y^3+z^3-3xyz}{2\left(x^2+y^2+z^2-xy-yz-zx\right)}\) tử = (x+y)3+z3 -3xy(x+y) - 3xyz =(x+y+z)(x2+2xy+y2-xz- yz+z2)-3xy(x+y+z) = (x+y+z)(x2+y2+z2-xy-yz-zx)
do đó D=\(\frac{x+y+z}{2}\)
Sửa đề\(2004\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2006\right)+1=A\)
Đặt \(2004\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2006\right)+1=A\)
Ta có:
\(A=2004\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2005+1\right)+1\)
\(=\left(2005-1\right)\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2005+1\right)+1\)
\(=2005\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2005+1\right)\)\(-\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2005+1\right)+1\)
\(=\left(2005^{2007}+2005^{2006}+2005^{2005}+...+2005^2+2005\right)\)\(-\left(2005^{2006}+2005^{2005}+2005^{2004}+...+2005+1\right)+1\)
\(=2005^{2007}⋮2005^{2007}\left(dpcm\right)\)
\(P=\frac{x\left(x+5\right)+y\left(y+5\right)+2\left(xy-3\right)}{x\left(x+6\right)+y\left(y+6\right)+2xy}\)
\(=\frac{x^2+5x+y^2+5y+2xy-6}{x^2+6x+y^2+6y+2xy}\)
\(=\frac{\left(x+y\right)^2+5\left(x+y\right)-6}{\left(x+y\right)^2+6\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(x+y+5\right)-6}{\left(x+y\right)\left(x+y+6\right)}\)
\(=\frac{2005\times\left(2005+5\right)-6}{2005\times\left(2005+6\right)}\)
\(=\frac{2005\times2010-6}{2005\times2011}\)
\(=\frac{2004}{2005}\)
Với mọi n thuộc N* ta có :
\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)
\(=\left(n^2+n+\frac{1}{2}\right)\left(n^2-n+\frac{1}{2}\right)\)
\(\Rightarrow N=\frac{\left(2^2+2+\frac{1}{2}\right)\left(2^2-2+\frac{1}{2}\right)...\left(2008^2+2008+\frac{1}{2}\right)\left(2008^2-2008+\frac{1}{2}\right)}{\left(1^2+1+\frac{1}{2}\right)\left(1^2-1+\frac{1}{2}\right)...\left(2007^2+2007+\frac{1}{2}\right)\left(2007^2-2007+\frac{1}{2}\right)}\)
\(=\frac{\left(2.3+\frac{1}{2}\right)\left(1.2+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)...\left(2008.2009+\frac{1}{2}\right)}{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)...\left(2007.2008+\frac{1}{2}\right)}\)
\(=\frac{2008.2009+\frac{1}{2}}{\frac{1}{2}}=8068145\)
bằng 1
cho cách làm đi bn