\(2004.2006.\left(2005^2+1\right)\)
Những câu hỏi liên quan
1/A2004.2006 B2005^22/Aleft(3+1right)left(3^2+1right)left(3^4+1right)left(3^8+1right)left(3^{16}+1right)va B3^{22}-13/ A2004.2006.2008^2vaa B2005.20072009^24/Aleft(a+1right)left(a^2+1right)left(a^4+1right)left(a^8+1right)...left(a^{2m}+1right) va Ba^{2m+1}-1vs a thuoc ℝ, m thuoc ℕ, m lon hon hoac 2004
Đọc tiếp
1/A=2004.2006 B=\(2005^2\)
2/A=\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)va B=\(3^{22}-1\)
3/ A=2004.2006.\(2008^2\)vaa B=2005.2007\(2009^2\)
4/A=\(\left(a+1\right)\left(a^2+1\right)\left(a^4+1\right)\left(a^8+1\right)...\left(a^{2m}+1\right)\) va B=\(a^{2m+1}-1\)vs a thuoc \(ℝ\), m thuoc \(ℕ\), m lon hon hoac = 2004
Gửi các bạn bài toán tính nhanh nè
a, A=\(\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{6}\right).\left(1+\dfrac{1}{10}\right).\left(1+\dfrac{1}{15}\right).\left(1-\dfrac{1}{21}\right).\left(1-\dfrac{1}{28}\right)\)
d, D=\(\dfrac{2006.2005-1}{2004.2006+2005}\)
XEM BẠN NÀO GIẢI DC TRƯỚC NÈ
mk chỉ giải đc câu d thôi nha ; bn thông cảm
d) \(D=\dfrac{2006.2005-1}{2004.2006+2005}=\dfrac{2006.2005-1}{2004.2006+2006-1}\)
\(D=\dfrac{2006.2005-1}{2006\left(2004+1\right)-1}=\dfrac{2006.2005-1}{2006.2005-1}=1\)
Đúng 0
Bình luận (0)
so sánh 2 số sau: A=2004.2006.(20052+1) và B=20054
A = (2005 - 1).(2005 +1).(20052 + 1) = (20052 - 1).(20052 + 1) = 20054 - 1 < 20054
=> A < B
Đúng 0
Bình luận (0)
5 tháng 8 2015 lúc 15:35
so sánh A= 2004.2006.(20052+1)
và B= 20054
A= 2004.2006.(20052+1)
=(2005-1)(2005+1)(20052+1)
=(20052-1)(20052+1)
=20054-1<20054
=> A<B
Đúng 0
Bình luận (0)
Tính hợp lí:
2006.2005-1/ 2004.2006+2005
\(\frac{2006.2005-1}{2004.2006+2005}\)
\(\Leftrightarrow\)\(\frac{2006.\left(2004+1\right)-1}{2004.2006+2005}\)
\(\Leftrightarrow\frac{2006.2004+2016-1}{2004.2006+2005}\)
\(\Leftrightarrow\frac{2006.2004+2005}{2004.2006+2005}\)
\(=1\)
Đúng 0
Bình luận (0)
giải dùm mình nha mn
(12+22+32+......+20052)-(1.3+2.4+3.5+.....+2004.2006)=?
Giải
Ta gọi T = (1^2+2^2+...+2005^2)-(1.3+2.4+3.5+...+2004.2006)
Đặt A = 1^2+2^2+3^2+...+2005^2
=> A = 1.1 + 2.2 +3.3 +...+ 2005.2005
=> A = 1.(2-1) + 2.(3-1) + 3.(4-1) +...+ 2005.(2006-1)
==> A = 1.2-1.1 + 2.3-1.2 + 3.4-1.3+...+2005.2006-1.2005
=> A = (1.2+2.3+3.4+...+2005.2006)-(1+2+3+...+2005)
Xét 1.2 +2.3+3.4+...+2005.2006
= 1/3.(1.2.3+2.3.3+...+2005.2006.3)
=1/3.[1.2.(3-0)+2.3.(4-1)+...+2005.2006.(2007-2004)]
=1/3.(1.2.3+2.3.4-1.2.3+...+2005.2006.2007-2004.2005.2006)
= 1/3 . 2005.2006.2007
= 2005.2006.2007/3 = 2690738070
Vậy A= 2690738070 - (1+3+5+...+2005)
=> A= 2690738070- [(2005-1):2+1].(2005+1)/2
=> A = 2690738070 - 1006009
=> A = 2689732061
Đắt B = 1.3+2.4+3.5+4.6+...+2003.2005 +2004.2006
=> B= (1.3+3.5+...+2003.2005)+(2.4+4.6+...+2004.2006)
=> 6B = (1.3.6+3.5.6+...+2003.2005.6)+(2.4.6+4.6.6+...+2004.2006.6)
=> 6B = [1.3.(5+1)+3.5.(7-1)+...+2003.2005.(2007-2001)] + [2.4.(6-0)+4.6.(8-2)+...+2004.2006.(2008-2002)]
=> 6B = (1.3.5+1.3.1+3.5.7-1.3.5+...+2003.2005.2007-2001.2003.2005)+(2.4.6+4.6.8-2.4.6+...+2004.2006.2008-2002.2004.2006)
=> 6B = 1.3.1+2003.2005.2007 + 2004.2006.2008
=> 6B = 16132350300
=> B = 16132350300/6 = 2688725050
Vì T = A - B = 2689732061-2688725050
=> T = 1007011
Đúng 0
Bình luận (0)
bạn nào giúp mk vs ạ!!
tính hợp lí:
A= 20054-2004.2006.(20052+1)
B=1999.(20002+2001)-2001.(20002-1999)
A= 20054-2004.2006.(20052+1)
=\(2005^4-\left(2005-1\right)\cdot\left(2005+1\right)\cdot\left(2005^2+1\right)\)
=\(2005^4-\left(2005^2-1\right)\cdot\left(2005^2+1\right)\)
=\(2005^4-\left(2005^4-1\right)\)
=1
Đúng 0
Bình luận (0)
B=1999.(20002+2001)-2001.(20002-1999)
=\(1999\cdot2000^2+1999\cdot2001-2001\cdot2000^2+2001\cdot1999\)
=\(2000^2\left(1999-2001\right)+2\cdot1999\cdot2001\)
=\(2000^2\cdot\left(-2\right)+2\cdot1999\cdot2001\)
=\(2000^2\cdot\left(-2\right)+2\left(2000-1\right)\left(2000+1\right)\)
=\(-2\cdot2000^2+2\left(2000^2-1\right)\)
=\(-2\cdot2000^2+2\cdot2000^2-2\)
=-2
Đúng 0
Bình luận (3)
Tính tổng : \(f\left(\frac{1}{2005}\right)+f\left(\frac{2}{2005}\right)+.....+\left(\frac{2004}{2005}\right)vớif\left(x\right)=\frac{100^x}{100^x+10}\)
chp a,b,c,x,y,z là các số nguyên dương thỏa \(x+y+z=a\) ;\(x^2+y^2+z^2=b\);\(a^2=b+4010\)
tính \(M=\sqrt[x]{\frac{\left(2005+y^2\right)\left(2005+z^2\right)}{\left(2005+x^2\right)}}+\sqrt[y]{\frac{\left(2005+x^2\right)\left(2005+z^2\right)}{2005+y^2}}\)\(+\sqrt[z]{\frac{\left(2005+x^2\right)\left(2005+y^2\right)}{2005+z^2}}\)
\(a^2=b+4010\Rightarrow\left(x+y+z\right)^2=x^2+y^2+z^2+4010\Rightarrow x^2+y^2+z^2+2xy+2yz+2xz=x^2+y^2+z^2+4010\)
\(\Rightarrow2xy+2yz+2xz=4010\Rightarrow xy+yz+xz=2005\)
\(x\sqrt{\frac{\left(2015+y^2\right)\left(2005+z^2\right)}{\left(2005+x^2\right)}}=x\sqrt{\frac{\left(xz+yz+xy+y^2\right)\left(xy+xz+yz+z^2\right)}{\left(xy+yz+x^2+xz\right)}}\)
\(=x\sqrt{\frac{\left(z\left(x+y\right)+y\left(x+y\right)\right)\left(x\left(y+z\right)+z\left(y+z\right)\right)}{\left(y\left(x+z\right)+x\left(x+z\right)\right)}}=x\sqrt{\frac{\left(y+z\right)^2\left(x+y\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=x\sqrt{\left(y+z\right)^2}=x\left(y+z\right)=xy+xz\)
tương tự : \(y\sqrt{\frac{\left(2015+x^2\right)\left(2015+z^2\right)}{2015+y^2}}=xy+yz;z\sqrt{\frac{\left(2005+x^2\right)\left(2005+y^2\right)}{2015+z^2}}=xz+yz\)
\(\Rightarrow M=xy+xz+xy+yz+xz+yz=2\left(xy+yz+xz\right)=2\cdot2005=4010\)
Đúng 0
Bình luận (0)