\(I=1+2-3-4+5+6-7-8+....+2009+2010\)
\(K=2011+2010\left(-4\cdot5^2+11+3^2\right)^{2009}\)
\(L=1+3-7+11-15+19\cdot23+...+99-103\)
help me !!!!!!!!!!!!!!
Nhanh nhất 1tick
bài 1 thực hiện phéptính
a) 2011+2010(-4x5^2+11x3^2)^2009
b)1+3-7+11-15+19-23+...+99-103
\(a,2011+2010\left(-4.5^2+11.3^2\right)^{2009}\)
\(=2011+2010\left(-1\right)=2011-2010\)
\(=1\)
tính tổng đại số
a.1+2-3-4+5+6-7-8+9+10-11-12+13+98-99-100+101+102
b.1-2-3+4+5-6-7+......+2009-2010-2011-2012
Giúp mình nhé!!!!!!
8 nha bạn
ht
Chứng minh : A=\(\frac{\left(2009^2+6\cdot2009+8\right)\cdot\left(2011^2+4\cdot2009+11\right)}{\left(2009\cdot2016+10\right)\left(2010\cdot2008+7\cdot2009+13\right)}=1\)
tính các dãy số sau:
-1-2+3+4-5-6+7+8-.....-2009-2010+2011
-1-2+3-4-5+6+....-2009+2010-2011
A=1-2+3+4-5-6+7+8-9-...+2007+2008-2009-2010
B=-1-2+3+4-5-6+7+8-9-10+11+12-...-2013-2014+2015+2016
C=(2+4+6+8+...+2014)-(3+5+7+9...+2011)
A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + ... + 2005 + 2006 - 2007 - 2008 + 2009 + 2010 ( có 2010 số )
A = ( 1 + 2 - 3 - 4 ) + ( 5 + 6 - 7 - 8 ) + .... + ( 2005 + 2006 - 2007 - 2008 ) + ( 2009 + 2010 )
A = ( - 4 ) + ( - 4 ) + ... + ( - 4 ) + 4019 ( có 503 số )
A = ( - 4 ) . 502 + 4019
A = - 2008 + 4019
A = 2011.
CHÚC LÀM BÀI VUI VẺ
Giải PT:
a, \(\dfrac{x^2+x+1}{x^2+x+2}+\dfrac{x^2+x+2}{x^2+x+3}=\dfrac{7}{6}\)
b, \(\dfrac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x\right)^2-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\dfrac{19}{49}\)
c, \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)
Help me!!! Mk cần gấp!!!
đkxđ với mọi x
đặt a=x2+x+1
\(\dfrac{a}{a+1}+\dfrac{a+1}{a+2}=\dfrac{7}{6}\)
<=> \(\dfrac{6a\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}+\dfrac{6\left(a+1\right)^2}{6\left(a+1\right)\left(a+2\right)}=\dfrac{7\left(a+1\right)\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}\)
=> 6a(a+2) +6(a+1)2 =7(a+1)(a+2)
<=> 6a2+12a +6a2 +12a+6 =a2 +21a+14
<=> 12a2 -a2+24a-21a+6-14=0
<=> 11a2+3a-8=0
<=> 11a2 +11a-8a-8=0
<=> (11a2 +11a)-(8a+8)=0
<=> 11a(a+1)-8(a+1)=0
<=> (a+1)(11a-8)=0
=> a=-1 và a=\(\dfrac{8}{11}\)
thay a=x2+x+1 ta đc
x2+x+1=-1
<=> x2+x+2 =0 (vô nghiệm)
và x2+x+\(\dfrac{3}{11}\) =0(vô nghiệm )
vậy pt trên vô nghiệm
c) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\left(2\right)\)ĐKXĐ : x # 0
( 2) <=> \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)
\(< =>8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right).\left(-2\right)=\left(x+4\right)^2\)
\(< =>8.\left[\left(x+\dfrac{1}{x}\right)^2-x^2-\dfrac{1}{x^2}\right]=\left(x+4\right)^2\)
\(< =>16=\left(x+4\right)^2\)
<=> x2 + 8x = 0
<=> x( x + 8) = 0
<=> x = 0 ( KTM ) hoặc x = - 8 ( TM )
Vậy,....
Giúp mk câu a, c thui nha!! Câu b mk làm đc rùi!!!
Nhã Doanh, ngonhuminh, nguyen thi vang, @hattori heiji, @Phùng Khánh Linh, ...
Tính :
A = 1 + 2 - 3 - 4 + 5 + 6 - 7 - 8 + 9 + 10 - 11 - 12 +....+ 2006 - 2007 - 2008 + 2009 - 2010 - 2011 - 2012
1Tinhs :A=1+2-3-4+5+6-7-8+.....+2005+2006-2007-2008+2009+2010-2011-2012+2014+2015-2016
HELP ME ! CÓ TICK
So sánh A và B biết : a, A=-2 /2011 -3/112 -5/113 - 7/114 ; B=-1/2011 -7/112 -5/113 - 3/114
b, A= 2006/2007 - 2007/2008 + 2008 / 2009 -2009/2010 B=1/2006.2007 -1/2008.2009