a)Tìm số hữu tỉ x biết:x+4/2005 + x+3/2006 = x+2/2007 +x+1/2008
b)Tìm x biết:(x-1)x+2 = (x-z)x+4
1,Tìm số hữu tỉ x biết\(\frac{x+4}{2005}+\frac{x+3}{2006}=\frac{x+2}{2007}+\frac{x+1}{2008}\)
2,tìm x biết:\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
1) \(\frac{x+4}{2005}\)\(+\)\(\frac{x+3}{2006}\)= \(\frac{x+2}{2007}\)\(+\)\(\frac{x+1}{2008}\)
\(\Leftrightarrow\) \(\frac{x+4}{2005}\)\(+\)1 \(+\)\(\frac{x+3}{2006}\)\(+\)1 = \(\frac{x+2}{2007}\)\(+\)1 \(+\)\(\frac{x+1}{2008}\)\(+\)1
\(\Leftrightarrow\)\(\frac{x+2009}{2005}\)+ \(\frac{x +2009}{2006}\)= \(\frac{x+2009}{2007}\)+\(\frac{x+2009}{2008}\)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006) = (x + 2009)(1/2007 + 1/2008)
\(\Leftrightarrow\)(x + 2009)(1/2005 + 1/2006 - 1/2007 - 1/2008) = 0
Ta thấy: 1/2005 + 1/2006 - 1/2007 - 1/2008 \(\ne\)0
\(\Leftrightarrow\)x + 2009 = 0
\(\Leftrightarrow\)x = -2009
\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=\left(x-1\right)\left(x-2\right)x=0\)
tìm đc x=0;1;2
ai giúp em với gấp lắm rồi: mong các bác cho lời giải ko ghi đáp án chống đối
1.Tìm các số hữu tỉ x,y,z biết:
a) x.(x-y+z)=11 ; y.(y-z-x)=25 ; z.(z+x-y)=35
b) (x+2)^2 + (y-3)^4 + (z-5)^6=0
2. So sánh A và B biết
a) A=-1/2011 - 3/11^2 - 5/11^3 - 7/11^4 và B= -1/2011 - 7/11^2 - 5/11^3 - 3/11^4
b) A= 2006/2007 - 2007/2008 + 2008/2009 - 2009/2010 và B= -1/2006.2007 - 1/2008.2009
mong mấy bạn giúp mình mai mình nộp rôì ko đùa đâu
ai lam guip toi cau nay voi mai toi nop bai roi
so sanh 2 phan so sau bang cach nahnh nhat: 2007/2008 voi 2008/2009
Tính bằng cách nhanh nhất :
a. ( 1-1/2 ) x ( 1 -1/3) x ( 1-1/4) x.... x ( 1-18)x ( 1-1/19 ) x ( 1 - 1/20 )
b. 3/2 x 4/3 x 5/4 x ..... x 2006/2005 x 2007/2006 x 2008/2007
giải hộ mih nhé
a)=1/2*2/3......*19/20
=1/20
b)=3/2*4/3......*2008/2007
=3/2007
Tìm x biết : x - 1 / 2009 + x - 2 / 2008 = x - 3 / 2007 + x- 4 / 2006
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)
\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\right)=0\)
\(\Leftrightarrow x=2010\)
Tìm x biết:
(x - 1)/2009 + (x - 2)/2008 = (x - 3)/2007 + (x - 4)/2006
\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=0\)
\(\Rightarrow x-2010=0\Rightarrow x=2010\)
Tìm x biết
a)\(x^4-30x^2+31x-30=0\)
b)\(\left(x^2+x\right)^2+4\times\left(x^2+x\right)=12\)
c)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
Nên x + 2009 = 0 => x = -2009
1.Tìm x biết:
a) 7x+2+7x+1+7/57=52x+52x+1+52x+3/131.
b) (4x-3)2=(4x-3)3.
c) |5(2x+3)|+|2(2x+3)|+|2x+3|=16.
d) |x2+|6x-2||=x2+4.
2.Tính tỉ số A/B,biết:
A=1/2+1/3+1/4+...+1/2007+1/2008+1/2009.
B=2008/1+2007/2+2006/3+...+2/2007+2/2008.
Câu 1: So sánh các số hữu tỉ:
A = 2006/2007 - 2007/2008 + 2008/2009 - 2009/2010 với B = -1/2006 x 2007 - (-1)/2007 x 2008
Tìm x ϵ Z biết:
x-1/2011 + x-2/2010 - x-3/2009 = x-4/2008
\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}-\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
<=> \(\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)-\left(\dfrac{x-3}{2009}-1\right)=\left(\dfrac{x-4}{2008}-1\right)\)
<=> \(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}-\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}=0\)
<=> \(\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
<=> x - 2012 = 0
<=> x = 2012