tim x biết |x+2008|+|x+2009|=1
Tìm XEZ biết
a)x+(x+1)+(x+2)+........+2008=2008
b)2009+2008+2007+........+(x+1)+x=2009
a)=> (2008+x).2008/2=2008
=>(2008+x)=2
=>x=-2006
Câu 2: (2 điểm) Tim x, biết:
1) (x−1)/2009+(x−2)/2008=(x−3)/2007+(x−4)/2006
2) (59−x)/41+(57−x)/43+(55−x)/45+(53−x)/47+(51−x)/49=−5
1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
Tìm x thuộc Z, biết :
a/ x + (x + 1) + (x + 2)+...+2008 = 2008
b/ 2009 + 2008 + 2007 +...+(x + 1) + x + 2009
a,Tìm x,y thuộc Z biết : 25-y^2=8(x-2009)^2
b,Tim x,y thuộc N biết : (2008x+3y+1).(2008^x+2008x+y)=225
Tim so tu nhien x biet:
1 phan 2+1 phan 6+1 phan 12+...+1 phan x(x+1)=2008 phan 2009
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(=>\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{x\left(x+1\right)}=\frac{2008}{2009}\)
\(=>\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2008}{2009}\)
\(=>1-\frac{1}{x+1}=\frac{2008}{2009}=>\frac{1}{x+1}=1-\frac{2008}{2009}=\frac{1}{2009}\)
=>x+1=2009
=>x=2008
Vậy x=2008
1/2+1/6+1/12+...+1/x*(x+1)=2008/2009
1/1*2+1/2*3+1/3*4+...+1/x*(x+1)=2008/2009
1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/(x+1)=2008/2009
1-1/x+1)=2008/2009
1/x+1=1-2008/2009
1/x+1=1/2009
nên x+1=2009
x=2009-1
x=2008 (tick nha)
\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)tim x
\(\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)
\(\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)
\(\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)
Vì \(\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)\ne0\)
=> x-2010 =0
=> x =2010
/x-2008/+/x-2009/+/x-2010/=2
Tim x
/ là dấu trị tuyệ đối nha
hãy so sánh
2008 x2009
2008 x 2009 – 1
Và
2008 x 2009
2009 x 2008
tìm x biết \(\left|x+\frac{1}{2009}\right|+\left|x+\frac{2}{2009}\right|+\left|x+\frac{3}{2009}\right|+...+\left|x+\frac{2008}{2009}\right|\) =2009x
\(\hept{\begin{cases}\left|x+\frac{1}{2009}\right|\ge0\\....\\\left|x+\frac{2008}{2009}\right|\ge0\end{cases}\Rightarrow\left|x+\frac{1}{2009}\right|+\left|x+\frac{2}{2009}\right|+....\left|x+\frac{2008}{2009}\right|\ge0}\)
\(\Rightarrow2009x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\hept{\begin{cases}\left|x+\frac{1}{2009}\right|=x+\frac{1}{2009}\\....\\\left|x+\frac{2008}{2009}\right|=x+\frac{2008}{2009}\end{cases}\Rightarrow x+\frac{1}{2009}+...+x+\frac{2008}{2009}}=2009x\)
\(2008x+201840=2009x\Rightarrow x=201840\)
p/s: cách làm thì khá ok, nhưng kq không chắc lắm nhé, có gì bn tính lại nha
Boul đẹp trai_tán gái đổ 100% sai 100%
Sao dòng cuối lại tek ? Các phân số ấy cộng vào không thể là 201840
Về hướng làm thì đúng nhưng chỉ đúng đến bước phá trị thôi
Tham khảo cách làm nhưg nhớ đổi đoạn cuối nhé !
a sorry cộng lại quên mẹ chia cho 2009 :> mà tính máy tính ko hiểu sao cx sai lun, đổi lại kq nha :>
\(x=\frac{2017036}{2009}=1004\)