\(A=\left|x-2001\right|+\left|x-1\right|\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
13 tháng 7 2018
\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
\(\frac{8\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}-\frac{8\left(x+2000\right)}{8\left(x+2000\right)\left(x+2007\right)}=\frac{7\left(x+2000\right)\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}\)
\(8x+8.2007-8x+8.2000=7\left(x^2+4007x+2000.2007\right)\)
\(8.7-7\left(x^2+4007x+2000.2007\right)=0\)
\(7\left(8-x^2-4007x-2000.2007\right)=0\)
\(8-x^2-4007x-2000.2007=0\)
\(x^2+4007x+4013992=0\)
\(\left(x^2+2008x\right)+\left(1999x+4013992\right)=0\)
\(\left(x+2008\right)\left(x+1999\right)=0\)
\(\hept{\begin{cases}x=-2008\\x=-1999\end{cases}}\)
13 tháng 7 2018
\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)
\(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+...+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)
\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
TN
19 tháng 6 2017
Áp dụng \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) rút gọn rồi quy đồng làm nốt
NM
31 tháng 1 2016
\(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
=> \(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+....+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)
<=> \(\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{8}\Leftrightarrow\left(x+2000\right)\left(x+2007\right)=8\)
=> x = -1999 hoặc x = - 2008
MT
3 tháng 8 2015
\(\text{ĐKXĐ: }x+1\ne0\text{ và }x-2001\ne0\)
\(\Leftrightarrow x\ne-1\text{ và }x\ne2001\)
\(\frac{\left(x^2-2000x-2001\right).2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{\left(x^2+x-2001x-2001\right).2001}{\left(x+1\right)\left(x-2001\right).2002}\)
\(=\frac{\left[x.\left(x+1\right)-2001\left(x+1\right)\right].2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{\left(x-2001\right)\left(x+1\right).2001}{\left(x+1\right)\left(x-2001\right).2002}=\frac{2001}{2002}\)
S
17 tháng 8 2017
Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\), ta có:
\(A=\left|x-2001\right|+\left|x-1\right|=\left|x-2001\right|+\left|1-x\right|\ge\left|x-2001+1-x\right|=\left|-2000\right|=2000\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2001\right)\left(1-x\right)\ge0\Rightarrow1\le x\le2001\)
Vậy...
HA
1
Y
29 tháng 6 2019
a) \(\left|x-2000\right|+\left|x-2002\right|=\left|x-2000\right|+\left|2002-x\right|\)
\(\ge\left|x-2000+2002-x\right|=2\) (1)
Dấu "=" \(\Leftrightarrow\left(x-2000\right)\left(2002-x\right)\ge0\)
\(\Leftrightarrow2000\le x\le2002\)
+ \(\left|x-2001\right|\ge0\forall x\). "=" \(\Leftrightarrow x=2001\) (2)
Từ (1) và (2) suy ra \(A\ge2\)
Dấu "=" \(\Leftrightarrow x=2001\)
b) \(B=\left|x-8\right|+\left|x-9\right|+\left|x-10\right|+\left|x+11\right|\)
+ \(\left|x-10\right|+\left|x+11\right|=\left|x+11\right|+\left|10-x\right|\)
\(\ge\left|x+11+10-x\right|=21\) (3)
Dấu "=" \(\Leftrightarrow\left(x+11\right)\left(10-x\right)\ge0\Leftrightarrow-11\le x\le10\)
+ \(\left|x-8\right|+\left|x-9\right|\ge\left|x-8+9-x\right|=1\) (4)
"=" \(\Leftrightarrow\left(x-8\right)\left(9-x\right)\ge0\Leftrightarrow8\le x\le9\)
Từ (3) và (4) suy ra \(B\ge22\)
"=" \(\Leftrightarrow8\le x\le9\)
26 tháng 5 2016
(x+4/2000 + 1)+(x+3/2001 + 1) = (x+2/2002 + 1)+(x+1/2003)+1
(x+2004/2000) + (x+2004/2001) = (x+2004/2002) + (x+2004/2003)
(x+2004).(1/2000+1/2001) = (x+2004).(1/2002+1/2003)
+ Với x+2004=0 suy ra x=-2004. Ta có 0.(1/2000+1/2001)=0.(1/2002+1/2003), đúng
+ Với x+2004 khác 0 thì (x+2004).(1/2000+1/2001) = (x+2004).(1/2002+1/2003)
1/2000+1/2001 = 1/2002+1/2003, vô lí vì 1/2000+1/2001 > 1/2002+1/2003
Vậy x=-2004
Ta có :
\(A=\left|x-2001\right|+\left|x-1\right|=\left|x-2001\right|+\left|1-x\right|\)
\(\Leftrightarrow A\ge\left|\left(x-2001\right)+\left(1-x\right)\right|\)
\(\Leftrightarrow A\ge\left|-2000\right|\)
\(\Leftrightarrow A\ge2000\)
Dấu "=" xảy ra khi :
\(\left(x-2001\right)\left(1-x\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2001\ge0\\1-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2001\le0\\1-x\le0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2001\\1\ge x\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2001\\1\le x\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2001\ge x\ge1\\x\in\varnothing\end{matrix}\right.\)
Vậy ..