A = |x - 2001| + |x - 1| > hoặc = |x -2001 + 1 - x| = 2000
Những câu hỏi liên quan
x +3*1/2=(-2000+-2000+-2000)*(-2001+-2001+-2001)*.......*(2000+2000+2000)
TÌM X ?
1/(x+2000)(x+2001)+1/(2001)(x+2002)+...+1/(x+2009)(x+2010)=10/11
\(\dfrac{1}{\left(x+2000\right)\left(x+2001\right)}+\dfrac{1}{\left(x+2001\right)\left(x+2002\right)}+...+\dfrac{1}{\left(x+2009\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2001}+\dfrac{1}{x+2001}-\dfrac{1}{x+2002}+...+\dfrac{1}{x+2009}-\dfrac{1}{x+2010}=\dfrac{10}{11}\)
\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{x+2010-x-2000}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\)
\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{10}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\left(x+2000\right)\left(x+2010\right)=11\\ \Leftrightarrow...\)
Đúng 2
Bình luận (0)
1) So sánh :
A = 2000/2001 + 2001/2002 và B = 2000+2001/2001+2002
2) Tìm cặp x,y thuộc Z, biết :
5/x + y/4 = 1/8
2) \(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)
\(\Rightarrow x\left(1-2y\right)=40\)
Vì \(1-2y\) luôn là số lẻ nên \(1-2y\in\left\{\pm1;\pm5\right\}\)
\(\Rightarrow y=\left\{0;1;-2;3\right\}\)
\(\Rightarrow x\in\left\{40;-40;8;-8\right\}\)
Vậy các cặp số x,y thỏa mãn là \(\left(0;40\right);\left(1;-40\right);\left(-2;8\right);\left(3;-8\right)\)
Đúng 0
Bình luận (0)
Ta có :
\(B=\dfrac{2000+2001}{2001+2002}=\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}\)
Mặt khác :
\(\dfrac{2000}{2001}>\dfrac{2000}{2001+2002}\)
\(\dfrac{2001}{2002}>\dfrac{2001}{2001+2002}\)
\(\Leftrightarrow A=\dfrac{2000}{2001}+\dfrac{2001}{2002}>\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}=\dfrac{2000+2001}{2001+2002}=B\)
\(\Leftrightarrow A>B\)
Đúng 0
Bình luận (0)
Ta có: B =20002001+2002 +20012001+2002
Mặt khác: 20002001 >20002001+2002
20012002 >20012001+2002
Suy ra 20002001 +20012002 >20002001+2002 +20012001+2002
hay A> B
Vậy A > B.
Đúng 0
Bình luận (0)
1/(x+2000)(x+2001) + 1/(x+2001)(x+2002) +1/(x+2002)(x+2003) +........+ 1/(x+2006)(x+2007)= 7/8
Tìm x:
x + 4/ 2000+ x +3 / 2001= x + 2 / 2002+ x + 1/2001
1. Tìm a, b thuộc Q biết a-b=2*(a+b)=a:b
2. Tìm x thuộc Q biết:
a, (x+4)/2000 + (x+3)/2001 = (x+2)/2002 + (x+1)/2001
cho x+2000/x-2000=y+2001/y-2001. chứng minh x/y=2000/2001
\(\frac{x+2000}{x-2000}=\frac{y+2001}{y-2001}\Rightarrow\left(x+2000\right)\left(y-2001\right)=\left(x-2000\right)\left(y+2001\right)\)
\(\Rightarrow\frac{x+2000}{y+2001}=\frac{x-2000}{y-2001}=\frac{x+2000+x-2000}{y+2001+y-2001}=\frac{2x}{2y}=\frac{x}{y}=\frac{x+2000-\left(x-2000\right)}{y+2001-\left(y-2001\right)}=\frac{2000}{2001}\)
=>đpcm
Đúng 0
Bình luận (0)
cho: x1+x2+x3+...+x2000+x2001+x2001=0 và x1+x2 + x3=x4+x5+x6=...=x1999+x2000+x2001=1
Tìm x2002
Sao cậu cho mình mà ko biết đó là gì? Bà bán hàng bọc lại rồi đưa cho cậu à. Cảm ơn, chào!
Đúng 0
Bình luận (0)
Tìm x, biết :a, left(frac{1}{1cdot2cdot3}+frac{1}{2cdot3cdot4}+...+frac{1}{98cdot99cdot100}right)x-3;b, left(frac{frac{2000}{1}+frac{1999}{2}+...+frac{1}{2000}}{frac{1}{2}+frac{1}{3}+...+frac{1}{2001}}right)xfrac{-1}{5}.c,left(frac{frac{2000}{1}+frac{1999}{2}+...+frac{1}{2000}+2000}{1+frac{1}{2}+frac{1}{3}+...+frac{1}{2001}}right):xfrac{-2001}{2002}.
Đọc tiếp
Tìm x, biết :
a, \(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{98\cdot99\cdot100}\right)x=-3\);
b, \(\left(\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}}\right)x=\frac{-1}{5}\).
c,\(\left(\frac{\frac{2000}{1}+\frac{1999}{2}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2001}}\right):x=\frac{-2001}{2002}\).