/x-2002/+/x-2001/
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NT
3
15 tháng 4 2018
\(\frac{2003\times4+1998+2001\times2002}{2002+2002\times1002+2002\times1003}\)
\(=\frac{2003\times4+2\times999+2001\times2\times1001}{2002.\left(1+1002+1003\right)}\)
\(=\frac{2\times\left(2003\times2+999+2001\times1001\right)}{1001\times2\times\left(1+1002+1003\right)}\)
\(=\frac{2003\times2+999+2001\times1001}{1001\times\left(1+1002+1003\right)}\)
\(=1\)
mk ko bít
AH
Akai Haruma
Giáo viên
10 tháng 1 2022
Lời giải:
Dãy $x,x+1, x+2,..., 2002$ có số số hạng là:
$\frac{2002-x}{1}+1=2003-x$
Tổng $x+(x+1)+....+2001+2002=\frac{(2002+x)(2003-x)}{2}$
Do đó:
$\frac{(2002+x)(2003-x)}{2}=2002$
$\Rightarrow (2002+x)(2003-x)=4004$
$2002.2003+x-x^2=4004$
$x^2-x-4006002=0$
$(x-2002)(x+2001)=0$
$\Rightarrow x=2002$ hoặc $x=-2001$
VN
1
11 tháng 5 2019
P=\(\frac{\left(2002+1\right)\times14+1988+2001\times2002}{2002\times\left(1+503+504\right)}\)
\(=\frac{2002\times14+2002+2001\times2002}{2002\times1008}\)
\(=\frac{2002\times\left(14+1+2001\right)}{2002\times1008}=\frac{2016}{1008}=2\)
TN
1
26 tháng 2 2022
\(\dfrac{1}{\left(x+2000\right)\left(x+2001\right)}+\dfrac{1}{\left(x+2001\right)\left(x+2002\right)}+...+\dfrac{1}{\left(x+2009\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2001}+\dfrac{1}{x+2001}-\dfrac{1}{x+2002}+...+\dfrac{1}{x+2009}-\dfrac{1}{x+2010}=\dfrac{10}{11}\)
\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{x+2010-x-2000}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\)
\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{10}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\left(x+2000\right)\left(x+2010\right)=11\\ \Leftrightarrow...\)
QD
6 tháng 8 2017
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 \(\left(0;40\right);\left(1;-40\right);\left(-2;8\right);\left(3;-8\right)\)
6 tháng 8 2017
Ta có :
\(B=\dfrac{2000+2001}{2001+2002}=\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}\)
Mặt khác :
\(\dfrac{2000}{2001}>\dfrac{2000}{2001+2002}\)
\(\dfrac{2001}{2002}>\dfrac{2001}{2001+2002}\)
\(\Leftrightarrow A=\dfrac{2000}{2001}+\dfrac{2001}{2002}>\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}=\dfrac{2000+2001}{2001+2002}=B\)
\(\Leftrightarrow A>B\)
4 tháng 4 2020
\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
<=> \(\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)
<=> \(\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)
<=> (x - 2004)(1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4) = 0
<=> x - 2004 = 0 (vì 1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4 khác 0)
<=> x = 2004
Vậy S = {2004}
I
4 tháng 4 2020
đề bài \(=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
\(\Leftrightarrow\frac{x}{2000}-\frac{4}{2000}+\frac{x}{2001}-\frac{3}{2001}+\frac{x}{2002}-\frac{2}{2002}=\frac{x}{2}-\frac{2002}{2}+\frac{x}{3}-\frac{2001\\}{3}+\frac{x}{4}-\frac{2000}{4}\)
\(\Leftrightarrow\frac{x}{2000}-\frac{1}{500}+\frac{x}{2001}-\frac{1}{667}+\frac{x}{2002}-\frac{1}{1001}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}+1001+667+500=0\)
\(\Leftrightarrow\left(\frac{x}{2000}+\frac{x}{2001}+\frac{x}{2002}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}\right)+\left(1001+667+500-\frac{1}{500}-\frac{1}{667}-\frac{1}{1001}\right)=0\)
=> x=1
/x-2002/+/x-2001/=/x-2002+x-2001/=/2x-4003/
\(\left|x-2002\right|+\left|x-2001\right|\)\(\le\)\(\left|x-2002+2001-x\right|=1\)
Dấu = xảy ra <=>
Xét cả hai trường hợp như sau :
TH1 : x - 2002 và 2001 - x < 0
\(\hept{\begin{cases}x-2002< 0\Rightarrow x=< 2002\\2001-x< 0\Rightarrow x>2001\end{cases}}\)
TH2 : x - 2002 và 2001 - x > 0
\(\hept{\begin{cases}x-2002>0\Rightarrow x>2002\\2001-x>0\Rightarrow x< 2001\end{cases}}\)\((\)loại\()\)
Vậy M\(_{min}\)= 1 khi x = 2002