/ x+201 /2001 +(y-202)2002=0
tìm x
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
19 tháng 2 2019
\(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1=\frac{1-x}{2002}+1+\left(\frac{x}{2003}-1\right)\)
\(\Leftrightarrow\frac{2-x+2001}{2001}=\frac{1-x+2002}{2002}+\frac{x-2003}{2003}\)
\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{x-2003}{2003}\)
\(\Leftrightarrow\left(x-2003\right)\left(\frac{1}{2003}+\frac{1}{2001}-\frac{1}{2002}\right)=0\)
\(\Leftrightarrow x-2003=0\)\(\left(v\text{ì}\frac{1}{2003}+\frac{1}{2001}-\frac{1}{2002}\ne0\right)\)
\(\Leftrightarrow x=2003\)
Vậy \(S=\left\{2003\right\}\)
HN
19 tháng 2 2019
d)Ta có : \(\frac{2-x}{2001}-1=\frac{1-x}{2002}-\frac{x}{2003}\)
\(\Leftrightarrow\frac{2-x}{2001}+1-2=\frac{1-x}{2002}+1+1-\frac{x}{2003}-2\)\(\Leftrightarrow\frac{2003-x}{2001}=\frac{2003-x}{2002}+\frac{2003-x}{2003}\)
\(\Leftrightarrow\frac{2003-x}{2001}-\frac{2003-x}{2002}-\frac{2003-x}{2003}=0\)\(\Leftrightarrow\left(2003-x\right)\left(\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow2003-x=0\Leftrightarrow x=2003\)
Vậy phương trình có tập nghiệm S = { 2003 }
H
3
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 :
\(\dfrac{2000}{2001}>\dfrac{2000}{2001+2002}\)
\(\dfrac{2001}{2002}>\dfrac{2001}{2001+2002}\)
\(\Leftrightarrow A=\dfrac{2000}{2001}+\dfrac{2001}{2002}>\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}=\dfrac{2000+2001}{2001+2002}=B\)
\(\Leftrightarrow A>B\)
6 tháng 7 2019
a) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
b) Sửa đề :
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow x=300\)
c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\)
\(\Leftrightarrow x=2004\)
Vậy....
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
có |x+201|^2001 > hoặc =0 với mọi x ( giá trj tuyệt đối)
(y-202)^2002 > hoặc = 0 với mọi y ( luỹ thừa bậc chẵn)
Suy ra |x+201|^2001 + (y-202)^2002 > hoặc = 0 với mọi x;y
Theo bài ra thì : |x+201|^2001 + (y-202)^2002 = 0
nên |x+201|^2001 + (y-202)^2002 = 0
<=> |x+201|^2001 = 0 <=> x+201=0 <=> x=-201
<=> (y-202)^2002 = 0 <=> y-202=0 <=> y=202
vậy (x;y) = (-201;202)
tick nha !! very