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a) 7x+4=3x+16\(\Leftrightarrow\)4x=12\(\Leftrightarrow\)x=3
b)(x+9)(3x-15)=0\(\Leftrightarrow\)x+9=0 hoặc 3x-15=0
\(\Rightarrow\)x\(\in\){-9;5}
c) |-5x|=2x+21
Nếu x\(\le\)0 thì -5x=2x+21\(\Leftrightarrow\)x=-3 (t/m)
Nếu x>0 thì -5x=-2x-21\(\Leftrightarrow\)x=7 (t/m)
Vậy x\(\in\){-3;7}
d) 3x-5>15-x\(\Leftrightarrow\)4x>20\(\Leftrightarrow\)x>5
e) \(\dfrac{x+1}{2001}+\dfrac{x+5}{2005}< \dfrac{x+9}{2009}+\dfrac{x+13}{2013}\)
\(\Leftrightarrow\dfrac{x+1}{2001}-1+\dfrac{x+5}{2005}-1< \dfrac{x+9}{2009}-1+\dfrac{x+13}{2013}-1\)
\(\Leftrightarrow\)\(\dfrac{x-2000}{2001}+\dfrac{x-2000}{2005}-\dfrac{x-2000}{2009}-\dfrac{x-2000}{2013}< 0\)
\(\Leftrightarrow\)(x-2000)(\(\dfrac{1}{2001}+\dfrac{1}{2005}-\dfrac{1}{2009}-\dfrac{1}{2013}\))<0
Vì \(\dfrac{1}{2001}+\dfrac{1}{2005}-\dfrac{1}{2009}-\dfrac{1}{2013}>0\) nên x-2000<0
\(\Leftrightarrow\)x<2000
a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8-12+20x=0\)
\(\Leftrightarrow21x-4=0\)
\(\Leftrightarrow21x=4\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)
Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!
\(\frac{x-3}{2011}+\frac{x-5}{2009}+\frac{x-7}{2007}+\frac{x-9}{2005}=4\)
\(\Leftrightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)+\left(\frac{x-9}{2005}-1\right)=0\)
\(\Leftrightarrow\frac{x-2014}{2011}+\frac{x-2014}{2009}+\frac{x-2014}{2007}+\frac{x-2014}{2005}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2009}+\frac{1}{2007}+\frac{1}{2005}\right)=0\)
|________________A________________|
Do A > 0
nên x - 2014 = 0
<=> x = 2014
Giải các phương trình:
\(\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}=4\)
\(\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}=4\)
\(\Leftrightarrow\dfrac{x-3}{2011}+\dfrac{x-5}{2009}+\dfrac{x-7}{2007}+\dfrac{x-9}{2005}-4=0\)
\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-7}{2007}-1\right)+\left(\dfrac{x-9}{2005}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2007}+\dfrac{x-2014}{2005}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2007}+\dfrac{1}{2005}\right)=0\)
\(\Leftrightarrow x-2014=0\) ( do \(\dfrac{1}{2011}+\dfrac{1}{2009}+\dfrac{1}{2007}+\dfrac{1}{2005}\ne0\))
\(\Leftrightarrow x=2014\)
Vậy phương trình có nghiệm S=\(\left\{2014\right\}\)
ta có (x+1/2009 +1) + ( x+3/2007 + 1)- (x+5/2005 +1) - (x+7/1993 + 1) = 0
=>(x +100/ 2009) + (x+100/2007) - (x+100/2005)-(x+100/1993)
=> (x +100) * (1/2009 + 1/2007+ 1/2005 + 1/1993) = 0
=> x = -100
Bạn cứ tinh ý để ý đến phần tử và mẫu cộng lại bằng 100. Khi bạn bỏ phần x + 100 ra thì còn lại như trên. Sau đó lược bỏ còn lại x = -100
Mạn phép mk không chép đề , mk làm luôn nhé
\(\dfrac{x+1}{2009}+1+\dfrac{x+3}{2007}+1=\dfrac{x+5}{2005}+1+\dfrac{x+7}{1993}+1\)
⇔ \(\dfrac{x+2010}{2009}+\dfrac{x+2010}{2007}-\dfrac{x+2010}{2005}-\dfrac{x+2010}{1993}=0\)
⇔( x + 2010 )\(\left(\dfrac{1}{2009}+\dfrac{1}{2007}-\dfrac{1}{2005}-\dfrac{1}{1993}\right)=0\)
Ta thấy : \(\dfrac{1}{2009}< \dfrac{1}{2007}< \dfrac{1}{2005}< \dfrac{1}{1993}\)
⇒ \(\dfrac{1}{2009}+\dfrac{1}{2007}-\dfrac{1}{2005}-\dfrac{1}{1993}< 0\)
⇒ x + 2010 = 0
⇒ x = -2010
KL....
\(\dfrac{x-4}{2001}\)- 1 +\(\dfrac{x-3}{2002}\)-1 + \(\dfrac{x-2}{2003}\)-1 =\(\dfrac{x-2003}{2}\)-1 + \(\dfrac{x-2002}{3}\)-1 +\(\dfrac{x-2001}{4}\)-1 <=> \(\dfrac{x-2005}{2001}\)+\(\dfrac{x-2005}{2002}\)+\(\dfrac{x-2005}{2003}\)-\(\dfrac{x-2005}{2}\)-\(\dfrac{x-2005}{3}\)-\(\dfrac{x-2005}{4}\)= 0 <=> (x-2005). (\(\dfrac{1}{2001}\)+\(\dfrac{1}{2002}\)+\(\dfrac{1}{2003}\)-\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)) =0 <=> x-2005=0 ( vì \(\dfrac{1}{2001}\) +\(\dfrac{1}{2002}\) +\(\dfrac{1}{2003}\)- \(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\) khác 0) =>x = 2005
x-4/2001+ x-3/2002 + x-2/2003= x-2003/2 + x-2002/3 + x-2001/4
<=>(x-4/2001 -1)+(x-3/2002 -1)+(x-2/2003 -1)-(x-2003/2 -1)+
(x-2002/3 -1)+(x-2001/4 -1) =0
<=>x-2005/2001+ x-2005/2002+ x-2005/2003- x-2005/2-
x-2005/3- x-2005/4 =0
<=>(x-2005).(1/2001+1/2002+1/2003- 1/2- 1/3- 1/4)=0
<=>x-2005=0 (vì 1/2001+1/2002+1/2003-1/2-1/3-1/4)
<=>x=2005
Vậy pt có nghiệm là x=2005
\(\dfrac{x+10}{2003}+\dfrac{x+6}{2007}+\dfrac{x+12}{2001}+3=0\)
\(\Leftrightarrow\dfrac{x+10}{2003}+1+\dfrac{x+6}{2007}+1+\dfrac{x+12}{2001}+1=0\)
\(\Leftrightarrow\dfrac{x+10+2003}{2003}+\dfrac{x+6+2007}{2007}+\dfrac{x+12+2001}{2001}=0\)
\(\Leftrightarrow\dfrac{x+2013}{2003}+\dfrac{x+2013}{2007}+\dfrac{x+2013}{2001}=0\)
\(\Leftrightarrow\left(x+2003\right)\left(\dfrac{1}{2003}+\dfrac{1}{2007}+\dfrac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
Vậy pt có nghiệm x = -2013
\(\dfrac{x+10}{2003}\)+\(\dfrac{x+6}{2007}\)+\(\dfrac{x+12}{2001}\)+3=0
<=> \(\dfrac{x+10}{2003}\)+1+\(\dfrac{x+6}{2007}\)+1+\(\dfrac{x+12}{2001}\)+1=0
<=> (\(\dfrac{x+10}{2003}\)+1) + (\(\dfrac{x+6}{2007}\)+1) + (\(\dfrac{x+12}{2001}\)+1)=0
<=> \(\dfrac{x+2013}{2003}\)+\(\dfrac{x+2013}{2007}\)+\(\dfrac{x+2013}{2001}\)=0
<=> (x+2013)(\(\dfrac{1}{2003}+\dfrac{1}{2007}+\dfrac{1}{2001}\))=0
<=> x+2013=0( Vì \(\dfrac{1}{2003}+\dfrac{1}{2007}+\dfrac{1}{2001}\)>0)
<=> x= -2013
Vậy S={-2013}
a: \(\Leftrightarrow\left(\dfrac{x+2001}{5}+1\right)+\left(\dfrac{x+1999}{7}+1\right)+\left(\dfrac{x+1997}{9}+1\right)+\left(\dfrac{x+1995}{11}+1\right)=0\)
=>x+2006=0
=>x=-2006
b: \(\Leftrightarrow\left(\dfrac{x-15}{100}-1\right)+\left(\dfrac{x-10}{105}-1\right)+\left(\dfrac{x-100}{5}-1\right)=\left(\dfrac{x-100}{15}-1\right)+\left(\dfrac{x-105}{10}-1\right)+\left(\dfrac{x-110}{5}-1\right)\)
=>x-105=0
=>x=105