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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....
\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}=\dfrac{x+2007}{3}+\dfrac{x+2006}{4}\)
\(\Leftrightarrow\dfrac{x+1}{2009}+1+\dfrac{x+2}{2008}+1=\dfrac{x+2007}{3}+1+\dfrac{x+2006}{4}+1\)
\(\Leftrightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}=\dfrac{x+2010}{3}+\dfrac{x+2010}{4}\)
\(\Rightarrow x+2010=0\)
\(\Rightarrow x=-2010\)
Vậy pt có nghiệm duy nhất \(x=-2010\)
Bài của bạn nè bạn gái!
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1=\dfrac{x-4}{2010}-1+\dfrac{x-5}{2009}-1+\dfrac{x-6}{2008}-1\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{1012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
mà \(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{10}{2008}\ne0\)
\(\Rightarrow x-2014=0\Rightarrow x=2014\)
vậy x=2014
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}+1+\dfrac{x-2}{2012}+1+\dfrac{x-3}{2011}+1-\dfrac{x-4}{2010}+1-\dfrac{x-5}{2009}+1-\dfrac{x-6}{2008}+1=0\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\right)=0\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
Vậy PT có nghiệm là \(x=2014\)
a)\(\dfrac{2x+1}{x-3}-\dfrac{x}{x+3}=0\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(x+3\right)-\left(x-3\right)x=0\)
\(\Leftrightarrow2x^2+7x+3-x^2+3x=0\)
\(\Leftrightarrow x^2+10x+3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5+\sqrt{22}\\x=-5-\sqrt{22}\end{matrix}\right.\)(tm)
b.
\(\left(x-4\right)\left(x-5\right)=12\)
\(\Leftrightarrow x^2-9x+20-12-0\)
\(\Leftrightarrow x^2-9x+8=0\)
\(\Leftrightarrow x^2-8x-x+8=0\)
\(\Leftrightarrow x\left(x-8\right)-\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)
Vậy........
\(\dfrac{x+2}{2011}+\dfrac{x+2017}{4}=\dfrac{x+4}{2009}+\dfrac{x+2019}{6}\)
\(\Leftrightarrow\dfrac{x+2013-2011}{2011}+\dfrac{x+2013+4}{4}=\dfrac{x+2013-2009}{2009}+\dfrac{x+2013+6}{6}\)
\(\Leftrightarrow\dfrac{x+2013}{2011}-1+\dfrac{x+2013}{4}+1=\dfrac{x+2013}{2009}-1+\dfrac{x+2013}{6}+1\)
\(\Leftrightarrow\dfrac{x+2013}{2011}+\dfrac{x+2013}{4}-\dfrac{x+2013}{2009}-\dfrac{x+2013}{6}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2011}+\dfrac{1}{4}-\dfrac{1}{2009}-\dfrac{1}{6}\right)=0\)
\(\Leftrightarrow x+2013=0\)
\(\Leftrightarrow x=-2013\)
Vậy PT trên có nghiệm là S={-2013}
\(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)
⇔\(\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)
⇔ \(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)
⇔ \(\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)
⇔ x+2009=0
⇔ x=-2009
vậy x=-2009 là nghiệm của pt
a) ( x2 + x )2 + 4( x2 + x ) = 12
<=> ( x2 + x )2 + 4( x2 + x ) + 4 - 16 = 0
<=> ( x2 + x + 2)2 - 16 = 0
<=> ( x2 + x + 2 + 4)( x2 + x + 2 - 4) = 0
<=> ( x2 + x + 6 )( x2 + x - 2) = 0
Do : x2 + x + 6
= x2 + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}+6-\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\) ≥ \(\dfrac{23}{4}\) > 0 ∀x
=> x2 + x - 2 = 0
<=> x2 - x + 2x - 2 = 0
<=> x( x - 1) + 2( x - 1) = 0
<=> ( x - 1)( x + 2 ) = 0
<=> x = 1 hoặc : x = - 2
KL.....
b) Kuroba kaito làm rùi nhé 
\(a.\dfrac{3x-2}{5}+\dfrac{x-1}{9}=\dfrac{14x-3}{15}-\dfrac{2x+1}{9}\\ \Leftrightarrow\dfrac{27x-18}{45}+\dfrac{5x-5}{45}=\dfrac{42x-9}{45}-\dfrac{10x+5}{45}\\ \Rightarrow27x-18+5x-5=42x-9-10x-5\\ \Leftrightarrow32x-23=32x-14\\ \Leftrightarrow0x=9\\ \Rightarrow Phươngtrìnhvônghiệm\\ \Rightarrow S=\phi\)
\(b.\dfrac{x+3}{2}-\dfrac{2-x}{3}-1=\dfrac{x+5}{6}\\ \Leftrightarrow\dfrac{3x-9}{6}-\dfrac{4-2x}{6}-\dfrac{6}{6}=\dfrac{x+5}{6}\\ \Rightarrow3x-9-4+2x-6=x+5\\ \Leftrightarrow5x-19=x+5\\ \Leftrightarrow4x=24\\ \Rightarrow x=6\\ \Rightarrow S=\left\{6\right\}\)
\(c.\dfrac{x+5}{2010}+\dfrac{x+4}{2011}+\dfrac{x+3}{2012}+\dfrac{x+2}{2013}=-4\\ \Leftrightarrow\dfrac{x+5}{2010}+1+\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1+\dfrac{x+2}{2013}+1=-4+4\\ \Rightarrow\dfrac{2015+x}{2010}+\dfrac{2015+x}{2011}+\dfrac{2015+x}{2012}+\dfrac{2015+x}{2013}=0\\ \Leftrightarrow\left(2015+x\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)=0\)
Do \(\dfrac{1}{2010}+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}>0\)
nên \(2015+x=0\Rightarrow x=-2015\)
Câu d tương tự...thêm rồi chuyển vế sang :v
\(\frac{x-3}{2011}+\frac{x-5}{2009}+\frac{x-7}{2007}+\frac{x-9}{2005}=4\)
\(\Leftrightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)+\left(\frac{x-9}{2005}-1\right)=0\)
\(\Leftrightarrow\frac{x-2014}{2011}+\frac{x-2014}{2009}+\frac{x-2014}{2007}+\frac{x-2014}{2005}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2009}+\frac{1}{2007}+\frac{1}{2005}\right)=0\)
|________________A________________|
Do A > 0
nên x - 2014 = 0
<=> x = 2014