NN
2
K
Khách
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D
2
MD
30 tháng 3 2017
\(\dfrac{2-x}{2007}\) - 1 = \(\dfrac{1-x}{2008}\) - \(\dfrac{x}{2009}\)
<=> \(\dfrac{2-x}{2009}\) +1 -1 +1 = \(\dfrac{1-x}{2008}\) +1 - \(\dfrac{x}{2009}\) +1
<=> \(\dfrac{2-x+2007}{2007}\) = \(\dfrac{1-x+2008}{2008}\) + \(\dfrac{-x+2009}{2009}\)
<=> \(\dfrac{2009-x}{2007}\) = \(\dfrac{2009-x}{2008}\) + \(\dfrac{2009-x}{2009}\)
<=> (2009-x)(\(\dfrac{1}{2007}\) - \(\dfrac{1}{2008}\) - \(\dfrac{1}{2009}\) ) = 0
<=> 2009 -x = 0
hoặc: \(\dfrac{1}{2007}\) - \(\dfrac{1}{2008}\) -\(\dfrac{1}{2009}\) = 0
Vì \(\dfrac{1}{2007}\) \(\ne\) \(\dfrac{1}{2008}\) + \(\dfrac{1}{2009}\)
=> \(\dfrac{1}{2007}\) - (\(\dfrac{1}{2008}\) + \(\dfrac{1}{2009}\) ) \(\ne\) 0
=> 2009 -x =0
<=> x =2009
KT
30 tháng 3 2017
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\\ \Leftrightarrow\dfrac{2009-x}{2007}-2=\dfrac{2009-x}{2008}-\dfrac{2009-x}{2009}-2\)
\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\right)=0\)
\(\Rightarrow2009-x=0\Leftrightarrow x=2009\)
VM
3
ND
3 tháng 6 2018
\(x^4+4x^2-5=0\)
\(\Leftrightarrow x^4-x^2+5x^2-5=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)+5\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=0\left(l\right)\\x=1\\x=-1\end{matrix}\right.\)
ND
3 tháng 6 2018
\(4\left(x+5\right)-3\left|2x-1\right|=0\)
\(\Leftrightarrow3\left|2x-1\right|=4\left(x+5\right)\)
\(\Leftrightarrow\left|2x-1\right|=\dfrac{4}{3}\left(x+5\right)\left(ĐK:x\ge-5\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{4}{3}\left(x+5\right)\\2x-1=-\dfrac{4}{3}\left(x+5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{4}{3}x+\dfrac{20}{3}\\2x-1=-\dfrac{4}{3}x-\dfrac{20}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{23}{3}\\\dfrac{2}{3}x=-\dfrac{17}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{2}\left(l\right)\\x=-\dfrac{17}{10}\left(n\right)\end{matrix}\right.\)
Vậy: \(x=-\dfrac{17}{10}\)
DM
13 tháng 2 2018
\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}=\dfrac{x+2007}{3}+\dfrac{x+2006}{4}\)
\(\Leftrightarrow\dfrac{x+1}{2009}+1+\dfrac{x+2}{2008}+1=\dfrac{x+2007}{3}+1+\dfrac{x+2006}{4}+1\)
\(\Leftrightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}=\dfrac{x+2010}{3}+\dfrac{x+2010}{4}\)
\(\Rightarrow x+2010=0\)
\(\Rightarrow x=-2010\)
Vậy pt có nghiệm duy nhất \(x=-2010\)
24 tháng 1 2023
\(C=\dfrac{\sqrt{x}-\sqrt{x+1}}{-1}+\dfrac{\sqrt{x+1}-\sqrt{x+2}}{-1}+...+\dfrac{\sqrt{x+2007}-\sqrt{x+2008}}{-1}\)
\(=-\sqrt{x}+\sqrt{x+1}-\sqrt{x+1}+\sqrt{x+2}-...-\sqrt{x+2007}+\sqrt{x+2008}\)\(=-\sqrt{x}+\sqrt{x+2008}\)
\(C=-\sqrt{\sqrt[2007]{2008}}+\sqrt{\sqrt[2007]{2008}+2008}\)
AH
Akai Haruma
Giáo viên
28 tháng 1 2021
Lời giải:
a)
PT \(\Leftrightarrow \frac{(x+2)^3}{8}-\frac{x^3+8}{2}=0\)
\(\Leftrightarrow (x+2)^3-4(x^3+8)=0\)
\(\Leftrightarrow (x+2)^3-4(x+2)(x^2-2x+4)=0\)
\(\Leftrightarrow (x+2)[(x+2)^2-4(x^2-2x+4)]=0\)
\(\Leftrightarrow (x+2)(-3x^2+12x-12)=0\)
\(\Leftrightarrow (x+2)(x^2-4x+4)=0\Leftrightarrow (x+2)(x-2)^2=0\Rightarrow x=\pm 2\)
b) Bạn kiểm tra lại xem có sai đề không?
DT
1
7 tháng 3 2017
Không chẳng có vấn đề gì cả. có thể sai so với cái đề nào đó "nội hàm nó đúng"
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{-x+4}{2006}+\dfrac{-x-2008}{6}\)
\(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right).x=\left(\dfrac{4}{2006}-\dfrac{2008}{6}-\dfrac{2}{2008}-\dfrac{3}{2007}\right)\)\(x=\dfrac{\left(\dfrac{4}{2006}-\dfrac{2008}{6}-\dfrac{2}{2008}-\dfrac{3}{2007}\right)}{\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}\right).}\)
Thích thì rút gọn chẳng thích thì kệ nó
BB
6 tháng 3 2017
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
6 tháng 3 2017
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\dfrac{x-1}{2013}+1+\dfrac{x-2}{2012}+1+\dfrac{x-3}{2011}+1-\dfrac{x-4}{2010}+1-\dfrac{x-5}{2009}+1-\dfrac{x-6}{2008}+1=0\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\ne0\right)=0\)
\(\Leftrightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
Vậy PT có nghiệm là \(x=2014\)
Giải:
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\)
\(\Leftrightarrow\dfrac{2-x}{2007}-1+2=\dfrac{1-x}{2008}-\dfrac{x}{2009}+2\)
\(\Leftrightarrow\dfrac{2-x}{2007}+1=\dfrac{1-x}{2008}+1-\dfrac{x}{2009}+1\)
\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}-\dfrac{x+2009}{2009}\)
\(\Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}-\dfrac{2009-x}{2009}\)
\(\Leftrightarrow\dfrac{2009-x}{2007}-\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}=0\)
\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\right)=0\)
Vì \(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\ne0\)
\(\Leftrightarrow2009-x=0\)
\(\Leftrightarrow x=2009\)
Vậy ...
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\)
\(\Leftrightarrow\left(\dfrac{2-x}{2007}+1\right)-\left(1+1\right)=\left(\dfrac{1-x}{2008}+1\right)-\left(\dfrac{x}{2009}+1\right)\)
\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}-\dfrac{x+2009}{2009}\)
\(\Leftrightarrow\dfrac{2-x+2007}{2007}=\dfrac{1-x+2008}{2008}+\dfrac{-x+2009}{2009}\)
\(\Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}\)
\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2009}\right)=0\)
\(\Leftrightarrow2009-x=0\)
\(\Leftrightarrow x=2009\)