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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?
`(x-1)/2013+(x-2)/2012+(x-3)/2011=(x-4)/2010+(x-5)/2009 +(x-6)/2008`
`<=> ((x-1)/2013-1)+((x-2)/2012-1)+((x-3)/2011-1)=( (x-4)/2010-1)+((x-5)/2009-1)+((x-6)/2008-1)`
`<=> (x-2014)/2013 +(x-2014)/2012+(x-2014)/2011=(x-2014)/2010+(x-2014)/2009+(x-2014)/2008`
`<=> x-2014=0` (Vì `1/2013+1/2012+1/2011-1/2010-1/2009-1/2008 \ne 0`)
`<=>x=2014`
Vậy `S={2014}`.
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2013}-1\right)+\left(\dfrac{x-2}{2012}-1\right)+\left(\dfrac{x-3}{2011}-1\right)=\left(\dfrac{x-4}{2010}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-6}{2008}-1\right)\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
\(\Leftrightarrow\left(x-2014\right).A=0\)
\(\text{Vì A }\ne0\)
\(\Rightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2014\right\}\)
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
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!
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+4}{2006}+\dfrac{x+2028}{6}=0\)
\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+4}{2006}+1\right)+\left(\dfrac{x+2028}{6}-3\right)=0\)
\(\Leftrightarrow\)\(\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2006}+\dfrac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}=0\right)\)
\(\Leftrightarrow x+2010=0\) vì \(\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}+\dfrac{1}{6}>0\right)\)
=> x=-2010
vậy.....
\(\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}=-4\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+4=0\\ \Rightarrow\dfrac{x+2}{2008}+\dfrac{x+3}{2007}+\dfrac{x+5}{2005}+\dfrac{x+4}{2006}+1+1+1+1=0\\ \Rightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}+\dfrac{x+2010}{2005}+\dfrac{x+2010}{2006}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2005}+\dfrac{1}{2006}\right)=0\)
mà `1/2008+1/2007+1/2005+1/2006≠ 0`
`=> x+2010=0`
`=>x=-2010`
\(\Leftrightarrow\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+4}{2006}+1\right)=0\)
=>x+2010=0
=>x=-2010
\(\dfrac{2-x}{2007}\) - 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
\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\\ \Leftrightarrow\dfrac{2009-x}{2007}-2=\dfrac{2009-x}{2008}-\dfrac{2009-x}{2009}-2\)
\(\Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}+\dfrac{1}{2009}\right)=0\)
\(\Rightarrow2009-x=0\Leftrightarrow x=2009\)
\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}=\dfrac{x+2007}{3}+\dfrac{x+2006}{4}\)
\(\Leftrightarrow\dfrac{x+1}{2009}+1+\dfrac{x+2}{2008}+1=\dfrac{x+2007}{3}+1+\dfrac{x+2006}{4}+1\)
\(\Leftrightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}=\dfrac{x+2010}{3}+\dfrac{x+2010}{4}\)
\(\Rightarrow x+2010=0\)
\(\Rightarrow x=-2010\)
Vậy pt có nghiệm duy nhất \(x=-2010\)
giải phương trình
x^2+x-2=0
vậy kết quả bằng mấy vậy