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Bài 15:
Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)
\(\Leftrightarrow\frac{x+2+2008}{2008}+\frac{x+3+2007}{2007}+\frac{x+4+2006}{2006}+\frac{x+2028-18}{6}=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}>0\)
nên x+2010=0
hay x=-2010
Vậy: x=-2010
Bài 17:
Ta có: \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+1+65}{65}+\frac{x+3+63}{63}=\frac{x+5+61}{61}+\frac{x+7+59}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
Vì \(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)
nên x+66=0
hay x=-66
Vậy: x=-66
a) Ta có: \(x\left(x+1\right)-\left(x+2\right)\left(x-3\right)=7\)
\(\Leftrightarrow\) \(x^2+x-\left(x^2-x-6\right)=7\)
\(\Leftrightarrow\) \(x^2+x-x^2+x+6=7\)
\(\Leftrightarrow\) \(2x+6=7\)
\(\Leftrightarrow\) \(2x=1\) \(\Leftrightarrow\) \(x=\frac{1}{2}\)(thỏa mãn)
Vậy phương trình có nghiệm duy nhất \(x=\frac{1}{2}\)
b) Ta có : \(\frac{x-3}{x+1}=\frac{x^2}{x^2-1}\) (1)
Điều kiện xác định của phương trình là \(x\) \(\ne\)\(1\) và \(x\) \(\ne\) \(-1\)
Khi đó (1) trở thành: \(\frac{\left(x-3\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{x^2}{\left(x+1\right)\left(x-1\right)}\)
\(\Leftrightarrow\) \(\left(x-3\right)\left(x-1\right)=x^2\) (Vì \(\left(x+1\right)\left(x-1\right)\ne0\))
\(\Leftrightarrow\) \(x^2-4x+3=x^2\)
\(\Leftrightarrow\) \(3-4x=0\)
\(\Leftrightarrow\) \(4x=3\) \(\Leftrightarrow\) \(x=\frac{3}{4}\)(thỏa mãn)
Vậy phương trình có nghiệm duy nhất \(x=\frac{3}{4}\)
a) x - 1 / x + 1 / x + 1 = 2x - 1/x^2 + x
ĐKXĐ:x khác 0;-1
<=>(x - 1)(x + 1) / x(x + 1) + x / x(x+1) = 2x + 1 / x(x + 1)
=>(x - 1)(x + 1) + x = 2x + 1
<=>x^2 - 1^2 + x = 2x + 1
<=>x^2 + x - 2x - 1 - 1 = 0
<=>x^2 - x - 2 = 0
<=>x^2 + x - 2x - 2 = 0
<=>x(x + 1) - 2(x + 1) = 0
<=>(x - 2)(x + 1) = 0
<=>x - 2 = 0
x + 1 = 0
<=>x = 2(t/m)
x = -1(loại vì ĐKXĐ: x khác 0;-1)
Vậy S = {2}
a)
PT <=> \(\left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)
<=> \(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)
<=> \(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\)
Mà \(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\ne0\)
<=> x - 2013 = 0
<=> x = 2013
KL: ...
b) PT <=> \(\left(x^4-5x^3\right)+\left(5x^3-25x^2\right)-\left(5x^2-25x\right)+\left(6x-30\right)=0\)
<=> \(x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)
<=> \(\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)
<=> \(\left(x-5\right)\left[\left(x^3+6x^2\right)-\left(x^2+6x\right)+\left(x+6\right)\right]=0\)
<=> \(\left(x-5\right)\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]=0\)
<=> \(\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)=0\)
<=> \(\left[{}\begin{matrix}x=5\\x=-6\\x=\varnothing\end{matrix}\right.\)
KL: ...
a) Ta có: \(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)
\(\Leftrightarrow\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}-2012=0\)
\(\Leftrightarrow\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1=0\)
\(\Leftrightarrow\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1\right)=0\)
mà \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+1>0\)
nên x-2013=0
hay x=2013
Vậy: Tập nghiệm S={2013}
b) Ta có: \(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+6x-5x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+6\right)-5\left(x+6\right)\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)(1)
Ta có: \(x^2-x+1\)
\(=x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)
hay \(x^2-x+1>0\forall x\)(2)
Từ (1) và (2) suy ra (x+6)(x-5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x+6=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=5\end{matrix}\right.\)
Vậy: Tập nghiệm S={-6;5}
\(\frac{x}{2008}+\frac{x+1}{2009}+...+\frac{x+4}{2012}=5\)
\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+...+\left(\frac{x+4}{2012}-1\right)=0\)
\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+...+\frac{x-2008}{2012}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)=0\)
Mà \(\left(\frac{1}{2008}+\frac{1}{2009}+..+\frac{1}{2012}\right)\ne0\)
Nên \(x-2008=0\)
\(\Leftrightarrow x=2008\)
Vậy : \(x=2008\)
\(\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}=5\)
\(\Leftrightarrow\frac{x}{2008}+\frac{x+1}{2009}+\frac{x+2}{2010}+\frac{x+3}{2011}+\frac{x+4}{2012}-5=0\)
\(\Leftrightarrow\left(\frac{x}{2008}-1\right)+\left(\frac{x+1}{2009}-1\right)+\left(\frac{x+2}{2010}-1\right)+\left(\frac{x+3}{2011}-1\right)+\left(\frac{x+4}{2012}-1\right)=0\)
\(\Leftrightarrow\frac{x-2008}{2008}+\frac{x-2008}{2009}+\frac{x-2008}{2010}+\frac{x-2008}{2011}+\frac{x-2008}{2012}=0\)
\(\Leftrightarrow\left(x-2008\right)\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\ne0\)
\(\Rightarrow x-2008=0\)\(\Leftrightarrow x=2008\)
Vậy \(x=2008\)
`(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}`.
\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}=\dfrac{x-4}{2010}+\dfrac{x-5}{2009}+\dfrac{x-6}{2008}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2013}-1\right)+\left(\dfrac{x-2}{2012}-1\right)+\left(\dfrac{x-3}{2011}-1\right)=\left(\dfrac{x-4}{2010}-1\right)+\left(\dfrac{x-5}{2009}-1\right)+\left(\dfrac{x-6}{2008}-1\right)\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}=\dfrac{x-2014}{2010}+\dfrac{x-2014}{2009}+\dfrac{x-2014}{2008}\)
\(\Leftrightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}-\dfrac{x-2014}{2010}-\dfrac{x-2014}{2009}-\dfrac{x-2014}{2008}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}-\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
\(\Leftrightarrow\left(x-2014\right).A=0\)
\(\text{Vì A }\ne0\)
\(\Rightarrow x-2014=0\)
\(\Leftrightarrow x=2014\)
\(\text{Vậy phương trình có tập nghiệm là }S=\left\{2014\right\}\)
\(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Rightarrow\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2010+18}{6}=0\)
\(\Rightarrow\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2010}{6}+3=0\)
\(\Rightarrow\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)+\left(\frac{x+4}{2006}+1\right)+\left(\frac{x+2010}{6}\right)=0\)
\(\Rightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+6\right)=0\)
Vì :\(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)
=> x + 2010 = 0
=> x = -2010
b) \(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)
\(\Rightarrow\frac{x-3}{2011}+\frac{x-2}{2012}-\frac{x-2012}{2}-\frac{x-2011}{3}=0\)
\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)-\left(\frac{x-2012}{2}-1\right)-\left(\frac{x-2011}{3}-1\right)=0\)
\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}-\frac{x-2014}{2}-\frac{x-2014}{3}=0\)
\(\Rightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)=0\)
Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\ne0\)
=> x - 2014 = 0
=> x = 2014
c) \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Rightarrow\frac{x+1}{65}+\frac{x+3}{63}-\frac{x+5}{61}-\frac{x+7}{59}=0\)
\(\Rightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)-\left(\frac{x+5}{61}+1\right)-\left(\frac{x+7}{59}+1\right)=0\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Rightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
Vì :\(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)
=> x + 66 = 0
=> x = -66