Giải phương trình :
a) \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
b) \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
c) \(6x^4-5x^3-38x^2-5x+6=0\)
K
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L
31 tháng 5 2017
câu 2 :
\(\Leftrightarrow\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}-\frac{x+4}{2005}-\frac{x+5}{2004}-\frac{x+6}{2003}\)=0
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x-2009}{2003}\)=0
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)
\(\Rightarrow x+2009=0\)
\(\Rightarrow x=-2009\)
12 tháng 7 2017
c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
Nên x + 2009 = 0 => x = -2009
QD
8 tháng 2 2017
b)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
<=>\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)\( \left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
mà \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
nên phương trình đó xảy ra khi và chỉ khi x+2009=0
<=>x=-2009
Vậy phương trình có no là x=-2009
D
8 tháng 2 2017
b) \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)\)=
\(\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\) \(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\)\(\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\) \(\left(x+2009\right)\)\(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)= 0
\(\Leftrightarrow\)\(x+2009=0\)
( vì \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\) \(\ne0\))
\(\Leftrightarrow\) \(x=-2009\)
Vậy x = -2009
NN
13 tháng 11 2020
\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+3=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}+3\)
\(\Leftrightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)\)
\(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)(1)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\ne0\)(2)
Từ (1) và (2) \(\Rightarrow x+2009=0\)\(\Rightarrow x=-2009\)
Vậy \(x=-2009\)
KH
11 tháng 7 2016
a) \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)
\(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)
\(\Leftrightarrow x+2010=0\) ( vì 1/2003 + 1/2006 -- 1/2011 -- 1/2015 \(\ne\)0)
\(\Leftrightarrow x=-2010\)
câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<
PN
30 tháng 3 2016
Câu \(1.\) Giải phương trình
\(a.\) \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\) \(\left(1\right)\)
Đặt \(y=x^2+x\) \(\left(2\right)\) thì khi đó, phương trình \(\left(1\right)\) sẽ có dạng:
\(y^2+4y=12\)
\(\Leftrightarrow\) \(y^2+4y-12=0\)
\(\Leftrightarrow\) \(y^2+4y+4-16=0\)
\(\Leftrightarrow\) \(\left(y+2\right)^2-4^2=0\)
\(\Leftrightarrow\) \(\left(y-2\right)\left(y+6\right)=0\)
\(\Leftrightarrow\) \(^{y-2=0}_{y+6=0}\) \(\Leftrightarrow\) \(^{y=2}_{y=-6}\)
Đến bước này, ta cần xét hai trường hợp sau:
\(\text{*)}\) \(TH_1:\) Với \(y=2\) thì phương trình \(\left(2\right)\) trở thành:
\(x^2+x=2\)
\(\Leftrightarrow\) \(x^2+x-2=0\)
\(\Leftrightarrow\) \(\left(x^2-1\right)+x-1=0\)
\(\Leftrightarrow\) \(\left(x-1\right)\left(x+1\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\) \(\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\) \(^{x-1=0}_{x+2=0}\) \(\Leftrightarrow\) \(^{x=1}_{x=-2}\) (dùng dấu ngoặc nhọn nhé bạn!)
\(\text{*)}\) \(TH_2:\) Với \(y=-6\) thì phương trình \(\left(2\right)\) trở thành:
\(x^2+x=-6\)
\(\Leftrightarrow\) \(x^2+x+6=0\)
\(\Leftrightarrow\) \(x^2+2.\frac{1}{2}.x+\frac{1}{4}+\frac{23}{4}=0\)
\(\Leftrightarrow\) \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\) \(\left(3\right)\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\) với mọi \(x\) \(\Rightarrow\) \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}>0\)
Do đó, phương trình \(\left(3\right)\) vô nghiệm!
Vậy, tập nghiệm của phương trình \(\left(1\right)\) là \(S=\left\{-1;2\right\}\)
PN
30 tháng 3 2016
Câu \(1.\) Giải phương trình!
\(b.\)
\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\) \(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\) \(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\) \(\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\) \(\left(4\right)\)
Do \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\) nên từ \(\left(4\right)\) suy ra
\(x+2009=0\) \(\Leftrightarrow\) \(x=-2009\)
Vậy, \(S=\left\{-2009\right\}\)
\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)
\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)
\(\text{Giải}\)
\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)