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\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)
\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)
\(\text{Giải}\)
\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)
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
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\)
\(\Rightarrow x+2009=0\)
\(\Rightarrow x=-2009\)
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
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
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 \(S=\left\{-1;2\right\}\)
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, \(S=\left\{-2009\right\}\)
Câu 6 :
a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)
=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)
=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)
=> \(15x+10x+x-1=15-9x+1-2x\)
=> \(15x+10x+x-1-15+9x-1+2x=0\)
=> \(37x-17=0\)
=> \(x=\frac{17}{37}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)
Bài 7 :
a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> \(x-23=0\)
=> \(x=23\)
Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)
c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x+2005=0\)
=> \(x=-2005\)
Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)
e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)
a ) \(4\left(x+5\right)-3\left|2x-1\right|=0\)
\(\Leftrightarrow3\left|2x-1\right|=4\left(x+5\right)\)
\(\Leftrightarrow\left|2x-1\right|=\frac{4}{3}\left(x+5\right)\left(ĐK:x\ge-5\right)\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=\frac{4}{3}\left(x+5\right)\\2x-1=-\frac{4}{3}\left(x+5\right)\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=\frac{4}{3}x+\frac{20}{3}\\2x-1=-\frac{4}{3}x-\frac{20}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x=-\frac{23}{3}\\\frac{2}{3}x=-\frac{17}{3}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{23}{2}\left(l\right)\\x=-\frac{17}{10}\left(n\right)\end{cases}}\)
Vậy \(x=-\frac{17}{10}\)
b ) \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)
\(\Leftrightarrow\frac{2-x}{2007}+1=\left(\frac{1-x}{2008}+1\right)+\left(1-\frac{x}{2009}\right)\)
\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}=\frac{2009-x}{2009}\)
\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)
\(\Leftrightarrow2009-x=0\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\right)\)
\(\Leftrightarrow x=2019\)
Vậy phương trình có nghiệm \(x=2019\)
c ) \(x^4+4x^2-5=0\)
\(\Leftrightarrow x^4-x^2+5x^2-5=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)+5\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2+5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+5=0\left(l\right)\\x=1\end{cases}}\)
\(x=-1\)
Vậy \(x=1\) hoặc \(x=-1\)
Chúc bạn học tốt !!!
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