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\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1998}=3\)
\(\Leftrightarrow\left(\frac{x-1}{2006}-1\right)+\left(\frac{x-10}{1997}-1\right)+\left(\frac{x-19}{1998}-1\right)=0\)
\(\Leftrightarrow\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1998}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)
Dễ thấy cái đằng sau luôn > 0 nên x-2007=0 <=> x=2007
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)
\(\Leftrightarrow\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1988}=0\)
\(\Leftrightarrow x=2007\)
✰ ღ๖ۣۜDαɾƙ ๖ۣۜBαηɠ ๖ۣۜSĭℓεηтღ✰
lắm tắt thế này đi thi ko đc điểm đâu nhóc =))
Đặt \(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}\left(1\right)\)
\(\left(1\right)\Leftrightarrow\frac{x-2007}{2006}=\frac{x-2007}{1997}=\frac{x-2007}{1998}=0\)
\(\Rightarrow x=2007\)
\(\)Sửa lại đề câu a:
\(a.\frac{x-13}{2006}+\frac{x-22}{1997}+\frac{x-21}{1998}=3\\ \Leftrightarrow\frac{x-13}{2006}-1+\frac{x-22}{1997}-1+\frac{x-21}{1998}-1=0\\\Leftrightarrow \frac{x-2019}{2006}+\frac{x-2019}{1997}+\frac{x-2019}{1998}=0\\ \Leftrightarrow\left(x-2019\right)\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1998}\right)=0\\\Leftrightarrow x-2019=0\left(Vi\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1998}\ne0\right)\\\Leftrightarrow x=2019\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{2019\right\}\)
Đặt \(y=x^2+x\) ta có:
\(y^2+4y=12\\\Leftrightarrow y^2+4y-12=0\\\Leftrightarrow y^2+4y+4-16=0\\ \Leftrightarrow\left(y+2\right)^2-4^2=0\\\Leftrightarrow \left(y+2-4\right)\left(y+2+4\right)=0\\ \Leftrightarrow\left(y-2\right)\left(y+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y-2=0\\y+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=2\\y=-6\end{matrix}\right.\)
Thay \(y=x^2+x\) vào ta có:
\(x^2+x=2\\ \Leftrightarrow x^2+x-2=0\\ \Leftrightarrow x^2-x+2x-2=0\\ \Leftrightarrow\left(x-1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(x^2+x=-6\\ \Rightarrow x^2+x+6\ge0\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{1;-2\right\}\)
a/Viết đề mà cx sai đc nữa: \(\left(\frac{x+2}{98}+1\right)\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)\left(\frac{x+5}{95}+1\right)\)
\(\Leftrightarrow\frac{x+100}{98}.\frac{x+100}{97}-\frac{x+100}{96}.\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)^2\left(\frac{1}{98.97}-\frac{1}{96.95}\right)=0\)
\(\Rightarrow x=-100\)
b/\(\Leftrightarrow\left(\frac{x+1}{1998}+1\right)+\left(\frac{x+2}{1997}+1\right)=\left(\frac{x+3}{1996}+1\right)+\left(\frac{x+4}{1995}+1\right)\)
\(\Leftrightarrow\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}=0\)
\(\Leftrightarrow\left(x+1999\right)\left(...\right)=0\Rightarrow x=-1999\)
b,\(\frac{x+1}{1998}+\frac{x+2}{1997}=\frac{x+3}{1996}+\frac{x+4}{1995}\)
=>\(\frac{x+1}{1998}+1\frac{x+2}{1997}+1=\frac{x+3}{1996}+1+\frac{x+4}{1995}+1\)
\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}=\frac{x+1999}{1996}+\frac{x+1999}{1995}\)
\(\Leftrightarrow\)\(\frac{x+1999}{1998}+\frac{x+1999}{1997}-\frac{x+1999}{1996}-\frac{x+1999}{1995}\)=0
\(\Leftrightarrow\)\(\left(x+1999\right)\left(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\right)\)=0
\(\Leftrightarrow\)x+1999=0(Vì \(\frac{1}{1998}+\frac{1}{1997}-\frac{1}{1996}-\frac{1}{1995}\ne0\))
\(\Leftrightarrow\)x=-1999
Vậy x=-1999
a, \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
\(\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
\(\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
\(=>x+36=0\)
\(=>x=36\)
b, Ta có : \(\frac{x-10}{1994}+\frac{x-8}{1996}+\frac{x-6}{1994}+\frac{x-4}{2000}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2000}{4}+\frac{x-1998}{6}+\frac{x-1996}{8}+\frac{x-1994}{10}\)
=> \(\frac{x-10}{1994}-1+\frac{x-8}{1996}-1+\frac{x-6}{1994}-1+\frac{x-4}{2000}-1+\frac{x-2}{2002}-1=\frac{x-2002}{2}-1+\frac{x-2000}{4}-1+\frac{x-1998}{6}-1+\frac{x-1996}{8}-1+\frac{x-1994}{10}-1\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1994}+\frac{x-2004}{2000}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{4}+\frac{x-2004}{6}+\frac{x-2004}{8}+\frac{x-2004}{10}\)
=> \(\frac{x-2004}{1994}+\frac{x-2004}{1996}+\frac{x-2004}{1994}+\frac{x-2004}{2000}+\frac{x-2004}{2002}-\frac{x-2004}{2}-\frac{x-2004}{4}-\frac{x-2004}{6}-\frac{x-2004}{8}-\frac{x-2004}{10}=0\)
=> \(\left(x-2004\right)\left(\frac{1}{1994}+\frac{1}{1996}+\frac{1}{1998}+\frac{1}{2000}+\frac{1}{2002}-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)=0\)
=> \(x-2004=0\)
=> \(x=2004\)
Vậy phương trình có tập nghiệm là \(S=\left\{2004\right\}\)
a) Sửa đề: \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
Ta có: \(\frac{x+1}{35}+\frac{x+3}{33}=\frac{x+5}{31}+\frac{x+7}{29}\)
\(\Leftrightarrow\frac{x+1}{35}+1+\frac{x+3}{33}+1=\frac{x+5}{31}+1+\frac{x+7}{29}+1\)
\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}=\frac{x+36}{31}+\frac{x+36}{29}\)
\(\Leftrightarrow\frac{x+36}{35}+\frac{x+36}{33}-\frac{x+36}{31}-\frac{x+36}{29}=0\)
\(\Leftrightarrow\left(x+36\right)\left(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\right)=0\)
Vì \(\frac{1}{35}+\frac{1}{33}-\frac{1}{31}-\frac{1}{29}\ne0\)
nên x+36=0
hay x=-36
Vậy: x=-36
Bạn sửa đề lại nha.
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)
=>\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}-3=0\)
=>\(\left(\frac{x-1}{2006}-1\right)+\left(\frac{x-10}{1997}-1\right)+\left(\frac{x-19}{1988}-1\right)=0\)
=>\(\frac{x-1-2006}{2006}+\frac{x-10-1997}{1997}+\frac{x-19-1988}{1988}=0\)
=>\(\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1988}=0\)
=>\(\left(x-2007\right).\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)
Vì \(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\ne0\)
=>x-2007=0
=>x=2007