Tìm x bt
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)
tìm x
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1998}=3\)
\(\Leftrightarrow\left(\frac{x-1}{2006}-1\right)+\left(\frac{x-10}{1997}-1\right)+\left(\frac{x-19}{1998}-1\right)=0\)
\(\Leftrightarrow\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1998}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)
Dễ thấy cái đằng sau luôn > 0 nên x-2007=0 <=> x=2007
tìm x
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}\)
Đặt \(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}\left(1\right)\)
\(\left(1\right)\Leftrightarrow\frac{x-2007}{2006}=\frac{x-2007}{1997}=\frac{x-2007}{1998}=0\)
\(\Rightarrow x=2007\)
Em kiểm tra lại đề bài nhé! Thiếu đề rồi.
tìm x thỏa mãn
\(\frac{x+1}{2006}\)+\(\frac{x-10}{1997}\)+\(\frac{x-19}{1988}\)= 3
tìm x:
x - \(\frac{1}{2006}\) + x - \(\frac{10}{1997}\) + x - \(\frac{19}{1988}\) = 3
tìm x, biết: (x-1)/2006+(x-10)/1997+(x-19)/1988=3
<=> (x-1/2006 - 1)+(x-10/1997 - 1)+(x-19/1988 - 1) = 0
<=> x-2007/2006 + x-2007/1997 + x-2007/1988 = 0
<=> (x-2007).(1/2006+1/1997+1/1988) = 0
<=> x-2007=0 ( vì 1/2006+1/1997+1/1988 > 0 )
<=> x=2007
Vậy x=2007
k mk nha
tìm x biết
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1998}=3\)
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
tim x thoa man
x-1\2006+x-10\1997+x-19\1988=3
\(\frac{x-1}{2016}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)
\(pt\Leftrightarrow\frac{x-1}{2016}-1+\frac{x-10}{1997}-1+\frac{x-19}{1988}-1=0\)
\(\Leftrightarrow\frac{x-2017}{2016}+\frac{x-2017}{1997}+\frac{x-2017}{1988}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)
Dễ thấy: \(\frac{1}{2016}+\frac{1}{1997}+\frac{1}{1988}\ne0\)
\(\Rightarrow x-2017=0\Rightarrow x=2017\)
\(\dfrac{x+1}{2006}+\dfrac{x+10}{1997}+\dfrac{x+19}{1988}=3\)
Tìm x nha
thanks you
Sửa đề :
\(\dfrac{x+1}{2006}+\dfrac{x+10}{1997}+\dfrac{x+19}{1988}=-3\)
\(\Leftrightarrow\left(\dfrac{x+1}{2006}+1\right)+\left(\dfrac{x+10}{1997}+1\right)+\left(\dfrac{x+19}{1988}+1\right)=0\)
\(\Leftrightarrow\dfrac{x+2007}{2006}+\dfrac{x+2007}{1997}+\dfrac{x+2007}{1988}=0\)
\(\Leftrightarrow\left(x+1007\right)\left(\dfrac{1}{2006}+\dfrac{1}{1997}+\dfrac{1}{1988}\right)=0\)
Mà \(\dfrac{1}{2006}+\dfrac{1}{1997}+\dfrac{1}{1988}\ne0\)
\(\Leftrightarrow x+2007=0\)
\(\Leftrightarrow x=-2007\)
Vậy..
Giải PT\(\frac{x-29}{1986}+\frac{x-27}{1988}+\frac{x-25}{1990}+\frac{x-23}{1992}+\frac{x-21}{1994}+\frac{x-19}{1996}=\frac{x-1986}{29}+\frac{x-1988}{27}+\frac{x-1990}{25}+\frac{x-1992}{23}+\frac{x-1994}{21}+\frac{x-1996}{19}\)
pạn -1 vào mỗi phân số là xong. Rùi ra x\(\frac{x-2015}{1986}\)+\(\frac{x-2015}{1988}\)+ \(\frac{x-2015}{1990}\)+...+\(\frac{x-2015}{x1996}\)-\(\frac{x-2015}{29}\)-\(\frac{x-2015}{27}\)-...\(\frac{x-2015}{19}\)=0
<=>(x-2015)(\(\frac{1}{1986}\)+\(\frac{1}{1988}\)+... -\(\frac{1}{19}\))=0...(mà \(\frac{1}{1986}\)+...- \(\frac{1}{19}\) khác 0)
=>x-2015=0
<=> x=2015