\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1998}=3\)
\(\Leftrightarrow\left(\frac{x-1}{2006}-1\right)+\left(\frac{x-10}{1997}-1\right)+\left(\frac{x-19}{1998}-1\right)=0\)
\(\Leftrightarrow\frac{x-2007}{2006}+\frac{x-2007}{1997}+\frac{x-2007}{1998}=0\)
\(\Leftrightarrow\left(x-2007\right)\left(\frac{1}{2006}+\frac{1}{1997}+\frac{1}{1988}\right)=0\)
Dễ thấy cái đằng sau luôn > 0 nên x-2007=0 <=> x=2007
Tìm x bt
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}=3\)
tìm x
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1988}\)
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
tìm x biết
\(\frac{x-1}{2006}+\frac{x-10}{1997}+\frac{x-19}{1998}=3\)
Giải PT\(\frac{x-29}{1986}+\frac{x-27}{1988}+\frac{x-25}{1990}+\frac{x-23}{1992}+\frac{x-21}{1994}+\frac{x-19}{1996}=\frac{x-1986}{29}+\frac{x-1988}{27}+\frac{x-1990}{25}+\frac{x-1992}{23}+\frac{x-1994}{21}+\frac{x-1996}{19}\)
Giải Phương trình
\(\left(2x-1\right)^3+\left(x+2\right)^3=\left(3x+1\right)^3\)
\(\frac{x-1988}{15}+\frac{x-1969}{17}+\frac{x-1946}{19}+\frac{x-1919}{21}=10\)
\(\frac{\left(2009-x\right)^2+\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}{\left(2009-x^2\right)-\left(2009-x\right)\left(x-2010\right)+\left(x-2010\right)^2}=\frac{19}{49}\)
\(\frac{x-1991}{9}+\frac{x-1993}{7}+\frac{x-1995}{5}+\frac{x-1997}{3}+\frac{x-1999}{1}=\frac{x-9}{1991}+\frac{x-7}{1993}+\frac{x-5}{1995}+\frac{x-3}{1997}+\frac{x-1}{1999}\)
Rút gọn các biểu thức:
\(\frac{x+1}{x^2-2x}.\frac{4-x}{x^2+x}\)
\(\frac{x^3}{x+2006}.\frac{2x+1975}{x+1}+\frac{x^3}{x+2006}.\frac{31-x}{x+1}\)
\(\frac{19x+8}{x-7}.\frac{5x-9}{x+2006}-\frac{19x+8}{x-7}.\frac{4x-2}{x+2006}\)
