\(\dfrac{x-5}{1990}\)+\(\dfrac{x-15}{1980}\)=\(\dfrac{x-1990}{5}\)+\(\dfrac{x-1980}{15}\)
Tìm x biết:
\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}=\dfrac{x-1980}{15}+\dfrac{x-1990}{5}\)
\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}=\dfrac{x-1980}{15}+\dfrac{x-1990}{5}\)
\(\Leftrightarrow(\dfrac{x-5}{1990}-1)+(\dfrac{x-15}{1980}-1)=(\dfrac{x-1980}{15}-1)+(\dfrac{x-1990}{5}-1)\)
\(\Leftrightarrow\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}-\dfrac{x-1995}{15}-\dfrac{x-1995}{5}=0\)
\(\Leftrightarrow\left(x-1995\right)\left(\dfrac{1}{1990}+\dfrac{1}{1980}-\dfrac{1}{15}-\dfrac{1}{5}\right)=0\)
\(\Leftrightarrow x-1995=0\)
\(\Leftrightarrow x=1995\)
mọi người ơi giúp mình câu giải phương trình này với:
\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}+\dfrac{x-25}{1970}+\dfrac{x-1990}{5}+\dfrac{x-1980}{15}+\dfrac{x-1970}{25}=0\)
\(\Leftrightarrow\left(\dfrac{x-5}{1990}-1\right)+\left(\dfrac{x-15}{1980}-1\right)+\left(\dfrac{x-25}{1970}-1\right)\\ +\left(\dfrac{x-1990}{5}-1\right)+\left(\dfrac{x-1980}{15}-1\right)+\left(\dfrac{x-1970}{25}-1\right)=0\\ \Leftrightarrow\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}+\dfrac{x-1995}{1970}+\dfrac{x-1995}{5}\\ +\dfrac{n-1995}{15}+\dfrac{n-1995}{25}=0\\ \Rightarrow\left(x-1995\right)\left(\dfrac{1}{1990}+\dfrac{1}{1980}+\dfrac{1}{1970}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}\right)=0\)
\(\Rightarrow x-1995=0\\ \Rightarrow x=1995\)
giải phương trình:
\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}+\dfrac{x-25}{1970}=\dfrac{x-1990}{5}+\dfrac{x-1980}{15}+\dfrac{x-1970}{25}\)
Ta có: \(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}\)
\(\Leftrightarrow\)\(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}-3=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}-3\)
\(\Leftrightarrow\)\(\frac{x-5}{1990}-1+\frac{x-15}{1980}-1+\frac{x-25}{1970}-1=\frac{x-1990}{5}-1+\frac{x-1980}{15}-1+\frac{x-1970}{25}-1\)\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)
\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)
\(\Leftrightarrow\)\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)
\(\Leftrightarrow\)\(x-1995=0\)
\(\Leftrightarrow\)\(x=1995\)
giải các phương trình sau:
1) \(\dfrac{x-11}{111}+\dfrac{x-12}{112}=\dfrac{x-23}{123}+\dfrac{x-24}{124}\)
2) \(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}=\dfrac{x-1980}{15}+\dfrac{x-1990}{5}\)
3) \(\dfrac{109-x}{91}+\dfrac{107-x}{93}+\dfrac{105-x}{95}+\dfrac{103-x}{97}=-4\)
\(\frac{x-5}{1990}+\frac{x-15}{1980}=\frac{x-1980}{15}+\frac{x-1990}{5}\)
pt <=> (x-5/1990 - 1) + (x-15/1980 - 1) = (x-1980/15 - 1) + (x-1990/5 - 1)
<=> x-1995/1990 + x-1995/1980 = x-1995/15 + x-1995/5
<=> x-1995/15 + x-1995/5 - x-1995/1990 - x-1995/1980 = 0
<=> (x-1995).(1/5+1/15-1/1990-1/1980) = 0
<=> x-1995 = 0 ( vì 1/5 + 1/15 - 1/1990 - 1/1980 > 0 )
<=> x = 1995
Vậy S={1995}
Tk mk nha
Ta có :
\(\frac{x-5}{1990}+\frac{x-15}{1980}=\frac{x-1980}{15}+\frac{x-1990}{5}\)
\(\Leftrightarrow\)\(\left(\frac{x-5}{1990}-1\right)+\left(\frac{x-15}{1980}-1\right)=\left(\frac{x-1980}{15}-1\right)+\left(\frac{x-1990}{5}-1\right)\)
\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}=\frac{x-1995}{15}+\frac{x-1995}{5}\)
\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)
\(\Leftrightarrow\)\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{15}+\frac{1}{5}\right)=0\)
Vì \(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{15}+\frac{1}{5}\ne0\)
Nên \(x-1995=0\)
\(\Rightarrow\)\(x=1995\)
Vậy \(x=1995\)
Chúc bạn học tốt ~
\(\Leftrightarrow\left(\frac{x-5}{1990}-1\right)+\left(\frac{x-15}{1980}-1\right)=\left(\frac{x-1980}{15}-1\right)+\left(\frac{x-1990}{5}-1\right)\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}=\frac{x-1995}{15}+\frac{x-1995}{5}\)
\(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\right)=0\)
\(\Leftrightarrow x-1995=0\)
\(\Leftrightarrow x=1995\)
Vậy pt có No Ià x=1995
\(\dfrac{x+3}{1980}+\dfrac{x+4}{2006}>\dfrac{x+1}{2009}+\dfrac{x+5}{2005}\)\(\dfrac{x+3}{1980}+\dfrac{x+4}{2006}>\dfrac{x+1}{2009}+\dfrac{x+5}{2005}\)
Tìm x:
d) x-5/1990 + x+5/1980 + x-25/1970=x-1990/5 + x-1980/15
d) x-5/1990 + x+5/1980 + x-25/1970=x-1990/5 + x-1980/15
\(\Leftrightarrow\left(\frac{x-5}{1990}-1\right)+\left(\frac{x-15}{1980}-1\right)+\left(\frac{x-25}{1970}-1\right)=\left(\frac{x-1990}{5}-1\right)+\left(\frac{x-1980}{15}-1\right)+\left(\frac{x-1970}{25}-1\right)\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\).
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}=0\)
\(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}+\frac{1}{15}+\frac{1}{25}\right)=0\)
\(\Leftrightarrow x-1995=0\).Do \(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}+\frac{1}{15}+\frac{1}{25}\ne0\)
\(\Leftrightarrow x=1995\)
\(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}+\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}=0\)
Tìm x
59-x/41 + 57-x/43 + 55-x/45 + 51-x/49 = -5
x-5/1990 + x-15/1980 + x-25/1970 = x-1990/5 + x-1980/15 + x-1970/25