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1. \(\Leftrightarrow\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{51-x}{49}+1=-5+5\)
\(\Leftrightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)
\(\Leftrightarrow x-100=0\Leftrightarrow x=100\)
2. \(\Leftrightarrow\frac{x-5}{1990}+1+\frac{x-15}{1980}+1+\frac{x-25}{1970}=\frac{x-1990}{5}+1+\frac{x-1980}{15}+1+\frac{x-1970}{25}+1\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)
\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)
\(\Leftrightarrow\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)
\(\Leftrightarrow x-1995=0\Leftrightarrow x=1995\)
\(\dfrac{x-5}{1990}+\dfrac{x-15}{1980}=\dfrac{x-1990}{5}+\dfrac{x-1980}{15}\\ =>\dfrac{x-5}{1990}-1+\dfrac{x-15}{1980}-1=\dfrac{x-1990}{5}-1+\dfrac{x-1980}{15}-1\\ =>\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}-\dfrac{x-1995}{5}-\dfrac{x-1995}{15}=0\\ =>\left(x-1995\right).\left(\dfrac{1}{1990}+\dfrac{1}{1980}-\dfrac{1}{5}-\dfrac{1}{15}\right)=0\\ =>x-1995=0\\ =>x=1995\)
P/s: Chuyển tất cả các hạng tử sang 1 vế rồi cộng thêm 1 vào các vế có dấu (+) đằng trước, cộng thêm -1 vào các hạng tử có dấu (-) phía trước rồi đặt nhân tử chung ra ngoài ta được:
\(Pt\Leftrightarrow\left(x-2004\right)\left(\frac{1}{1979}-\frac{1}{1980}-\frac{1}{1981}-\frac{1}{1982}-\frac{1}{25}+\frac{1}{24}+\frac{1}{23}+\frac{1}{22}\right)=0\)
\(\Leftrightarrow x-2004=0\)
\(\Rightarrow x=2004\)
Vậy x = 2004
https://olm.vn/hoi-dap/detail/263823966145.html?pos=616279814817
a)\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1980}{1975}\right|+\left|z-2004\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{1980}{1975}\right|=0\\\left|z-2004\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{19}{5}\\y=-\dfrac{1980}{1975}\\z=2004\end{matrix}\right.\)
b) \(\left|\dfrac{3}{4}+x\right|+\left|-\dfrac{1}{5}+y\right|+\left|x+y+z\right|\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left|\dfrac{3}{4}+x\right|=0\\\left|-\dfrac{1}{5}+y\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)
x−17/1990+x−21/1986+x+1 1004=4⇔x−17/1990−1+x−21/1986−1+x+1/1004−2=0
⇔x−2007/1990+x−2007/1986+x−2007/1004=0
⇔(x−2007)(1/1990+1/1986+1/1004)=0
⇔x−2007=0⇔x=2007
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 \(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}+\frac{1}{15}+\frac{1}{25}\ne0\)
\(\Leftrightarrow x=1995\)