DD
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Khách
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DD
3
14 tháng 2 2019
\(\frac{x-4}{2000}+\frac{x-3}{2001}+\frac{x-2}{2002}=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)
\(\Rightarrow\left(\frac{x-4}{2000}-1\right)+\left(\frac{x-3}{2001}-1\right)+\left(\frac{x-2}{2002}-1\right)=\left(\frac{x-2002}{2}-1\right)+\left(\frac{x-2001}{3}-1\right)+\left(\frac{x-2000}{4}-1\right)\)\(\Rightarrow\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)
\(\Rightarrow\left(x-2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}\right)=\left(x-2004\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
Với \(x-2004\ne0\)
\(\Rightarrow\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\left(KTM\right)\)
Với \(x-2004=0\)
\(\Rightarrow x=2004\)
NL
0
Y
9 tháng 2 2023
\(\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}=\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}\) \(\left(dkxd:x\ne-2\right)\)
\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}-\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(3x+1\right)-x\right]=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(6x+2-x\right)=0\)
Bỏ vế đằng trước \(x^2+x+1=0\) do vô nghiệm
\(\Leftrightarrow6x+2-x=0\)
\(\Leftrightarrow5x=-2\)
\(\Leftrightarrow x=-\dfrac{2}{5}\left(tmdk\right)\)
Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)
BT
1
21 tháng 7 2018
\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
\(\Rightarrow3x^2-3^2-3x^2+15x=1\)
\(\Rightarrow3x^2-9-3x^2+15x=1\)
\(\Rightarrow-9+15x=1\)
\(\Rightarrow15x=-8\)
\(\Rightarrow x=\frac{-8}{15}\)
19 tháng 2 2021
\(\dfrac{x-2}{2001}+\dfrac{x}{2003}=1+\dfrac{1-x}{2002}\Leftrightarrow\dfrac{x-2}{2001}+\dfrac{x}{2003}-\dfrac{x-1}{2002}-1=0\)
\(\Leftrightarrow\dfrac{x-2}{2001}-1+\dfrac{x}{2003}-1-\dfrac{x-1}{2002}+1=0\)
\(\Leftrightarrow\dfrac{x-2003}{2001}+\dfrac{x-2003}{2003}-\left(\dfrac{x-2003}{2002}\right)=0\)
\(\Leftrightarrow\left(x-2003\right)\left(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}\right)=0\) \(\Leftrightarrow x=2003\) vì \(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}>0\)Vậy...
19 tháng 2 2021
Ta có: \(\dfrac{x-2}{2001}+\dfrac{x}{2003}=1+\dfrac{1-x}{2002}\)
\(\Leftrightarrow\dfrac{x-2}{2001}+\dfrac{x}{2003}-1+\dfrac{1-x}{2002}=0\)
\(\Leftrightarrow\dfrac{x-2}{2001}-1+\dfrac{x}{2003}-1+\dfrac{1-x}{2002}+1=0\)
\(\Leftrightarrow\dfrac{x-2003}{2001}+\dfrac{x-2003}{2003}-\dfrac{x-2003}{2002}=0\)
\(\Leftrightarrow\left(x-2003\right)\left(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}\right)=0\)
mà \(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}\ne0\)
nên x-2003=0
hay x=2003
Vậy: S={2003}
NH
0