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(x + 1)(x2 - x + 1) - x(x2 - 2) = 19
<=> x3 - x2 + x + x2 - x + 1 - x3 + 2x = 19
<=> 2x + 1 = 19
<=> 2x = 19 - 1
<=> 2x = 18
<=> x = 9
=> x = 9
\(\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2-2\right)=19\)
\(\Leftrightarrow x^3-x^2+x+x^2-x+1-x^3+2x=19\)
\(\Leftrightarrow2x+1=19\)
\(\Leftrightarrow2x=19-1\)
\(\Leftrightarrow2x=18\Leftrightarrow x=\frac{18}{2}=9\)
\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-19\)
\(\Rightarrow x^3+2x^2+x+x^2+2x+1-x\left(x^2-2x+1\right)+\left(x^2-2x+1\right)-6x^2+12x-6=-19\)
\(\Rightarrow x^3+2x^2+x+x^2+2x+1-x\left(x^2-2x+1\right)+x^2-2x+1-6x^2+12x-6=-19\)
\(\Rightarrow x^3-2x^2+13x-4-x\left(x^2-2x+1\right)=-19\)
\(\Rightarrow x^3-2x^2+13x-4-x^3+2x^2-x=-19\)
\(\Rightarrow12x-4=-19\)
\(\Rightarrow12x=-15\)
\(\Rightarrow x=\frac{-5}{4}\)
<=> (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
\(\frac{x+2}{10}+\frac{x+2}{13}+\frac{x+2}{16}+\frac{x+2}{19}=0\)
\(\Leftrightarrow\left(x+2\right)\left(\frac{1}{10}+\frac{1}{13}+\frac{1}{16}+\frac{1}{19}\right)=0\)
Mà \(\frac{1}{10}+\frac{1}{13}+\frac{1}{16}+\frac{1}{19}\ne0\)
\(\Rightarrow x+2=0\Rightarrow x=-2\)
Vậy \(x=-2\)
\(\frac{x-1}{2018}+\frac{x-10}{2009}+\frac{x-19}{2000}=3\)
\(\frac{x-1}{2018}+\frac{x-10}{2009}+\frac{x-19}{2000}-3=0\)
\(\left(\frac{x-1}{2018}-1\right)+\left(\frac{x-10}{2009}-1\right)+\left(\frac{x-19}{2000}-1\right)=0\)
\(\frac{x-1-2018}{2018}+\frac{x-10-2009}{2009}+\frac{x-19-2000}{2000}=0\)
\(\frac{x-2019}{2018}+\frac{x-2019}{2009}+\frac{x-2019}{2000}=0\)
\(\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2009}+\frac{1}{2000}\right)=0\)
Vì \(\left(\frac{1}{2018}+\frac{1}{2009}+\frac{1}{2000}\right)\ne0\)do đó :
\(x-2019=0\)
\(x=2019\)
\(\frac{x-1}{2018}+\frac{x-10}{2009}+\frac{x-19}{2000}=3.\)
\(\Leftrightarrow\frac{x-1}{2018}-1+\frac{x-10}{2009}-1+\frac{x-19}{2000}-1=0\)
\(\Leftrightarrow\frac{x-2019}{2018}+\frac{x-2019}{2009}+\frac{x-2019}{2000}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2009}+\frac{1}{2000}\right)=0\)
\(\Leftrightarrow x-2019=0\Leftrightarrow x=2019\)
x + 19 =1476
x = 1476 - 19
x = 1457
x - 1223 =432
x = 432 + 1223
x = 1655
k mình nha
\(x+19=123x12\)
\(x+19=1476\)
\(x=1476-19=1457\)