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17 tháng 3 2023
=>3x+3+5x-5=3x^2-3
=>3x^2-3=8x-2
=>3x^2-8x-1=0
=>\(x=\dfrac{4\pm\sqrt{19}}{3}\)
18 tháng 7 2023
\(x^2+2x-10=0\)
\(\Leftrightarrow x^2+2x+1-9=0\)
\(\Leftrightarrow\left(x+1\right)^2-9=0\\\)
\(\Leftrightarrow\left(x+1\right)^2=9\)
\(\Leftrightarrow\left(x+1\right)^2=\pm\sqrt{9}\)
\(\Leftrightarrow\left(x+1\right)^2=\left(\pm3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3-1\\x=-3-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
Vậy S={2;-4}
NT
0
HT
1
27 tháng 6 2017
Ta có : 9(x + 1)2 - (3x - 2)(3x + 2) = 10
=> 9(x2 + 2x + 1) - (9x2 - 4) = 10
=> 9x2 + 18x + 9 - 9x2 + 4 = 10
=> 9x2 - 9x2 + 18x + 13 = 10
=> 18x = 10 - 13
=> 18x = -3
=> x = \(-\frac{1}{6}\)
HT
1
27 tháng 6 2017
Ta có : 9(x + 1)2 - (3x - 2)(3x + 2) = 10
=> 9(x2 + 2x + 1) - (9x2 - 4) = 10
=> 9x2 + 18x + 9 - 9x2 + 4 = 10
=> 9x2 - 9x2 + 18x + 13 = 10
=> 18x = 10 - 13
=> 18x = -3
=> x = \(-\frac{1}{6}\)
TN
1
26 tháng 2 2022
\(\dfrac{1}{\left(x+2000\right)\left(x+2001\right)}+\dfrac{1}{\left(x+2001\right)\left(x+2002\right)}+...+\dfrac{1}{\left(x+2009\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2001}+\dfrac{1}{x+2001}-\dfrac{1}{x+2002}+...+\dfrac{1}{x+2009}-\dfrac{1}{x+2010}=\dfrac{10}{11}\)
\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{x+2010-x-2000}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\)
\(\Leftrightarrow\dfrac{1}{x+2000}-\dfrac{1}{x+2010}=\dfrac{10}{11}\\ \Leftrightarrow\dfrac{10}{\left(x+2000\right)\left(x+2010\right)}=\dfrac{10}{11}\\ \Leftrightarrow\left(x+2000\right)\left(x+2010\right)=11\\ \Leftrightarrow...\)
DT
0
\(\frac{x-1}{2018}+\frac{x-10}{2009}+\frac{x-19}{2000}=3\)
\(\frac{x-1}{2018}+\frac{x-10}{2009}+\frac{x-19}{2000}-3=0\)
\(\left(\frac{x-1}{2018}-1\right)+\left(\frac{x-10}{2009}-1\right)+\left(\frac{x-19}{2000}-1\right)=0\)
\(\frac{x-1-2018}{2018}+\frac{x-10-2009}{2009}+\frac{x-19-2000}{2000}=0\)
\(\frac{x-2019}{2018}+\frac{x-2019}{2009}+\frac{x-2019}{2000}=0\)
\(\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2009}+\frac{1}{2000}\right)=0\)
Vì \(\left(\frac{1}{2018}+\frac{1}{2009}+\frac{1}{2000}\right)\ne0\)do đó :
\(x-2019=0\)
\(x=2019\)
\(\frac{x-1}{2018}+\frac{x-10}{2009}+\frac{x-19}{2000}=3.\)
\(\Leftrightarrow\frac{x-1}{2018}-1+\frac{x-10}{2009}-1+\frac{x-19}{2000}-1=0\)
\(\Leftrightarrow\frac{x-2019}{2018}+\frac{x-2019}{2009}+\frac{x-2019}{2000}=0\)
\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2009}+\frac{1}{2000}\right)=0\)
\(\Leftrightarrow x-2019=0\Leftrightarrow x=2019\)