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\(\Leftrightarrow x\left(x+2\right)-x\left(x+3\right)=0.\)
\(\Leftrightarrow x\left(x+2-x-3\right)=0\)
\(\Leftrightarrow-x=0\)
\(\Leftrightarrow x=0\)
(x-1)/2015 + x/2014 + 1/503 - (x-3)/2013 - x/2012 - 1/1007 =0
(x-2016)/2015 + (x-2016)/2014 - (x-2016)/2012 - (x-2016)/2013 = 0
(x-2016) ( 1/2015 + 1/2016 - 1/2013 - 1/2012) = 0
Mà 1/2015 + 1/2016 - 1/2013 - 1/2012 khác 0
Suy ra x -2016=0
x=2016
Chỗ nào thắc mắc nhớ hỏi mik nhe!
\(Đk:\) \(x\ne1,x\ne2,x\ne3\)
\(\Rightarrow\dfrac{x+4}{\left(x-2\right)\left(x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(x-1\right)}=\dfrac{2x+5}{\left(x-3\right)\left(x-1\right)}\)
\(\Rightarrow\dfrac{\left(x+4\right)\cdot\left(x-3\right)+\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)\left(x-3\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(x-3\right)\left(x-1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Rightarrow0x-14=x-10\)
\(\Rightarrow x=-4\left(tmđk\right)\)
1) \(\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3-2\right)\left(x-3+2\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
2) \(x^2-2x=24\)
\(\Leftrightarrow x^2-2x-24=0\)
\(\Leftrightarrow x^2+4x-6x-24=0\)
\(\Leftrightarrow x\left(x+4\right)-6\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)
\(\left(3x-1\right)\left(2x-3\right)\left(2x-3\right)\left(x+5\right)=0\)
Th1 : \(3x-1=0=>x=\frac{1}{3}\)
Th2 : \(2x-3=0=>x=\frac{3}{2}\)
TH3 : \(x+5=0=>x=-5\)
Mik tl mà chẳng có ai T kì quá z
\(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\\ \Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\\ \Leftrightarrow3x^2-12x+12+9x-9-3x^2-3x+9=0\\ \Leftrightarrow-6x+12=0\\ \Leftrightarrow x=2\)
\(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)
\(\Leftrightarrow3x^2-12x+12+9x-9-3x^2-2x+9=0\)
\(\Leftrightarrow-6x-6=0\)
\(\Leftrightarrow-6\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy phương trình có nghiệm là \(-1\)