a )
\(5x\left(x-3\right)+7\left(x-3\right)=0\)
\(\Rightarrow\left(5x+7\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x+7=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-7\\x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{7}{5}\\x=3\end{cases}}}\)
Vậy ...
b )
\(x^{2017}=x^{2018}\)
\(\Rightarrow x^{2017}-x^{2018}=0\)
\(\Rightarrow x^{2017}\left(1-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^{2017}=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy ...
c )
\(2x^2=x\)
\(\Rightarrow2x^2:x=1\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy ...
e )
\(x^5=x^4\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}\)( làm tương tự như phần b )
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d )
\(x^3+x^2=0\)
\(\Rightarrow x^2\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
Vậy ...
Trả lời
\(a)5x\left(x-3\right)+7\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x+7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\5x+7=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0+3\\5x=-7\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=\frac{-7}{5}\end{cases}}}\)
Vậy x=3; x=\(\frac{-7}{5}\)
\(b)x^{2017}=x^{2018}\)
\(\Leftrightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left[0;1\right]\)
\(c)2x^2=x\)
\(\Leftrightarrow x=0\)
Vậy x=0
\(d)x^3+x^2=0\)
Vì \(x^2\ge0\)
Để \(x^3+x^2=0\Leftrightarrow x^3=x^2\left(=0\right)\)
\(\Leftrightarrow x=0\)
Vậy x=0
\(e)x^5=x^4\)
\(\Leftrightarrow x\in\left\{0;1\right\}\)
Vậy \(x\in\left\{0;1\right\}\)
