a) TH1 : \(x-1=0\)
\(\Rightarrow x=1\)
TH2 : \(x-1\ne0\)
\(\Rightarrow5x\left(x-1\right)=1.\left(x-1\right)\)
\(5x=1\)
\(x=\frac{1}{5}\)
Vậy ...
b) \(2\left(x+5\right)-x^2-5x=0\)
\(2\left(x+5\right)-\left(x^2+5x\right)=0\)
\(2\left(x+5\right)-x\left(x+5\right)=0\)
\(\left(2-x\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+5=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)
a) 5x(x - 1) = x - 1
=> 5x(x - 1)
b) 2(x + 5) - x2 - 5x = 0
2(x + 5) + (-x2 - 5x) = 0
=> 2(x + 5) - x(x + 5) = 0
=> (x + 5) (2 - x) = 0
=> x + 5 = 0 => x = -5
=> 2 - x = 0 => x = 2
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