Question

Tìm x:

a) 5x(x-1) = x-1

b) 2(x+5) - x² - 5x = 0

mong mọi người sẽ giúp đỡ

Answer 1

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) \(2\left(x+5\right)-x^2-5x=0\)

\(2\left(x+5\right)-\left(x^2+5x\right)=0\)

\(2\left(x+5\right)-x\left(x+5\right)=0\)

\(\left(2-x\right)\left(x+5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2-x=0\\x+5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

Answer 2

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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