a, x.( x - 2 ) + 2x - 4 = 0
<=> (x-2)(x+2)=0
<=> x=2 V x=-2
b, 5x.(x - 3 ) - x + 3 = 0
<=> (x-3)(5x-1)=0
<=> x=3 V x=1/5
a ) \(x.\left(x-2\right)+2x-4=0\)
\(\Leftrightarrow x^2-2x+2x-4=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
b ) \(5x.\left(x-3\right)-x+3=0\)
\(\Leftrightarrow5x.\left(x-3\right)+\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x+1=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-\frac{1}{5}\end{array}\right.\)
Vậy ............
a, x.( x - 2 ) + 2x - 4 = 0
=>x2+2x+2x-4=0
=>x2+4x-4=0
=>(x-2)2=0
=>x=2
b, 5x.(x - 3 ) - x + 3 = 0
=>5x2-15x-x+3=0
=>x(5x-1)-3(5x-1)=0
=>(x-3)(5x-1)=0
\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=3\end{array}\right.\)
Cho sửa lại .
a ) \(x.\left(x-2\right)+2x-4=0\)
\(\Leftrightarrow x^2-2x+2x-4=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
Vậy ........................
b ) \(5x.\left(x-3\right)-x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x-1=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\5x=1\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{1}{5}\end{array}\right.\)
Vạy........
