a) \(x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)
b) \(x^2-3x+2=0\Leftrightarrow x^2-3x+\frac{9}{4}-\frac{1}{4}=0\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{1}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{2}=\sqrt{\frac{1}{4}}=\frac{1}{2}\\x-\frac{3}{2}=-\sqrt{\frac{1}{4}}=-\frac{1}{2}\end{cases}}\)
Giải tiếp nha
c) \(x^2+5x+6=0\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}\)
b , x2 - 3x + 2 = 0
\(\Leftrightarrow\) x2 - x - 2x + 2 =0
\(\Leftrightarrow\)(x-1) (x-2) =0
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
#mã mã#
a, x^2 - 2x + 1 = 0
=> (x-1)^2 = 0
=> x - 1 = 0
=> x = 1
b, x^2 - 3x + 2 = 0
=> x^2 - x - 2x + 2 = 0
=> x(x-1)-2(x-1) = 0
=> (x-2)(x-1) = 0
=> x - 2 = 0 hoặc x - 1 = 0
=> x = 2 hoặc x = 1
c, x^2 + 5x + 6 = 0
=> x^2 + 2x + 3x + 6 = 0
=> x(x+2) + 3(x + 2) = 0
=> (x+3)(x+2) = 0
=> x + 3 =0 hoặc x + 2 = 0
=> x = 3 hoặc x = -2
