1,
a, = 2x.(x-2)
b, = (x^2+y^2+2xy)-(2x+2y)
= (x+y)^2-2.(x+y)
= (x+y).(x+y-2)
2,
a,<=> x^2-1-x^2-2x = 3
<=> -2x-1=3
<=> -2x=4
<=> x=4 : (-2) = -2
b, <=>(x^2-4x+4)-7=0
<=>(x-2)^2-7=0
<=> (x-2)^2=7
=> x-2=+-\(\sqrt{7}\)
<=> x=2+-\(\sqrt{7}\)
k mk nha
a, \(2x-4x\)
\(=-2x\)
b, \(x^2+y^2+2xy-2x-2y\)
\(=\left(x+y\right)^2-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-2\right)\)
a, \(\left(x+1\right)\left(x-1\right)-x\left(x+2\right)=3\)
\(\Leftrightarrow x^2-1-x^2-2x=3\)
\(\Leftrightarrow-2x=4\)
\(\Leftrightarrow x=-2\)
b,\(x^2-4x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
1)
a)\(2x-4x=x\left(2-4\right)=-2x\)
b)\(x^2+y^2+2xy-2x-2y\)
\(\Leftrightarrow\left(x+y\right)^2-2\left(x+y\right)\)
\(\Leftrightarrow\left(x+y\right)\left(x+y-2\right)\)
2)
a)\(\left(x+1\right)\left(x-1\right)-x\left(x+2\right)=3\)
\(\Leftrightarrow\left(x^2-1\right)-x\left(x+2\right)=3\)
\(\Leftrightarrow\left(x^2-1\right)-x^2-2x=3\)
\(\Leftrightarrow x^2-1-x^2-2x=3\)
\(\Leftrightarrow-1-2x=3\)
\(\Leftrightarrow-2x=4\)
\(\Leftrightarrow=-2\)
b)\(x^2-4x+3=0\)
\(\Leftrightarrow x^2-3x-x+3=0\)
\(\Leftrightarrow x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
P/ s tham khảo nha
