x2 - 3x - 1 0 = 0
<=> x2 - 2x + 5x - 10 = 0
<=> x(x - 2) + 5(x - 2) = 0
<=> (x + 5)(x - 2) = 0
<=> x + 5 = 0 hoặc x - 2 = 0
<=> x = - 5 hoặc x = 2
\(\Leftrightarrow x^2-5x+2x-10=0\)
\(\Leftrightarrow x\left(x-5\right)+2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)
Vậy...
\(x^2-3x-10=0\)
\(=>x.\left(x-3\right)=10\)
\(=>x;x-3\inƯ\left(10\right)\)
\(=>x;x-3\in\left\{-10;-5;-2;-1;1;2;5;10\right\}\)
\(=>x\in\left\{....\right\}\)
Vậy ...
P/s : cách lớp 6 , ko chắc :v
\(x^2-3x-10=0\)
\(\Leftrightarrow x^2+2x-5x-10=0\)
\(\Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=5\end{cases}}\)
Vậy phương trình có tập nghiệm \(S=\left\{-2;5\right\}\)
\(x^2-3x-10=0\)
\(\Leftrightarrow x^2-5x+2x-10=0\)
\(\Leftrightarrow x\left(x-5\right)+2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=5\end{cases}}}\)
\(x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)