Cho f(x) = 0 thì ta có: \(x^2+7x+10=0\Leftrightarrow\left(x^2+2x\right)+\left(5x+10\right)=0\)
\(\Leftrightarrow x\left(x+2\right)+5\left(x+2\right)=0\Leftrightarrow\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-5\end{cases}}\).Vậy...
f(x) = x^2 + 7x + 10
= x^2 + 2x + 5x + 10
= x(x + 2) + 5(x + 2)
= (x + 5)(x+2)
xét f(x) = 0
=> (x+5)(x+2) = 0
=> x + 5 = 0 hoặc x + 2 = 0
=> x = -5 hoặc x = -2
vậy_
\(F\left(x\right)=x^2+7x+10=\left(x^2+2x.\frac{7}{2}+\left(\frac{7}{2}\right)^2\right)-\frac{49}{4}+10\)
\(\left(x+\frac{7}{2}\right)^2-\frac{9}{4}=0\)
\(\left(x+\frac{7}{2}\right)^2=\left(\frac{3}{2}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{7}{2}=\frac{3}{2}\\x+\frac{7}{2}=-\frac{3}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{4}{2}=-2\\x=-\frac{10}{2}=-5\end{cases}}}\)
Vậy...
