19 tháng 9 2022 lúc 17:27
Xét hai số \(x_1;x_2\) sao cho \(x_1>x_2\ge-1\)
Ta có:
\(f\left(x_1\right)-f\left(x_2\right)=\left(x_1^2+2x_1\right)-\left(x_2^2+2x_2\right)=x_1^2-x_2^2+2x_1-2x_2\)
\(=\left(x_1-x_2\right)\left(x_1+x_2\right)+2\left(x_1-x_2\right)\)
\(=\left(x_1-x_2\right)\left(x_1+x_2+2\right)\)
Do \(x_1>x_2\ge-1\Rightarrow\left\{{}\begin{matrix}x_1-x_2>0\\x_1+x_2+2>0\end{matrix}\right.\)
\(\Rightarrow\left(x_1-x_2\right)\left(x_1+x_2+2\right)>0\)
\(\Rightarrow f\left(x_1\right)-f\left(x_2\right)>0\)
\(\Rightarrow f\left(x_1\right)>f\left(x_2\right)\)
Hay hàm số đã cho đồng biến với \(x\ge-1\)
Đúng 1
Bình luận (0)
