Ta có:
\(f\left(x_1\right)=2x_1+3\)
\(f\left(x_2\right)=2x_2+3\)
\(\Rightarrow f\left(x_1\right)+f\left(x_2\right)=\left(2x_1+3\right)+\left(2x_2+3\right)\)
\(\Rightarrow f\left(x_1\right)+f\left(x_2\right)=\left(2x_1+2x_2\right)+\left(3+3\right)\)
\(\Rightarrow f\left(x_1\right)+f\left(x_2\right)=2\times\left(x_1+x_2\right)+6\)
\(\Rightarrow f\left(x_1\right)+f\left(x_2\right)=2\times5+6\)
\(\Rightarrow f\left(x_1\right)+f\left(x_2\right)=10+6\)
\(\Rightarrow f\left(x_1\right)+f\left(x_2\right)=16\)
Vậy \(f\left(x_1\right)+f\left(x_2\right)=16\).
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