Question

giúp mình với

 

Answer 1

Ta có: \(29a+2c=3b\Rightarrow29a=3b-2c\)

Thay \(x=2\) vào biểu thức \(f\left(x\right)\) có:

\(f\left(2\right)=2^2a+2b+c\)

\(f\left(2\right)=4a+2b+c\)

Thay \(x=-5\) vào biểu thức \(f\left(x\right)\) có:

\(f\left(-5\right)=5^2a-5b+c\)

\(f\left(-5\right)=25a-5b+c\)

Cộng \(f\left(2\right)\) và \(f\left(-5\right)\), có:

\(f\left(2\right)+f\left(-5\right)=4a+2b+c+25a-5b+c\)

\(f\left(2\right)+f\left(-5\right)=29a-3b+2c\)

Mà \(29a=3b-2c\)

\(\Rightarrow f\left(2\right)+f\left(-5\right)=3b-2c-3b+2c=0\)

\(\Rightarrow f\left(2\right)+f\left(-5\right)=0\)

\(\Rightarrow f\left(2\right)+f\left(-5\right)\le0\left(đpcm\right)\)