Ta có \(f\left(x\right)=ax^2+bx^4+x+3+11\)
\(=>f\left(-2\right)=a\left(-2\right)^2+b\left(-2\right)^4+\left(-2\right)+3+11=3\)
\(=>f\left(-2\right)=4a+16b+12=3\)
Ta có \(f\left(2\right)=a\left(2^2\right)+b\left(2^4\right)+2+3+11\)
\(=>f\left(2\right)=4a+16b+16\)
\(=>f\left(2\right)=4a+16b+12+4\)
Mà \(f\left(-2\right)=4a+16b+12=3\)
\(=>f\left(2\right)=4a+16b+12+4\)
\(=>f\left(2\right)=3+4\)
\(=>f\left(2\right)=7\)
Đúng 0
Bình luận (0)
