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Ta có: \(f\left(0\right)=a.0^2+b.0+c=0+0+c=c\) mà \(f\left(0\right)=1\)\(\Rightarrow c=1\)
\(f\left(1\right)=a.1^2+b.1^2+c=a+b+1\)mà \(f\left(1\right)=2\)\(\Rightarrow a+b+1=2\)\(\Rightarrow a+b=1\)
\(f\left(2\right)=a.2^2+2.b+c=4a+2b+1\)mà \(f\left(2\right)=8\)\(\Rightarrow4a+2b+1=8\)\(\Rightarrow4a+2b=7\)\(\Rightarrow2\left(2a+b\right)=7\)\(\Rightarrow2a+b=3,5\)\(\Rightarrow a+\left(a+b\right)=3,5\)\(\Rightarrow a+1=3,5\)\(\Rightarrow a=2,5\)
Lại có: \(a+b=1\)\(\Rightarrow2,5+b=1\)\(\Rightarrow b=1-2,5=-1,5\)
Ta có: \(f\left(-2\right)=a.\left(-2\right)^2+b.\left(-2\right)+c=2,5.4+\left(-1.5\right).\left(-2\right)+1=10+3+1=14\)
Bài 1 : làm tương tự với bài 2;3 nhé
Ta có : \(f\left(0\right)=c=2010;f\left(1\right)=a+b+c=2011\)
\(\Rightarrow f\left(1\right)=a+b=1\)
\(f\left(-1\right)=a-b+c=2012\Rightarrow f\left(-1\right)=a-b=2\)
\(\Rightarrow a+b=1;a-b=2\Rightarrow2a=3\Leftrightarrow a=\dfrac{3}{2};b=\dfrac{3}{2}-2=-\dfrac{1}{2}\)
Vậy \(f\left(-2\right)=4a-2b+c=\dfrac{4.3}{2}-2\left(-\dfrac{1}{2}\right)+2010=6+1+2010=2017\)
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 \(f\left(-2\right)=4a+16b+12=3\)
\(=>f\left(2\right)=4a+16b+12+4\)
\(=>f\left(2\right)=3+4\)
\(=>f\left(2\right)=7\)
\(g\left(x\right)=ax^3-bx\)
\(f\left(x\right)=g\left(x\right)-15\)
\(f\left(-x\right)=-g\left(x\right)-15\)
\(f\left(x\right)+f\left(-x\right)=-30\)
\(f\left(5\right)+f\left(-5\right)=-30\Rightarrow f\left(-5\right)=-30-5=-35\)
\(f\left(-5\right)=-35\)
Ta có : \(f\left(0\right)=c=1\)
\(f\left(1\right)=a+b+c=2\)
\(f\left(2\right)=4a+2b+c=8\)
\(\Rightarrow c=1,a=\frac{5}{2},b=\frac{-3}{2}\)
Vì vậy mà \(f\left(x\right)=\frac{5}{2}x^2-\frac{3}{2}x+1\)
nên \(f\left(-2\right)=\frac{5}{2}.\left(-2\right)^2-\frac{3}{2}.\left(-2\right)+1=14\)