Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Với mọi x thỏa mãn: f( a + b ) = f (ab)
=>f( 0 ) = f( -1/2 . 0 ) = f ( -1/2 + 0 ) = f( -1/2 ) = -1/2
=> f ( 2006 ) = f ( 2006 + 0 ) = f(2006 . 0 ) = f(0 ) = -1/2
Xét đa thức \(F\left(x\right)=ax^2+bx+c\)
\(F\left(0\right)=c=2016\)
\(F\left(1\right)=a+b+c=2017\Rightarrow a+b=1\) (1)
\(F\left(-1\right)=a-b+c=2018\Rightarrow a-b=2\) (2)
Từ (1), (2)
\(\Rightarrow\hept{\begin{cases}a+b-a+b=-1\\a+b+a-b=3\end{cases}}\Rightarrow\hept{\begin{cases}2b=-1\\2a=3\end{cases}}\Rightarrow\hept{\begin{cases}b=-0,5\\a=1,5\end{cases}}\)
\(\Rightarrow F\left(2\right)=1,5.2^2-0,5.2+2016=2021\)
Vậy \(F\left(2\right)=2021\).
a)\(F\left(0\right)=2016-3.0\left|2.0-5\right|\)
\(F\left(0\right)=2016\)
\(F\left(3\right)=2016-3.3\left|2.3-5\right|\)
\(F\left(3\right)=2016-9\)
\(F\left(3\right)=2007\)
Ta có
\(F\left(0\right)=2016\)
\(\Leftrightarrow a\cdot0^2+b\cdot0+c=2016\)
\(\Leftrightarrow0+0+c=2016\)
\(\Leftrightarrow c=2016\)
\(F\left(1\right)=2016\)
\(\Leftrightarrow a\cdot1^2+b\cdot1+c=2017\)
\(\Leftrightarrow a+b+c=2017\)
\(\Leftrightarrow a+b+2016=2017\)
\(\Leftrightarrow a+b=1\) \(\left(1\right)\)
\(F\left(-1\right)=2018\)
\(\Leftrightarrow a\cdot\left(-1\right)^2+b\cdot\left(-1\right)+c=2018\)
\(\Leftrightarrow a-b+c=2018\)
\(\Leftrightarrow a-b+2016=2018\)
\(\Leftrightarrow a-b=2\) \(\left(2\right)\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow a=\left(1+2\right)\div2=3\div2=1.5\)
\(\Rightarrow b=1-1.5=-0.5\)
Vậy \(F\left(x\right)=1.5x^2-0.5x+2016\)
\(\Leftrightarrow F\left(2\right)=1.5\cdot2^2-0.5\cdot2+2016\)
\(=1.5\cdot4-0.5\cdot2+2016\)
\(=6-1+2016=2021\)
Vậy \(F\left(2\right)=2021\)
nhớ k nha
Theo bài ra ta có:
\(\hept{\begin{cases}c=2016\\a+b+c=2017\\a-b+c=2018\end{cases}\Leftrightarrow2a+2c=4035\Leftrightarrow2a=4035-2016.2=3}\)
\(\Leftrightarrow a=\frac{3}{2}\)
thay vào ta tính dc b nha
\(f\left(1\right)=3\cdot1^2+1+1=5\\ f\left(-\dfrac{1}{3}\right)=3\cdot\left(-\dfrac{1}{3}\right)^2-\dfrac{1}{3}+1=\dfrac{1}{3}-\dfrac{1}{3}+1=1\\ f\left(\dfrac{2}{3}\right)=3\cdot\left(\dfrac{2}{3}\right)^2+\dfrac{2}{3}+1=\dfrac{4}{3}+\dfrac{2}{3}+1=3\\ f\left(-2\right)=3\left(-2\right)^2-2+1=12-2+1=11\\ f\left(-\dfrac{4}{3}\right)=3\cdot\left(-\dfrac{4}{3}\right)^2-\dfrac{4}{3}+1=\dfrac{16}{3}-\dfrac{4}{3}+1=5\)
\(f\left(1\right)=3\cdot1^2+1+1=5\)
\(f\left(-\dfrac{1}{3}\right)=3\cdot\dfrac{1}{9}-\dfrac{1}{3}+1=1\)
\(f\left(\dfrac{2}{3}\right)=3\cdot\dfrac{4}{9}+\dfrac{2}{3}+1=3\)