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Có \(c=2a+4b\). Ta tính f ( -1 ) và f ( 2 )
\(f\left(-1\right)=a-b+c=a-b+2a+4b=3a+3b=3\left(a+b\right)\)
\(f\left(2\right)=4a+2b+c=4a+2b+2a+4b=6a+6b=6\left(a+b\right)\)
\(\Rightarrow f\left(-1\right).f\left(2\right)=3\left(a+b\right).6\left(a+b\right)=18\left(a+b\right)^2\)
Có \(\left(a+b\right)^2\ge0\forall x\Leftrightarrow18\left(a+b\right)^2\ge0\forall x\left(đpcm\right)\)
\(f\left(-1\right)=a\left(-1\right)^2+b.\left(-1\right)+c\)
\(=a-b+c\)
\(f\left(2\right)=a.2^2+b.2+c\)
\(=4a+2b+c\)
\(\Rightarrow f\left(2\right)-2.f\left(-1\right)=\left(4a+2b+c\right)-2\left(a-b+c\right)\)
\(=2a+4b-c=0\)
\(\Rightarrow f\left(2\right)=2.f\left(-1\right)\)
\(\Rightarrow f\left(2\right)\)và \(2.f\left(-1\right)\)cùng dấu
\(\Rightarrow f\left(2\right)\)và \(f\left(-1\right)\)cùng dấu
\(\Rightarrow f\left(2\right).f\left(-1\right)\ge0\)(đpcm)
Ta có :\(f\left(-1\right)=a.\left(-1\right)^2+b.\left(-1\right)+c=a-b+c\)
\(f\left(2\right)=a.2^2+b.2+c=4a+2b+c\)
\(\implies\) \(f\left(2\right)-2f\left(-1\right)=\left(4a+2b+c\right)-2.\left(a-b+c\right)\)
\(\implies\) \(f\left(2\right)=2.f\left(-1\right)\)
\(\implies\) \(f\left(-1\right).f\left(2\right)=f\left(-1\right).2f\left(-1\right)=f\left(-1\right)^2.2\) \(\geq\) \(0\)
\(\implies\) \(f\left(-1\right).f\left(2\right)\) \(\geq\) \(0\) \(\left(đpcm\right)\)
\(f\left(x\right)=ax^2+bx+c\Rightarrow\hept{\begin{cases}f\left(0\right)=c\\f\left(1\right)=a+b+c\\f\left(2\right)=4a+2b+c\end{cases}}\)
\(f\left(0\right)\) nguyên \(\Rightarrow c\) nguyên \(\Rightarrow\hept{\begin{cases}2a+2b\\4a+2b\end{cases}}\) nguyên
\(\Rightarrow\left(4a+2b\right)-\left(2a+2b\right)=2a\)(nguyên)
\(\Rightarrow2b\) nguyên
\(\Rightarrowđpcm\)
f(-1)=a-b+c
f(3)=9a+3b+c
f(3)-f(-1)=8a+b=4(2a+b)
Mà 2a+b=0 =) f(3)-f(-1)=0
=) f(3)=f(-1) =) f(3).f(-1)=(a-b+c)^2
Mà (a-b+c)^2 >= 0 =) f(-1).f(3)>=0
Ta có : f(x) = ax2 + bx + c
=> f( -1 ) = a - b + c
f(3) = 9a + 3b + c
=> f(3) - f( -1 ) = 8a + 4b = 4 ( 2a + b ) = 4.0 = 0
=> f(3) = f( -1 )
=> f( -1 ). f(3) = f(3). f(3) = [ f(3) ]2 \(\ge\) 0
=> đpcm
Study well ! >_<