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\(f\left(1\right)=1+1+1^2+...+1^{2013}=1.2014=2014\)
\(f\left(-1\right)=1-1+1-1+1-1+...+1-1=0+0+0+...+0=0\)
đúng nha
4. (3/4-81)(3^2/5-81)(3^3/6-81)....(3^6/9-81).....(3^2011/2014-81)
mà 3^6/9-81=0 => (3/4-81)(3^2/5-81)....(3^2011/2014-81)=0
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:\(f\left(x\right)=0\Rightarrow|3x-1|=0\Rightarrow3x-1=0\)
\(3x=0+1=1\)
\(x=1:3=\dfrac{1}{3}\)
\(f\left(x\right)=1\Rightarrow|3x-1|=1\Rightarrow3x-1=\pm1\)
*Với \(3x-1=1\Rightarrow3x=1+1=2\)
\(x=2:3=\dfrac{2}{3}\)
*Với \(3x-1=-1\Rightarrow3x=-1+1=0\)
\(x=0:3=0\)
\(f\left(x\right)=\dfrac{1}{2}\Rightarrow|3x-1|=\dfrac{1}{2}\Rightarrow3x-1=\pm\dfrac{1}{2}\)
*Với \(3x-1=\dfrac{1}{2}\Rightarrow3x=\dfrac{1}{2}+1=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}:3=\dfrac{3}{2}.\dfrac{1}{3}=\dfrac{1}{2}\)
*Với \(3x-1=-\dfrac{1}{2}\Rightarrow3x=-\dfrac{1}{2}+1=\dfrac{1}{2}\)
\(x=\dfrac{1}{2}:3=\dfrac{1}{2}.\dfrac{1}{3}=\dfrac{1}{6}\)
\(f\left(x\right)=-\dfrac{2010}{2011}\Rightarrow|3x-1|=-\dfrac{2010}{2011}\Rightarrow x\in\varnothing\)
\(f\left(0\right)=2010\Rightarrow a.0^2+b.0+c=2010\Rightarrow c=2010\)
\(f\left(1\right)=2011\Rightarrow a.1^2+b.1+c=2011\Rightarrow a+b+c=2011\)
\(\Rightarrow a+b+2010=2011\Rightarrow a+b=1\) (1)
\(f\left(-1\right)=2012\Rightarrow a.\left(-1\right)^2+b.\left(-1\right)+c=2012\)
\(\Rightarrow a-b+c=2012\Rightarrow a-b+2010=2012\)
\(\Rightarrow a-b=2\Rightarrow a=b+2\)
Thế vào (1) \(\Rightarrow b+2+b=1\Rightarrow2b=-1\Rightarrow b=-\dfrac{1}{2}\)
\(\Rightarrow a=b+2=-\dfrac{1}{2}+2=\dfrac{3}{2}\)
\(\Rightarrow f\left(x\right)=\dfrac{3}{2}x^2-\dfrac{1}{2}x+2010\)
\(\Rightarrow f\left(-2\right)=\dfrac{3}{2}.\left(-2\right)^2-\dfrac{1}{2}.\left(-2\right)+2010=2017\)
Ta có: \(f\left(1\right)=1+1+1+....+1\)
=> \(f\left(1\right)=2012\)
Ta lại có: \(f\left(-1\right)=1-1+1-1+...+1-1\) = 0
ta có : \(f\)(1) = \(1+1+1+1+.....+1+1\) = 1 + 2011 = \(2012\)
: \(f\)(-1) = \(1-1+1-1+.....+1-1\) = 0