f(0) = 1

\(\Rightarrow\) a.0² + b.0 + c = 1 

\(\Rightarrow\) c = 1

Vậy hệ số a = 0; b = 0; c = 1

f(1) = 2

\(\Rightarrow\) a.1² + b.1 + c = 2

\(\Rightarrow\) a + b + c = 2

Vậy hệ số a = 1; b = 1; c = 1

f(2) = 4

\(\Rightarrow\) a.2² + b.2 + c = 4

\(\Rightarrow\) 4a + 2b + c = 4

Vậy hệ số a = 4; b = 2; c = 1

Chúc bn học tốt! (chắc vậy :D)

Ta có: f(0) = a.0² + b.0 + c = 2

=> c = 2

  f(1) = a.1² + b.1 + c  = 1

=> a + b + c = 1 => a + b = 1 - c = 1 - 2 = -1 (1)

f(-2) = a.(-2)² + b.(-2) + c = 2

=> 4a - 2b = 2 - c =  2 - 2 = 0

=> 2a - b = 0 (2)

Từ (1) và (2) cộng vế theo vế:

(a + b) + (2a - b) = -1

=> 3a = -1

=> a = -1/3

=> b = -1 - a = -1 + 1/3 = -2/3

Vậy ....

\(f\left(-1\right)=2\Rightarrow-a+b-c+d=2\\ f\left(0\right)=1\Rightarrow d=1\\ f\left(1\right)=7\Rightarrow a+b+c+d=7\\ f\left(\dfrac{1}{2}\right)=3\Rightarrow\dfrac{1}{8}a+\dfrac{1}{4}b+\dfrac{1}{2}c+d=3\)

\(d=1\Rightarrow-a+b-c=1;a+b+c=6\\ \Rightarrow2b=7\\ \Rightarrow b=\dfrac{7}{2}\\ \Rightarrow\dfrac{1}{8}a+\dfrac{7}{8}+\dfrac{1}{2}c=2\\ \Rightarrow\dfrac{1}{2}\left(\dfrac{1}{4}a+\dfrac{7}{4}+c\right)=2\\ \Rightarrow\dfrac{1}{4}a+\dfrac{7}{4}+c=4\\ \Rightarrow a+7+4c=16\\ \Rightarrow a+4c=9;a+c=6-\dfrac{7}{2}=\dfrac{5}{2}\\ \Rightarrow3c=\dfrac{13}{2}\Rightarrow c=\dfrac{13}{6}\\ \Rightarrow a=\dfrac{5}{2}-\dfrac{13}{6}=\dfrac{1}{3}\)

Vậy \(\left(a;b;c;d\right)=\left(\dfrac{1}{3};\dfrac{7}{2};\dfrac{13}{6};1\right)\)

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt[3]{ax+1}-\sqrt[]{1-bx}}{x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{ax}{\sqrt[3]{\left(ax+1\right)^2}+\sqrt[3]{ax+1}+1}+\dfrac{bx}{1+\sqrt[]{1-bx}}}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{a}{\sqrt[3]{\left(ax+1\right)^2}+\sqrt[3]{ax+1}+1}+\dfrac{b}{1+\sqrt[]{1-bx}}\right)=\dfrac{a}{3}+\dfrac{b}{2}\)

Hàm liên tục tại \(x=0\) khi:

\(\dfrac{a}{3}+\dfrac{b}{2}=3a-5b-1\Leftrightarrow8a-11b=3\)

\(f\left(x\right)=ax^2+bx+c\)

\(\Rightarrow f\left(x-1\right)=a\left(x-1\right)^2+b\left(x-1\right)+c\)

\(\Rightarrow f\left(x\right)-f\left(x-1\right)=ax^2+bx+c-ax^2+2ax-a-bx+b-c=x\)

\(\Leftrightarrow2ax-a+b-x=0\)

\(\Leftrightarrow\left(2a-1\right)x+b-a=0\)

\(\Leftrightarrow\hept{\begin{cases}2a-1=0\\b-a=0\end{cases}\Leftrightarrow}a=b=\frac{1}{2}\)

\(\)và Hàm số đúng với mọi giá trị của \(c\)

Vậy \(a=b=\frac{1}{2};c\in R\)

f(0)=a0+b0+c=2010

=>c=2010

f(1)=a1+b1+c=a1+b1+2010

=>a+b=1 (1)

f(-1)=a1+(-b1)+c=a1-b1+2010

=>a-b=2 (2)

Từ (1) và (2) => a=(2+1):2=1,5

                        b=(1-2):2=-0,5

Vậy f(2)=1,5.2+(-0,5)x2+2010=2014

\(f\left(-1\right)=\lim\limits_{x\rightarrow-1^-}f\left(x\right)=\lim\limits_{x\rightarrow-1^-}\left(2-ax\right)=2+a\)

\(\lim\limits_{x\rightarrow-1^+}f\left(x\right)=\lim\limits_{x\rightarrow-1^+}\left(x^2-bx+2\right)=3+b\)

\(f\left(1\right)=\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\left(4x+a\right)=4+a\)

\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(x^2-bx+2\right)=3-b\)

Hàm liên tục trên R khi và chỉ khi:

\(\left\{{}\begin{matrix}2+a=3+b\\4+a=3-b\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0\\b=-1\end{matrix}\right.\)