Chọn \(f\left(x\right)=5x+5\)

Khi đó: \(\lim\limits_{x\rightarrow1}\dfrac{5x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{20x+29}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{5\left(\sqrt{x}+1\right)}{\sqrt{20x+29}+3}=\dfrac{10}{7+3}=1\)

\(\left(x-1\right).f\left(x\right)=\left(x+5\right).f\left(x+3\right)\) (*)

Thay x = 1 vào (*) ,có :

Lời giải:

\(f(1)=f(-1)\)

\(\Leftrightarrow a_4+a_3+a_2+a_1+a_0=a_4-a_3+a_2-a_1+a_0\)

\(\Leftrightarrow 2(a_3+a_1)=0\Leftrightarrow a_3+a_1=0(1)\)

\(f(2)=f(-2)\)

\(\Leftrightarrow 16a_4+8a_3+4a_2+2a_1+a_0=16a_4-8a_3+4a_2-2a_1+a_0\)

\(\Leftrightarrow 16a_3+4a_1=0\Leftrightarrow 4a_3+a_1=0(2)\)

Từ \((1);(2)\Rightarrow a_3=a_1=0\)

Do đó:
\(f(x)=a_4x^4+a_2x^2+a_0\)

\(\Rightarrow f(-x)=a_4(-x)^4+a_2(-x)^2+a_0=a_4x^4+a_2x^2+a_0\)

Vậy $f(x)=f(-x)$.

Do \(\lim\limits_{x\rightarrow-1}\dfrac{2f\left(x\right)+1}{x+1}=5\) hữu hạn nên \(2f\left(x\right)+1=0\) phải có nghiệm \(x=-1\)

\(\Leftrightarrow2f\left(-1\right)=-1\Leftrightarrow f\left(-1\right)=-\dfrac{1}{2}\)

Đoạn dưới tự hiểu là \(\lim\limits_{x\rightarrow-1}\) (vì kí tự lim rất rắc rối)

\(I=\dfrac{\left[4f\left(x\right)+3\right]\left[\sqrt{4f^2\left(x\right)+2f\left(x\right)+4}-2\right]+2\left[4f\left(x\right)+3\right]-2}{x^2-1}\)

\(=\dfrac{\left[4f\left(x\right)+3\right]\left[4f^2\left(x\right)+2f\left(x\right)\right]}{\left(x+1\right)\left(x-1\right)\left[\sqrt{4f^2\left(x\right)+2f\left(x\right)+4}+2\right]}+\dfrac{4\left[2f\left(x\right)+1\right]}{\left(x+1\right)\left(x-1\right)}\)

\(=\dfrac{2f\left(x\right)+1}{x+1}.\dfrac{f\left(x\right).\left[4f\left(x\right)+3\right]}{x-1}+\dfrac{2f\left(x\right)+1}{x+1}.\dfrac{4}{x-1}\)

\(=5.\dfrac{f\left(-1\right).\left[4f\left(-1\right)+3\right]}{-2}+5.\dfrac{4}{-2}=\dfrac{5.\left(-\dfrac{1}{2}\right)\left(-2+3\right)}{-2}+5.\dfrac{4}{-2}=...\)

Lời giải:
a. 

$f(-1)=a-b+c$

$f(-4)=16a-4b+c$

$\Rightarrow f(-4)-6f(-1)=16a-4b+c-6(a-b+c)=10a+2b-5c=0$

$\Rightarrow f(-4)=6f(-1)$

$\Rightarrow f(-1)f(-4)=f(-1).6f(-1)=6[f(-1)]^2\geq 0$ (đpcm)

b.

$f(-2)=4a-2b+c$

$f(3)=9a+3b+c$

$\Rightarrow f(-2)+f(3)=13a+b+2c=0$

$\Rightarrow f(-2)=-f(3)$

$\Rightarrow f(-2)f(3)=-[f(3)]^2\leq 0$ (đpcm)

a. 

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