\(f\left(x\right)=\left(m-4\right)x^2+\left(m+1\right)x+2m-1\)

\(f\left(x\right)< 0,\forall x\in R\Leftrightarrow\left\{{}\begin{matrix}a< 0\\\Delta< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m-4< 0\\\left(m+1\right)^2-4\left(m-4\right)\left(2m-1\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\m^2+2m+1-4\left(2m^2-m-8m+4\right)< 0\end{matrix}\right.\)

\(\Leftrightarrow m^2+2m+1-8m^2+36m-16< 0\)

\(\Leftrightarrow-7m^2+38m-15< 0\)

\(\Leftrightarrow\left\{{}\begin{matrix}m< 4\\\left[{}\begin{matrix}m< \dfrac{3}{7}\\m>5\end{matrix}\right.\end{matrix}\right.\)

\(KL:m\in\left(5;+\infty\right)\)

\(\sqrt{1+\dfrac{1}{x^2}+\dfrac{1}{\left(x+1\right)^2}}=\sqrt{\dfrac{x^2+\left(x+1\right)^2+x^2\left(x+1\right)^2}{x^2\left(x+1\right)^2}}=\sqrt{\dfrac{x^2\left(x+1\right)^2+2x^2+2x+1}{x^2\left(x+1\right)^2}}\)

\(=\sqrt{\dfrac{\left(x^2+x\right)^2+2\left(x^2+x\right)+1}{\left(x^2+x\right)^2}}=\sqrt{\dfrac{\left(x^2+x+1\right)^2}{\left(x^2+x\right)^2}}=\dfrac{x^2+x+1}{x^2+x}\)

\(=1+\dfrac{1}{x}-\dfrac{1}{x+1}\)

\(\Rightarrow f\left(1\right).f\left(2\right)...f\left(2020\right)=5^{1+1-\dfrac{1}{2}+1+\dfrac{1}{2}-\dfrac{1}{3}+...+1+\dfrac{1}{2020}-\dfrac{1}{2021}}\)

\(=5^{2021-\dfrac{1}{2021}}\)

\(\Rightarrow\dfrac{m}{n}=2021-\dfrac{1}{2021}=\dfrac{2021^2-1}{2021}\)

\(\Rightarrow m-n^2=2021^2-1-2021^2=-1\)

Giải:

Vì \(f\left(x_1x_2\right)=f\left(x_1\right).f\left(x_2\right)\) nên ta có:

\(f\left(4\right)=f\left(2.2\right)=f\left(2\right).f\left(2\right)=5.5=25\)

Mà:

\(f\left(2\right)=5\)

\(\Leftrightarrow f\left(8\right)=f\left(4.2\right)=f\left(4\right).f\left(2\right)=25.5=125\)

Vậy: \(f\left(8\right)=125\)

Thay x vào f (x)

\(\Rightarrow f\left(2\right)=-m.2+7=3\)

\(\Rightarrow-2m=-4\Rightarrow m=2\)

Vậy m= 2

Thay x = -5 vào f (x)

\(\Rightarrow f\left(-5\right)=\left(2\right).\left(-5\right)+7\)

\(\Rightarrow f\left(-5\right)=-3\)

Giả sử \(f\left(x\right)\) có bậc k \(\Rightarrow f'\left(x\right)\) có bậc \(k-1\) và \(f''\left(x\right)\) có bậc \(k-2\)

\(\Rightarrow f''\left(x\right)+3x^2-5\) có bậc lớn nhất bằng \(max\left\{k-2;2\right\}\)

\(\Rightarrow k-1=max\left\{k-2;2\right\}\Rightarrow k-1=2\) (do \(k-1\ne k-2\) với mọi k)

\(\Rightarrow f\left(x\right)\) là đa thức bậc 3 có dạng: \(y=ax^3+bx^2+cx-5\) với \(a\ne0\)

\(3ax^2+2bx+c=6ax+2b+3x^2-5\)

\(\Leftrightarrow3ax^2+2bx+c=3x^2+6ax+2b-5\)

Đồng nhất 2 vế: \(\left\{{}\begin{matrix}3a=3\\2b=6a\\c=2b-5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=1\\b=3\\c=1\end{matrix}\right.\)

\(\Rightarrow f\left(x\right)=x^3+3x^2+x-5\)

Đặt \(sin^2x=t\Rightarrow0\le t\le1\)

\(f\left(t\right)=0\Leftrightarrow t^3+3t^2+t-5=0\)

\(\Leftrightarrow\left(t-1\right)\left(t^2+4t+5\right)=0\Rightarrow t=1\)

\(\Rightarrow sin^2x=1\Leftrightarrow cosx=0\)

\(\Rightarrow x=\frac{\pi}{2}+k\pi\)

\(\Rightarrow\frac{\pi}{2}+k\pi\le2020\Rightarrow k\le\frac{4040-\pi}{2\pi}\)

\(\Rightarrow k_{max}=642\Rightarrow x_{max}=\frac{\pi}{2}+642\pi\)

a) f(3)-5f(3)=\(x^2\)

-4f(3)=\(x^2\)

f(3)=\(-\frac{x^2}{4}\)

b)f(-1)+f(1/-1)+f(1)=6

2f(-1)=6-f(1)

vay f(-1)=\(\frac{6-f\left(1\right)}{2}\)

Ta có: y=f(x)=(a+2)x-3a+2

=> (.) f(5)=(a+2)*5-3a+2=11

=>5a+10-3a+2=11

=>2a+12=11

=>2a=-1

=>a=-1/2

(.) f(-1)=(a+2)*(-1)-3a+2

=> -a-2-3a+2=6

=>-4a=6

=>a=-6/4=-3/2

câu đầu tiên đề sai thì phải???