\(F\left(x\right)=\int\left(e^x.ln\left(ax\right)+\dfrac{e^x}{x}\right)dx=\int e^xln\left(ax\right)dx+\int\dfrac{e^x}{x}dx=\int e^xlnxdx+\int\dfrac{e^x}{x}dx+\int e^x.lna.dx\)

Xét \(I=\int e^xlnxdx\)

Đặt \(\left\{{}\begin{matrix}u=lnx\\dv=e^xdx\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}du=\dfrac{dx}{x}\\v=e^x\end{matrix}\right.\)

\(\Rightarrow I=lnx.e^x-\int\dfrac{e^x}{x}dx\)

\(\Rightarrow F\left(x\right)=e^x.lnx+e^x.lna+C\)

\(F\left(\dfrac{1}{a}\right)=e^{\dfrac{1}{a}}ln\left(\dfrac{1}{a}\right)+e^{\dfrac{1}{a}}.lna+C=0\Rightarrow C=0\)

\(F\left(2020\right)=e^{2020}ln\left(2020\right)+e^{2020}.lna=e^{2020}\)

\(\Rightarrow ln\left(2020a\right)=1\Rightarrow a=\dfrac{e}{2020}\)

Do \(\left(x+3\right)^{2020}\ge0\) và \(\left(y-2\right)^{2020}\ge0\) với mọi \(x,y\)

Để \(\left(x+3\right)^{2020}+\left(y-2\right)^{2020}=0\) thì \(x+3=0\) và \(y-2=0\)

Vậy \(x=-3,y=2\)

(x+3)^2020>=0

(y-2)^2020>=0

=>(x+3)^2020+(y-2)^2020>=0 với mọi x,y

Dấu = xảy ra khi x=-3 và y=2

x² + 2y² + z² - 2xy - 2y - 4z + 5 = 0

<=> ( x² - 2xy + y² ) + ( y² - 2y + 1 ) + ( z² - 4z + 4 ) = 0

<=> ( x - y )² + ( y - 1 )² + ( z - 2 )² = 0

Vì \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-1\right)^2\ge0\\\left(z-2\right)^2\ge0\end{cases}}\forall x;y;z\)=> ( x - y )² + ( y - 1 )² + ( z - 2 )²\(\ge\)0\(\forall\)x ; y ; z

Dấu "=" xảy ra <=>\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\)<=>\(\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)( 1 )

Thay ( 1 ) vào A , ta được :

\(A=\left(1-1\right)^{2020}+\left(1-2\right)^{2020}+\left(2-3\right)^{2020}=0+1+1=2\)

Vậy A = 2

Ta có: \(x^2+2y^2+z^2-2xy-2y-4z+5=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(y^2-2y+1\right)+\left(z^2-4z+4\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-1\right)^2+\left(z-2\right)^2=0\)

Mà \(VT\ge0\left(\forall x,y,z\right)\) nên dấu "=" xảy ra khi:

\(\hept{\begin{cases}\left(x-y\right)^2=0\\\left(y-1\right)^2=0\\\left(z-2\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=y=1\\z=2\end{cases}}\)

\(\left(x+3\right)^{2020}+\left(y-2\right)^{2020}=0\)

Vì \(\left(x+3\right)^{2020}\ge0\forall x;\left(y-2\right)^{2020}\ge0\forall y\)

\(\Rightarrow\left(x+3\right)^{2020}+\left(y-2\right)^{2020}\ge0\forall x;y\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}\left(x+3\right)^{2020}=0\\\left(y-2\right)^{2020}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+3=0\\y-2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=2\end{cases}}}\)

Vậy ....

(x−6)2020+2(y−3)2020=0

Ta có : (�−6)2020≥0∀�(x−6)2020≥0∀x

            2(�+3)2020≥0∀�2(y+3)2020≥0∀y

        =>(�−6)2020+2(�+3)2020≥0∀�,�(x−6)2020+2(y+3)2020≥0∀x,y

Dấu "=" xảy ra <=>{�−6=0�+3=0<=>{�=6�=−3{x−6=0y+3=0​<=>{x=6y=−3​

[nay]

Xét \(\left(x^2+2020\right)\left(x-10\right)=0\)

Vì \(x^2\ge0\forall x\)\(\Rightarrow x^2+2020\ge2020\forall x\)

\(\Rightarrow\left(x^2+2020\right)\left(x-10\right)=0\)\(\Leftrightarrow x-10=0\)\(\Leftrightarrow x=10\)

Ta thấy: trong biểu thức \(P=\left(x^2-1\right)\left(x^2-2\right)\left(x^2-3\right)......\left(x^2-2020\right)\)có chứa thừa số \(x^2-100\)

Thay \(x=10\)vào thừa số \(x^2-100\)ta được: \(10^2-100=100-100=0\)

\(\Rightarrow P=0\)

Vậy \(P=0\)

Theo đề bài, ta có: (x^2+2020)(x-10)=0

Vì x^2 luôn lớn hơn hoặc bằng 0 nên x^2+2020>0

=> x-10=0

Khi đó P=(x^2-1)(x^2-2)...(x^2-100)(x^2-101)...(x^2-2020)

 => P=(10^2-1)(10^2-2)...(10^2-100)(10^2-101)...(10^2-2020)

=> P=0 < Vì 10^2-100=0>

Vậy P=0