a: Ta có: \(-\left(x+5\right)^2\le0\forall x\)

\(\Leftrightarrow-\left(x+5\right)^2+2021\le2021\forall x\)

Dấu '=' xảy ra khi x=-5

a) Quy luật là gì ??

b) 

Đặt

 \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\\\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\\ \Rightarrow2A-A=1-\dfrac{1}{2^{2020}}\Rightarrow A=1-\dfrac{1}{2^{2020}}\)

Suy ra , phương trình trở thành :

213 -x  =13

<=> x=200

Địt con cụ

Dễ thấy x càng lớn thì A càng lớn

vậy ko có Max

Tìm Min \(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)+2020\)

\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)+2020\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)+2020\)

Đặt \(x^2+5x=a\)

\(\Rightarrow A=\left(a-6\right)\left(a+6\right)+2020\)

\(=a^2-6a+6a-36+2020\)

\(=a^2+1984\ge1984\left(a^2\ge0\right)\)

Vậy Min A = 1984 

Dấu "=" xảy ra khi \(a=0\Leftrightarrow x^2+5x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x+5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

nguoif bí ẩn ko có tên ko đc nói bậy

αi nhanh mình sẽ Tick ạ.

A = \(\dfrac{3^{100}.\left(-2\right)+3^{101}}{\left(-3\right)^{101}-3^{100}}\) 

A = \(\dfrac{3^{100}.\left(-2\right)+3^{100}.3}{\left(-3\right)^{100}.\left(-3\right)-3^{100}}\)

A = \(\dfrac{3^{100}.\left(-2+3\right)}{3^{100}.\left(-3\right)-3^{100}}\)

A = \(\dfrac{3^{100}.1}{3^{100}.\left(-3-1\right)}\)

A = \(\dfrac{3^{100}}{3^{100}}\) . \(\dfrac{1}{-4}\)

A = - \(\dfrac{1}{4}\)

1/4

b) \(x^3+6x^2+9x=0\)

\(\Leftrightarrow x^3+3x^2+3x^2+9x=0\)

\(\Leftrightarrow x^2\left(x+3\right)+3x\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2+3x\right)=0\)

\(\Leftrightarrow\left(x+3\right)x\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)^2x=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=0\\x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=0\end{cases}}}\)

Vậy \(x\in\left\{-3;0\right\}\)

a) \(2x\left(x-2\right)+x^2=4\)

\(\Leftrightarrow2x\left(x-2\right)+x^2-4=0\)

\(\Leftrightarrow2x\left(x-2\right)+\left(x-2\right)\left(x+2\right)=0\)

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

\(\Leftrightarrow\left(x-2\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\3x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{-2}{3}\end{cases}}}\)

Vậy \(x\in\left\{\frac{-2}{3};2\right\}\)

c) \(5x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow5x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left(5x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2020=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{1}{5}\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{5};2020\right\}\)

\(1,\Rightarrow2^b\left(2^{a-b}-1\right)=256=2^8\left(a>b\right)\)

Do \(2^b\) chẵn, \(2^{a-b}-1\) lẻ, \(2^8\) chẵn nên \(2^{a-b}-1=1\Leftrightarrow2^{a-b}=2\Leftrightarrow a-b=1\)

\(\Leftrightarrow2^b\cdot1=2^8\Leftrightarrow b=8\Leftrightarrow a=9\)

Vậy \(\left(a;b\right)=\left(8;9\right)\) 

Bài 1:

Từ đkđb hiển nhiên $a>b\Rightarrow a-b\geq 1$

$2^a-2^b=256$ 

$\Leftrightarrow 2^b(2^{a-b}-1)=256=2^8$

$\Leftrightarrow 2^{a-b}-1=2^{8-b}$

Với $a-b\geq 1$ thì $2^{a-b}$ chẵn, kéo theo $2^{a-b}-1$ lẻ

$\Rightarrow 2^{8-b}$ lẻ. Điều này xảy ra khi $8-b=0$

$\Leftrightarrow b=8$. Khi đó: $2^{a-b}-1=2^0=1$

$\Leftrightarrow 2^{a-b}=2=2^1\Leftrightarrow a-b=1$

$\Leftrightarrow a=b+1=9$ 

Vậy $(a,b)=(9,8)$

Bài 2: Không mất tổng quát giả sử $x\geq y$

$2020^x+2020^y=2020^{x+y}$

$\Leftrightarrow 2020^y(2020^{x-y}+1-2020^x)=0$

$\Leftrightarrow 2020^{x-y}+1-2020^x=0$

$\Rightarrow 2020^x=2020^{x-y}+1>1\Rightarrow x>0$

$\Rightarrow 2020^{x-y}+1\vdots 2020$

$\Rightarrow 2020^{x-y}\not\vdots 2020$

$\Rightarrow x-y=0$. Mà $2020^0+1=2\not\vdots 2020$ nên loại 

Vậy không tồn tại $x,y$ thỏa mãn.