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

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.

Bài 1.

Ta có:\(\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)=x^2+2020-x^2=2020\)

\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)\)

\(\Rightarrow y+\sqrt{y^2+2020}=\sqrt{x^2+2020}-x\)

\(\Rightarrow x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}\)   (1)

Ta có:\(\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)=y^2+2020-y^2=2020\)

\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)\)

\(\Rightarrow x+\sqrt{x^2+2020}=\sqrt{y^2+2020}-y\)

\(\Rightarrow x+y=\sqrt{y^2+2020}-\sqrt{x^2+2020}\)          (2)

Cộng vế với vế của (1) và (2) ta có:

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

\(\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)

Bài 2: 

Ta có: (2a+1)(2b+1)=9

nên \(2b+1=\dfrac{9}{2a+1}\)

\(\Leftrightarrow2b=\dfrac{9}{2a+1}-\dfrac{2a+1}{2a+1}=\dfrac{8-2a}{2a+1}\)

\(\Leftrightarrow b=\dfrac{8-2a}{4a+2}=\dfrac{4-a}{2a+1}\)

\(\Leftrightarrow b+2=\dfrac{4-a+4a+2}{2a+1}=\dfrac{3a+6}{2a+1}\)

Ta có: \(A=\dfrac{1}{a+2}+\dfrac{1}{b+2}\)

\(=\dfrac{1}{a+2}+\dfrac{2a+1}{3a+6}\)

\(=\dfrac{3+2a+1}{3a+6}\)

\(=\dfrac{2a+4}{3a+6}=\dfrac{2}{3}\)

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\(bx^2=ay^2\Leftrightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\Leftrightarrow\left(\dfrac{x^2}{a}\right)^{1010}=\left(\dfrac{y^2}{b}\right)^{1010}\\ \Leftrightarrow\dfrac{x^{2020}}{a^{1010}}=\dfrac{y^{2020}}{a^{1010}}\)

Áp dụng t/c dtsbn:

\(\dfrac{x^{2020}}{a^{1010}}=\dfrac{y^{2020}}{b^{1010}}=\dfrac{x^{2020}+y^{2020}}{a^{1010}+b^{1010}}\left(3\right)\)

Đặt \(\dfrac{x^2}{a}=\dfrac{y^2}{b}=k\Leftrightarrow x^2=ak;y^2=bk\)

\(x^2+y^2=1\Leftrightarrow ak+bk=1\Leftrightarrow k\left(a+b\right)=1\Leftrightarrow a+b=\dfrac{1}{k}\)

\(\Leftrightarrow\dfrac{2}{\left(a+b\right)^{1010}}=\dfrac{2}{\left(\dfrac{1}{k}\right)^{1010}}=2:\dfrac{1}{k^{1010}}=k^{1010}\left(1\right)\)

Mà \(\dfrac{x^{2020}}{a^{1010}}=\dfrac{\left(x^2\right)^{1010}}{a^{1010}}=\dfrac{a^{1010}k^{1010}}{a^{1010}}=k^{1010}\left(2\right)\)

Từ \(\left(1\right)\left(2\right)\left(3\right)\) ta được đpcm

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

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\right)=\sqrt{6}\)

\(y=\sqrt{\left(\sqrt{6}-1\right)^2}=\sqrt{6}-1\)

\(\Rightarrow x-y=1\Rightarrow P=1\)

\(B=x-2020-\sqrt{x-2020}+\dfrac{1}{4}+\dfrac{8079}{4}\)

\(B=\left(\sqrt{x-2020}-\dfrac{1}{2}\right)^2+\dfrac{8079}{4}\ge\dfrac{8079}{4}\)

\(B_{min}=\dfrac{8079}{4}\) khi \(x=\dfrac{8081}{4}\)

`(x+sqrt{x^2+2020})(sqrt{x^2+2020}-x)=x^2+2020-x^2=2020`

`=>y+sqrt{y^2+2020}=sqrt{x^2+2020}-x`

`<=>x+y=sqrt{x^2+2020}-sqrt{y^2+2020}`

Tương tự:`x+y=sqrt{y^2+2020}-sqrt{x^2+2020}`

Cộng từng vế

`=>2(x+y)=0`

`<=>S=0+2020=2020`

Gt\(\Leftrightarrow\left(x+\sqrt{x^2+2020}\right)\left(x-\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)

\(\Leftrightarrow\left(x^2-x^2-2020\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)

\(\Leftrightarrow-y-\sqrt{y^2+2020}=x-\sqrt{x^2+2020}\) (1)

Gt\(\Leftrightarrow\left(x+\sqrt{x^2+2020}\right)\left(y-\sqrt{y^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(y-\sqrt{y^2+2020}\right)\)

\(\Leftrightarrow\left(y^2-y^2-2020\right)\left(x+\sqrt{x^2+2020}\right)=2020\left(y-\sqrt{y^2+2020}\right)\)

\(\Leftrightarrow-x-\sqrt{x^2+2020}=y-\sqrt{y^2+2020}\) (2)

Từ (1) (2) cộng vế với vế \(\Rightarrow-\left(x+y\right)-\left(\sqrt{y^2+2020}+\sqrt{x^2+2020}\right)=x+y-\left(\sqrt{y^2+2020}+\sqrt{x^2+2020}\right)\)

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

\(\Leftrightarrow x+y=0\)

\(S=x+y+2020=2020\)