cho e xin in4

(y:3)x8=8x3198

(y:3)x8=25584

y:3=25584:8

y:3=3198

y=3198x3

y=9594

( y : 3 ) x 8 = 8 x 3198

=> y : 3 = 3198 ( vì cùng thừa số là 8 )

     y       = 3198 x 3 = 9594

y : 3  = 8 x 3198 : 8 

y : 3 = 3198

y = 3198 x 3

y= 9594

Vậy y = 9594

Nhớ k cho mình nhé! Thank you!!!

X=9594

TK NHA

cách 1

( y : 3 ) x 8 = 25584 

y  : 3 = 25584 : 8 

y : 3 = 3198 

y = 3198 : 3 

y = 1066 

\(Có: x=2010.2010\implies x=2010^2\)

        \(y=2000.2020\implies y=(2010-10).(2010+10)\implies 2010^2-10^2\)

Mà 2010² và 10² lớn hơn 0.

\(\implies 2010^2>2010^2-10^2\)

\(\implies x>y\)

Vậy x>y.

_Học tốt_

y=  19197

x= 9594

k mk nhé bạn iu

\(6\times\left(y+9845\right)=29042\times6\)

            \(y+9845\) \(=29042\times6\div6\)

           \(y+9845\)  \(=29042\)

           \(y\)                    \(=29042-9845\)

           \(y\)                    \(=19197\)

\(\left(x\div3\right)\times8=8\times3198\)

   \(x\div3\)        \(=8\times3198\div8\)

   \(x\div3\)        \(=3198\)

   \(x\)                  \(=3198\times3\)

   \(x\)                  \(=9594\)

(x:3)*8=25584

x:3=25584:8

x:3=3198

x=3198*3

x=9594

Bài này dễ mak bn~

\(\left(x:3\right).8=3198.8\)

\(\left(x:3\right).8=25584\)

\(x:3=25584:8=3198\)

\(\Rightarrow x=3198.3=9594\)

( x : 3 ) x 8 = 3198 x 8

x : 3 = 3198

x = 3198 x 3

x = 9594

 Tk nha!

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

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

\(\Leftrightarrow x+y=0\)

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

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

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.

2020 x 2020 > 2019 x 2021

\(2019\times2021=\left(2020-1\right)\left(2020+1\right)=2020^2-1< 2020^2=2020\times2020\)

2020x2020>2019x2021