\(\dfrac{a}{b}=\dfrac{a\left(b+2021\right)}{b\left(b+2021\right)}=\dfrac{ab+2021a}{b\left(b+2021\right)}\\ \dfrac{a+2021}{b+2021}=\dfrac{ab+2021b}{b\left(b+2021\right)}\)

Vì \(b>0\Rightarrow b\left(b+2021\right)>0\)

Nếu \(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2021}{b+2021}\)

Nếu \(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2021}{b+2021}=1\)

Nếu \(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2021}{b+2021}\)

\(a-b=2\left(a+b\right)=\dfrac{a}{b}\\ \Leftrightarrow a-b=2a+2b\\ \Leftrightarrow a+3b=0\Leftrightarrow a=-3b\)

\(a-b=\dfrac{a}{b}\Leftrightarrow-4b=\dfrac{-3b}{b}=-3\Leftrightarrow b=\dfrac{3}{4}\)

\(\Leftrightarrow a=-3\cdot\dfrac{3}{4}=-\dfrac{9}{4}\)

Theo bài,ta có

a-b=2(a+b)

a-b=2a+2b

-a=3b

=>a=-3b

a:b=-3b/b=-3

=>a-b=-3

-3b-b=-3

giúp mk vs

\(a=1+2+2^2+...+2^{2021}\)

\(\Rightarrow2a=2+2^2+2^3+...+2^{2022}\)

\(\Rightarrow2a-a=2+2^2+2^3+...+2^{2022}-1-2-2^2-...-2^{2021}\)

\(\Rightarrow a=2^{2022}-1\)

\(\Rightarrow a=2^{2022}-1=b\)

\(a=1+2+2^2+2^3+...+2^{2021}\)

\(2a=2+2^2+2^3+2^4...+2^{2021}+2^{2022}\)

\(2a-a=\)\(\left(2+2^2+2^3+2^4...+2^{2021}+2^{2022}\right)-\left(1+2+2^2+2^3+...+2^{2021}\right)\)

\(a=2^{2022}-1\)

⇒ a=b

\(a-b=\frac{2}{3}\left(a+b\right)\Leftrightarrow a-b=\frac{2}{3}a+\frac{2}{3}b\Leftrightarrow\frac{1}{3}a=\frac{5}{3}b\Leftrightarrow a=5b\Rightarrow a:b=5\)

\(\Rightarrow a-b=\frac{2}{3}a+\frac{2}{3}b=5\Leftrightarrow\hept{\begin{cases}a-b=5\\a+b=\frac{15}{2}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{25}{4}\\b=\frac{5}{4}\end{cases}}\)

1. Ta có a - b =2 (a+b)=2a+3b

<=> a-2a =2b+b

<=>a=3b<=> =2b+b

Thay a =-3b <=> -3b

=> a : b =-3b : b = 3

=>a-b=3

2(a+b)=-3<=>a+b=\(-\frac{3}{2}\)(Phân số nghịc đảo -)

Khi đó a= \(\frac{\left(a+b\right)+\left(a-b\right)}{2}=\frac{\left(-\frac{3}{2}\right)+\left(-3\right)}{2}=\frac{9}{4}\)

b=\(\frac{\left(a+b\right)-\left(a-b\right)}{2}=\frac{\left(-\frac{3}{2}\right)+\left(-3\right)}{2}=\frac{3}{4}\)

Thay a - b  (a+1)

a : b =a-b 

<=> b - 1  = -1 

a-b=ab 

=> a +b = 1

a-b = ab hay = a+1=-a

=>2a-1

=>\(\frac{1}{2}\)

xIn lỗi chữa lại :

a + b =ab =>a=ab-b =b(a-1)

Thay a = b - 1 v a 

=> a + b  = a : b 

\(a+b=\frac{a\left(b-1\right)}{b}=a-1\)

=> a = b =a - 1

=> a + b -a = -1 =>b=-1

v:

a.b=a+b

<=> -1.a =a+--1 

=> -a =a-1 =>a- -a =-1 < a =-1 => a \(\frac{1}{2}\)

T mới dừng cách này còn nữa :) cách này cx là cách 2 nha :)

\(a=-1\)