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

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

$\textbf{Bài 1}$

a) Vì $313{,}9543\ldots<a<314{,}1762\ldots$

nên chọn $a=314.$

b) Vì $-35{,}2475\ldots<a<-34{,}9628\ldots$ nên chọn $a=-35.$

Bài 2

a) Ta có $0,(37)=\dfrac{37}{99},\qquad 0,(62)=\dfrac{62}{99}.$

Do đó $0,(37)+0,(62)=\dfrac{37+62}{99}=\dfrac{99}{99}=1.$

b) Ta có $0,(33)=\dfrac{33}{99}=\dfrac13.$

Suy ra $0,(33)\cdot3=\dfrac13\cdot3=1.$

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

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

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

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

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

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

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

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

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

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

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

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

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

$\textbf{1)}$

Đặt $a-b=\dfrac23(a+b)=\dfrac ab=k.$

Khi đó $a=kb\ (b\ne0).$

Từ $a-b=k$ suy ra $kb-b=k\Leftrightarrow b(k-1)=k.$

Mặt khác, $\dfrac23(a+b)=k\Leftrightarrow2(kb+b)=3k.$

Thay $b=\dfrac{k}{k-1}$ vào:

$\dfrac{2k(k+1)}{k-1}=3k$

$\Leftrightarrow2(k+1)=3(k-1)$

$\Leftrightarrow k=5.$

Suy ra $b=\dfrac54,\qquad a=kb=\dfrac{25}{4}.$

Vậy $a=\dfrac{25}{4},\qquad b=\dfrac54.$

$\textbf{2)}$

Đặt $2(a+b)=5ab=\dfrac{20a}{b}=k.$

Khi đó $a=\dfrac{k}{20}b\ (b\ne0).$

Từ $\dfrac{20a}{b}=k$ suy ra $a=\dfrac{k}{20}b.$

Lại có $5ab=k\Leftrightarrow\dfrac{k}{4}b^2=k.$

Vì $k\ne0$ nên $b^2=4\Rightarrow b=\pm2.$

Từ $2(a+b)=k$ ta được $\dfrac{k}{10}b+2b=k.$

- Với $b=2$: $\dfrac{k}{5}+4=k\Rightarrow k=5,$ suy ra $a=\dfrac{5}{20}\cdot2=\dfrac12.$

- Với $b=-2$: $-\dfrac{k}{5}-4=k\Rightarrow k=-\dfrac{10}{3},$

suy ra $a=\dfrac{-10/3}{20}\cdot(-2)=\dfrac13.$

Vậy có hai cặp số: $\boxed{a=\dfrac12,\ b=2}$ hoặc $\boxed{a=\dfrac13,\ b=-2}.$

Đặt $a+b=ab=\dfrac ab=k.$

Khi đó $a=kb\ (b\ne0).$

Thay vào $a+b=ab$:

$kb+b=kb^2\Leftrightarrow b(k+1)=kb^2\Leftrightarrow b=\dfrac{k+1}{k}.$

Suy ra $a=kb=k+1.$

Lại có $(k+1)+\dfrac{k+1}{k}=k.$

Nhân hai vế với $k$:

$k(k+1)+k+1=k^2\Leftrightarrow2k+1=0\Leftrightarrow k=-\dfrac12.$

Do đó $a=k+1=\dfrac12,\qquad b=\dfrac{k+1}{k}=-1.$

Kiểm tra:

$a+b=\dfrac12-1=-\dfrac12,\quad ab=\dfrac12\cdot(-1)=-\dfrac12,\quad\dfrac ab=\dfrac{\frac12}{-1}=-\dfrac12.$

Vậy $a=\dfrac12,\ b=-1.$

Đặt $a+b=ab=\dfrac ab=k.$

Khi đó $a=kb\ (b\ne0).$

Thay vào $a+b=ab$:

$kb+b=kb^2\Leftrightarrow b(k+1)=kb^2\Leftrightarrow b=\dfrac{k+1}{k}.$

Suy ra $a=kb=k+1.$

Lại có $(k+1)+\dfrac{k+1}{k}=k.$

Nhân hai vế với $k$:

$k(k+1)+k+1=k^2\Leftrightarrow2k+1=0\Leftrightarrow k=-\dfrac12.$

Do đó $a=k+1=\dfrac12,\qquad b=\dfrac{k+1}{k}=-1.$

Kiểm tra:

$a+b=\dfrac12-1=-\dfrac12,\quad ab=\dfrac12\cdot(-1)=-\dfrac12,\quad\dfrac ab=\dfrac{\frac12}{-1}=-\dfrac12.$

Vậy $a=\dfrac12,\ b=-1.$