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

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

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 $\boxed{a=\dfrac13,\ b=-2}.$

Ta có:

\(a:b=2\dfrac{3}{3}:\dfrac{9}{10}=3:\dfrac{9}{10}=3\times\dfrac{10}{9}=\dfrac{30}{9}=\dfrac{10}{3}\)

Vậy, tỉ số của a và b là `10/3`

Vì \(\frac{ab}{2}=8\cdot\frac{a}{b}\)nên\(\Leftrightarrow a\cdot\frac{b}{2}=8\cdot\frac{a}{b}\)

\(\Leftrightarrow\frac{\left(\frac{ab}{2}\right)}{\left(\frac{a}{b}\right)}=8\Leftrightarrow\frac{ab}{2}\cdot\frac{b}{a}=8\Leftrightarrow\frac{ab^2}{2a}=8\)

\(\Leftrightarrow\frac{b^2}{2}=8\Leftrightarrow b^2=8\cdot2=16\Leftrightarrow b=\sqrt{16}=4\)

Vì \(a+\frac{b}{2}=\frac{ab}{2}\)(1)mà \(b=4\) nên thay b vào biểu thức (1)được:

\(a+\frac{4}{2}=\frac{a4}{2}\Leftrightarrow a+2=a\cdot2\)

\(\Leftrightarrow2=a\)

Vậy \(a=2;b=4\)để thỏa mãn \(a+\frac{b}{2}=\frac{ab}{2}=8\cdot\frac{a}{b}\)

a/ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}=\sqrt{\frac{1}{\left(b+c\right)^2}+\frac{1}{b^2}+\frac{1}{c^2}}\)

\(=\sqrt{\frac{\left(b+c\right)^2.b^2+\left(b+c\right)^2.c^2+b^2.c^2}{\left(b+c\right)^2.b^2.c^2}}\)

\(=\sqrt{\frac{\left(b^2+bc+c^2\right)^2}{\left(b+c\right)^2.b^2.c^2}}\)

\(=\left|\dfrac{b^2+bc+c^2}{\left(b+c\right).b.c}\right|\)

Vậy \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}\)là số hữu tỉ

b/ \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\sqrt{\frac{1}{\left(b+c\right)^2}+\frac{1}{b^2}+\frac{1}{\left(2b+c\right)^2}}\)

\(=\sqrt{\frac{\left(b+c\right)^2.b^2+\left(b+c\right)^2.\left(2b+c\right)^2+\left(2b+c\right)^2.b^2}{\left(b+c\right)^2.\left(2b+c\right)^2.b^2}}\)

\(=\sqrt{\frac{\left(3b^2+3bc+c^2\right)^2}{\left(b+c\right)^2.\left(2b+c\right)^2.b^2}}\)

\(=\left|\dfrac{3b^2+3bc+c^2}{\left(b+c\right).\left(2b+c\right).b}\right|\)

Vậy \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}\) là số hữu tỉ

Ta có:

a - b = 2(a + b)

=> a - b = 2a + 2b

=> a - 2a = 2b + b

=> -a = 3b

\(\Rightarrow\frac{a}{b}=-3\); \(a=-3b\)

Laị có:

a - b = \(3.\frac{a}{b}\)

=> -3b - b = 3.(-3)

=> -4b = -9

\(\Rightarrow b=\frac{-9}{-4}=\frac{9}{4}\)

\(\Rightarrow a=\frac{9}{4}.\left(-3\right)=\frac{-27}{4}\)

Vậy \(a=\frac{-27}{4};b=\frac{9}{4}\)