1/x-y/6=1/3

ĐKXĐ x khác 0

Ta có

      1/x - y/6 = 1/3

<=> 6 - xy = 2x ( nhân cả 2 vế với 6x khác 0 )

<=> 2x + xy = 6

<=> x(2 + y) = 6

Do x, y ∈ Z => x và  2 + y ∈ Z 

=> x|6 và 2+y | 6

=> (x , 2+y) = ( -1 ; -6) , (-6 ; -1), ( -2 ; - 3), ( - 3 ; - 2); (1 , 6) , ( 6, 1), (2 , 3) , (3,2)

=> (x,y) = (-1;-8), (-6;-3), (-2;-5), (-3;-4), (1;4), (6;-1), (2;1),(3; 0) ( TM ĐKXĐ)

$\frac{x}{2}+\frac{3}{y}=\frac{5}{4}$

$\Rightarrow \frac{xy+6}{2y}=\frac{5}{4}\Rightarrow 4xy+24=10y$

$\Rightarrow 10y-4xy=24$

$\Rightarrow 2y.\left(5-2x\right)=24$

$\Rightarrow y.\left(5-2x\right)=12$

$\Rightarrow y;5-2x\in Ư\left(12\right)$

$\Rightarrow y;5-2x\in \left\{1;2;3;4;6;12\right\}$

a) \(5,4+\left(-23,2\right)+4,6+\left(-0,8\right)\)

\(=\left(5,4+4,6\right)+\left(-23,2-0,8\right)=10-24=-14\)

b) \(\left(-14,3\right)+5,1+4,9+\left(-15,7\right)\)

\(=\left(5,1+4,9\right)+\left(-14,3-15,7\right)=10-30=-20\)

c) \(12,3+3,7+\left(-24,2\right)+\left(-12,3\right)+24,2\)

\(=\left(12,3-12,3\right)+\left(-24,2+24,2\right)+3,7=3,7\)

d) \(\left(-16,5\right).3,4+3,4+\left(-3,5\right)\)

\(=3,4\left(-16,5-3,5\right)=3,4\left(-20\right)=-68\)

\(c,TH1:\frac{2}{3}x-\frac{4}{9}=0=>x=\frac{2}{3}\)

\(TH2:\frac{1}{2}+-\frac{3}{7}:x=0=>x=\frac{6}{7}\)

Vậy \(x\in\left\{\frac{2}{3};\frac{6}{7}\right\}\)

c) \(\left(\frac{2}{3}.x-\frac{4}{9}\right)\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}.x-\frac{4}{9}=0\\\frac{1}{2}+\frac{-3}{7}:x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}.x=\frac{4}{9}\\\frac{-3}{7}:x=-\frac{1}{2}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{6}{7}\end{cases}}\)

\(10a+b+10b+c+10c+a=100a+10b+c\)

\(89a-b=10c\)

\(c\le9\Rightarrow10c\le90\Rightarrow89a-b\le90\Rightarrow a=1\)

\(\Rightarrow89-b=10c\)

\(10c⋮10\Rightarrow89-b⋮10\Rightarrow b=9\Rightarrow89-9=10c\Rightarrow c=8\)

Ta có \(19+98+81=198\)

\(\orbr{\begin{cases}x+2=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)

\(\left(x+2\right).\left(x+3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)

Vậy \(x\in\left\{-2;-3\right\}\)