Do x, y \(\in\) Z mà (x-3).(y+4) = -7 nên x-3 và y+4 là ước nguyên của -7.
Ư(-7) = {1; -1; 7; -7}
Ta có bảng giá trị:

Vậy (x; y) \(\in\) {(4;-11);(2;3);(10;-5);(-4;-3)}

ĐKXĐ: \(x+3\ne0\Leftrightarrow x\ne-3\)

\(\dfrac{x+3}{-4}=\dfrac{-9}{x+3}\\ \Leftrightarrow\left(x+3\right)\left(x+3\right)=\left(-9\right).\left(-4\right)\\ \Leftrightarrow\left(x+3\right)^2=6^2\\ \Rightarrow\left[{}\begin{matrix}x+3=6\\x+3=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-9\left(tm\right)\end{matrix}\right.\)

Vậy ...

\(\dfrac{x+3}{-4}=\dfrac{-9}{x+3}\)

\(\left(x+3\right)\left(x+3\right)=\left(-9\right)\cdot\left(-4\right)\)

\(\left(x+3\right)^2=36\)

\(\left(x+3\right)^2=6^2\)

\(x+3=6\)

\(x=3\)

-5/14 - 2/14 + 1/8 + 1/8=-1/4

-5/22 - 1 + 3/2 - 6/22=0

-1/2 + 1/4 + 2/8 - 11/77=-1/7

2/5 + 5/4 + (-2/5+5/4)-5/2=0

mình trả lời thế có đúng không nhỉ? 🤔

- \(\dfrac{5}{14}\) - \(\dfrac{2}{14}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)

= -(\(\dfrac{5}{14}\) + \(\dfrac{2}{14}\)) + (\(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)) 

= - \(\dfrac{7}{14}\) + \(\dfrac{2}{8}\)

= - \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\)

= - \(\dfrac{1}{4}\)

Bài 1:

a) 

\(\dfrac{1}{-5}=\dfrac{1:-1}{-5:-1}=\dfrac{-1}{5}\)

\(\dfrac{-3}{-15}=\dfrac{-3:-3}{-15:-3}=\dfrac{1}{5}\)

Vì \(-1< 1\) nên\(\dfrac{-1}{5}< \dfrac{1}{5}\)

Vậy \(\dfrac{1}{-5}< \dfrac{-3}{-15}\)

b) 

\(\dfrac{25}{-10}=\dfrac{25:-5}{-10:-5}=\dfrac{-5}{2}\)

\(\dfrac{15}{-2}=\dfrac{15:-1}{-2:-1}=\dfrac{-15}{2}\)

Vì \(-5>-15\) nên\(\dfrac{-5}{2}>\dfrac{-15}{2}\)

Vậy \(\dfrac{25}{-10}>\dfrac{15}{-2}\)

Bài 2:

a)

\(\dfrac{300}{540}=\dfrac{300:60}{540:60}=\dfrac{5}{9}\)

b)
\(\dfrac{-38}{95}=\dfrac{-38:19}{95:19}=\dfrac{-2}{5}\)

c)
\(\dfrac{-68}{-85}=\dfrac{-68:-17}{-85:-17}=\dfrac{4}{5}\)

Bài 3:

a)

\(\dfrac{x}{3}=-\dfrac{4}{6}\)

\(\dfrac{x}{6}=-\dfrac{4}{6}\)

\(x=-4\)

b) \(x-\dfrac{1}{3}=\dfrac{8}{6}\)

\(x=\dfrac{8}{6}+\dfrac{1}{3}\)

\(x=\dfrac{5}{3}\)

c) \(\dfrac{x}{9}=\dfrac{4}{x}\)

\(x^2=36\)

\(x=6\)

d) mình có giải câu hỏi trước rồi nhé

Sửa câu c)

\(x^2=36\)

\(\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

ĐKXĐ: \(\left\{{}\begin{matrix}x-5\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne5\\x\ne-2\end{matrix}\right.\)

\(\dfrac{3}{x-5}=\dfrac{-4}{x+2}\\ \Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\\ \Leftrightarrow3x+6=-4x+20\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\left(tm\right)\)

Vậy x = 2

-2 - 5 + x = -1 + 7

=> -7 + x = 6

=> x = 6 + 7 

=> x = 13

Theo đề bài ta có:

=> -7 + x = -1 +7

=> -7 + x = 6

=> x         = 6 - (-7)

=> x         = 6 + 7

=> x         = 13