\(A=\frac{\left(x+y\right)\left[\left(x+y\right)^2-3xy\right]}{\left(xy\right)^3}=\frac{4.\left(16-6\right)}{8}=5\)

\(\frac{17^2}{x}=\frac{7x^2}{-119}\)

=\(17^2.\left(-119\right)=7x^2.x\)

=\(-34391=7x^3\)

=\(-4913=x^3\)

x=\(\sqrt[3]{-4913}=-17\)

_ Số nguyên x đó là -2 nhé.

1.

\(2\left|x\right|+3\left|y\right|=13\Rightarrow\left|x\right|=\dfrac{13-3\left|y\right|}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}\left|y\right|\le\dfrac{13}{3}\\\left|y\right|\text{ là số lẻ}\end{matrix}\right.\)  \(\Rightarrow\left|y\right|=\left\{1;3\right\}\)

- Với \(\left|y\right|=1\Rightarrow\left|x\right|=5\Rightarrow\) có 4 cặp

- Với \(\left|y\right|=3\Rightarrow\left|x\right|=2\) có 4 cặp

Tổng cộng có 8 cặp số nguyên thỏa mãn

2.

\(x\left(y+3\right)=7y+21+1\)

\(\Leftrightarrow x\left(y+3\right)-7\left(y+3\right)=1\)

\(\Leftrightarrow\left(x-7\right)\left(y+3\right)=1\)

\(\Rightarrow\left(x;y\right)=\left(6;-4\right);\left(8;-2\right)\) có 2 cặp

B. \(2-\frac{13}{3}< x< 1-2,4\)

\(-\frac{7}{3}< x< -\frac{7}{5}\)

\(\Rightarrow x=-\frac{7}{4}\)

C. 13x + 350 = 1000

13x = 650

x = 50 

D. \(\frac{4}{7}x-\frac{5}{8}=\frac{17}{24}\)

\(\frac{4x}{7}=\frac{4}{3}\)

\(\Rightarrow12x=28\)

\(\Rightarrow x=\frac{7}{3}\)

E. \(\frac{3}{7}x=5\)

\(x=5:\frac{3}{7}=\frac{5.7}{3}=\frac{35}{3}\)

Vì \(x\in Z\Rightarrow x\in O\)

G. 10 

bn đưa ra câu hỏi thế này

mik đọc mõi mắt lw

đọc không nổi

-1.4

\(\frac{37-x}{x+13}=\frac{3}{7}\)

=>7.(37-x)=3.(x+13)

<=>259-7x=3x+39

<=>3x+7x=259-39

<=>10x=220

<=>x=22

\(\frac{37-x}{x+13}=\frac{3}{7}\Leftrightarrow\left(37-x\right).7=\left(x+13\right).3\Leftrightarrow259-7x=3x+39\)(nhân chéo)

\(\Leftrightarrow3x+7x=259-39\Rightarrow10x=220\Rightarrow x=220:10\Rightarrow x=22\)

Vậy x=22

1/  196

2/  5/4

3/  1/3