a, \(\frac{17}{y}=\frac{-7}{11}\)

\(\Rightarrow17\cdot11=-7\cdot y\)

\(\Rightarrow187=-7\cdot y\)

\(\Rightarrow\frac{187}{-7}=y\)

b, \(\frac{-8}{3x-1}=\frac{4}{7}\)

\(\Rightarrow\frac{-8}{3x-1}=\frac{-8}{-14}\)

\(\Rightarrow3x-1=-14\)

\(\Rightarrow3x=-14+1\)

\(\Rightarrow3x=-13\)

\(\Rightarrow x=\frac{-13}{3}\)

c, \(\frac{x}{-3}=\frac{-3}{x}\)

\(\Rightarrow x\cdot x=-3\cdot\left(-3\right)\)

\(\Rightarrow x^2=9\)

\(\Rightarrow x^2=\left(\pm3\right)^2\)

\(\Rightarrow x=\pm3\)

d, \(\frac{-4}{y}=\frac{x}{2}\)

\(\Rightarrow-4\cdot2=x\cdot y\)

\(\Rightarrow-8=x\cdot y\)

\(\Rightarrow x;y\inƯ\left(-8\right)=\left\{-1;1;-2;2;-4;4;-8;8\right\}\)

ta có bảng :

a)\(\frac{14}{y}\)\(=\)   \(\frac{-7}{11}\)

\(\Rightarrow\)\(14\cdot11=y\cdot\left(-7\right)\)

                   \(y=\)\(\frac{14\cdot11}{-7}\)

                   \(y=22\)

c) \(\frac{x}{-3}\) =    \(\frac{-3}{x}\)

\(\Rightarrow\) \(x\cdot x=\left(-3\right)\cdot\left(-3\right)\)

\(\Rightarrow\)\(x^2=9\)

\(\Rightarrow\)\(x^2=9\)hoặc  \(x^2=-9\)

\(TH1:\) \(x^2=9\)

     \(\Rightarrow\)\(x=3\)

\(TH2:\)\(x^2=-9\)

     \(\Rightarrow\)\(x=-3\)

\(\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3}{6}+\frac{2}{6}+\frac{1}{6}=\frac{6}{6}=1\)

\(\frac{13}{14}+\frac{14}{8}=\frac{13.4}{14.4}+\frac{14.7}{8.7}=\frac{52}{56}+\frac{98}{56}=\frac{150}{56}\simeq2,68\)

Như vậy: \(1\le x\le2,68\)

Mà x thuộc N => x=1 và x=2

Đáp số: x=1 và x=2

Câu a:

- \(\frac12\)(3 - 2\(x\)) - 7 = 5 - \(\frac13\)(\(x\) - \(\frac45\))

- \(\frac32\) + \(x\) - 7 = 5 - \(\frac13x\) + \(\frac{4}{15}\)

\(x\) + \(\frac13x\) = 5 + \(\frac{4}{15}\) + 7 + \(\frac32\)

(1 + \(\frac13\))\(x\) = \(\frac{150}{30}\) + \(\frac{8}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)

\(\frac43x\) = \(\frac{158}{30}\) + \(\frac{210}{30}\) + \(\frac{45}{30}\)

\(\frac43x\) = \(\frac{368}{30}\) + \(\frac{45}{30}\)

\(\frac43x\) = \(\frac{413}{30}\)

\(x\) = \(\frac{413}{30}\) : \(\frac43\)

\(x\) = \(\frac{413}{40}\)

Vậy \(x=\frac{413}{40}\)

Câu b:

(5 - 3x/2) : - 1 3/8 = - 7 1/3

(5 - 3x/2) : (-11/8) = - 22/3

5 - 3x/2 = - 22/3 x (-11/8)

5 - 3x/2 = 121/12

3x/2 = 5 - 121/12

3x/2 = - 61/12

x = - 61/12 : 3/2

x = -61/18

Vậy x = - 61/18

Câu a:

\(\frac{-8}{3x-1}\) = \(\frac{4}{-7}\)

-8.(-7) = 4.(3\(x\) - 1)

56 = 12\(x\) - 4

12\(x\) = 56+ 4

12\(x\) = 60

\(x\) = 60 : 12

\(x\) = 5

Vậy \(x\) = 5

Câu b:

\(\frac{x}{-3}\) = \(\frac{-3}{x}\)

\(x^2\) = (-3)\(^2\)

\(\left[\begin{array}{l}x=-3\\ x=3\end{array}\right.\)

Vậy \(x\in\left\lbrace-3;3\right\rbrace\)

Câu c:

\(-\frac{4}{y}=\frac{x}{2}\)

-4.2 = \(x.y\)

\(xy=-8\)

Ư(8) = (-8; -4; -2; -1; 1; 2; 4; 8}

Vậy (\(x;y\)) = (-8; 1); (-4; 2); (-2; 4); (-1; 8); (1; -8); (2; -4); (4; -2); (8; -1)

Câu 2:

(\(x-1)\)(y + 2) = 7

Ư(7) = {-7; -1; 1; 7}

Lập bảng ta có:

Theo bảng trên ta có:

(\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; - 1)

Vậy (\(x;y\)) = (-6; -3); (0; -9); (2; 5); (8; -1)

Thèo đề bài, ta có:

\(\frac{x^3}{2^3}=\frac{y^3}{4^3}=\frac{z^3}{6^3}=\frac{x}{2}=\frac{y}{4}=\frac{z}{6}=\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)

x ; y ; z thì bạn tự tìm nhé , chắc cái này không khó đâu nhỉ ??

\(\frac{x^3}{8}=\frac{y^3}{64}=\frac{z^3}{216}\Rightarrow\frac{x}{2}=\frac{y}{4}=\frac{z}{6}\Rightarrow\frac{x^2}{4}=\frac{y^2}{16}=\frac{z^2}{36}\) \(=\frac{x^2+y^2+z^2}{4+16+36}=\frac{14}{56}=\frac{1}{4}\)

\(\frac{x}{2}=\frac{1}{4}\Rightarrow x=\frac{1}{2}\)

\(\frac{y}{4}=\frac{1}{4}\Rightarrow y=1\)

\(\frac{z}{6}=\frac{1}{4}\Rightarrow z=\frac{3}{2}\)