a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)

Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)

Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)

Suy ra \(-15\le x\le-2\), x ϵ Z

b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)

Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)

Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)

Suy ra \(-1\le x\le0\), x ϵ Z

b, -4\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) < \(x\) < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))

   - \(\dfrac{13}{3}\).\(\dfrac{1}{3}\) < \(x\) < - \(\dfrac{2}{3}\).(-\(\dfrac{11}{12}\))

    - \(\dfrac{13}{9}\) < \(x\) < \(\dfrac{11}{18}\)

     \(x\) \(\in\) { -1; 0; 1}

a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)

  - \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) <   \(x\)   < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)

   -  \(\dfrac{46}{3}\)     <  \(x\) < - \(\dfrac{13}{7}\) 

          \(x\) \(\in\) {-15; -14;-13;..; -2}

a: \(\Leftrightarrow\left(x+1\right)^2=3^2=9\)

=>x+1=3 hoặc x+1=-3

=>x=2 hoặc x=-4

b: \(\Leftrightarrow\left(x-1\right)^2=16\)

=>x-1=4 hoặc x-1=-4

=>x=5 hoặc x=-3

a) \(\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

⇔ \(\left(x+1\right)^2=9\)

⇒ \(\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\)

⇒ \(\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vây ...

b) Tương tự câu a

Đề thiếu rồi bạn

a, \(\dfrac{x}{2}=-\dfrac{5}{y}\Rightarrow xy=-10\Rightarrow x;y\inƯ\left(-10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

c, \(\dfrac{3}{x-1}=y+1\Rightarrow\left(y+1\right)\left(x-1\right)=3\Rightarrow x-1;y+1\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

b: =>xy=12

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

\(\dfrac{1}{2}x=\dfrac{2}{3}y=\dfrac{3}{4}z\)

\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{\dfrac{3}{2}}=\dfrac{z}{\dfrac{4}{3}}\) 

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2}=\dfrac{y}{\dfrac{3}{2}}=\dfrac{x-y}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30\) 

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=30\Rightarrow x=60\\\dfrac{y}{\dfrac{3}{2}}=30\Rightarrow y=45\\\dfrac{z}{\dfrac{4}{3}}=30\Rightarrow z=40\end{matrix}\right.\)

1.

=2/5 x 12/3 + 2/5 x 15/3 + 2/5 x 1

= 2/5 x (12/3 + 15/3 + 1)

=2/5 x 1

=2/5

2.a=1;2

a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)

\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)

\(\Leftrightarrow x-4=25\)

\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)

b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)

\(\Leftrightarrow x\left(x+1\right)=18.4\)

\(\Leftrightarrow x\left(x+1\right)=72\)

vì \(72=8.9=\left(-8\right).\left(-9\right)\)

\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)

c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)

\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)

\(\Leftrightarrow2x+3-2x-8⋮x+4\)

\(\Leftrightarrow-5⋮x+4\)

\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)

\(a.\)

\(\dfrac{-28}{35}=\dfrac{16}{x}\)

\(\Rightarrow x=\dfrac{35\cdot16}{-28}=\dfrac{5\cdot7\cdot4\cdot4}{-7\cdot4}=-20\)

\(b.\)

\(\dfrac{x+7}{15}=\dfrac{-24}{36}\)

\(\Rightarrow x+7=\dfrac{15\cdot-24}{36}=\dfrac{5\cdot3\cdot-12\cdot2}{12\cdot3}=-10\)

\(\Leftrightarrow x=-17\)