Lời giải:Để $y$ nguyên thì $x^3+1\vdots x^4+1$

$\Leftrightarrow x^4+x\vdots x^4+1$

$\Leftrightarrow x^4+1+x-1\vdots x^4+1$

$\Leftrightarrow x-1\vdots x^4+1$

Nếu $x-1=0$ thì điều trên đúng. Kéo theo $y=1$

Nếu $x-1\neq 0$ thì $|x-1|\geq x^4+1(*)$

Cho $x>1$ thì $(*)\Leftrightarrow x-1\geq x^4+1$

$\Leftrightarrow x(1-x^3)-2\geq 0$ (vô lý với mọi $x>1$)

Cho $x< 1$ thì $(*)\Leftrightarrow 1-x\geq x^4+1$

$\Leftrightarrow x^4+x\leq 0$

$\Leftrightarrow x(x^3+1)\leq 0$

$\Leftrightarrow -1\leq x\leq 0$. Do $x$ nguyên nên $x=-1$ hoặc $x=0$

Với $x=-1$ thì $y=0$

Với $x=0$ thì $y=1$

Vậy..........

\(x+3⋮x+1\)

\(\Rightarrow\left(x+1\right)+2⋮x+1\)

\(\Rightarrow2⋮x+1\)

\(\Rightarrow x+1\in\left\{-1;+1:-2+2\right\}\)

\(\Rightarrow x\in\left\{-2;0;-3;1\right\}\)

\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}.4\frac{1}{2}-2.2\frac{1}{3}\right):\frac{7}{4}\)

\(=\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}.\frac{9}{2}-2.\frac{7}{3}\right):\frac{7}{4}\)

\(=\frac{59}{15}-\left[\frac{7}{3}\left(\frac{9}{2}-2\right)\right]:\frac{7}{4}\)

\(=\frac{59}{15}-\frac{35}{6}:\frac{7}{4}\)

\(=\frac{59}{15}-\frac{10}{3}\)

\(=\frac{9}{15}=\frac{3}{5}\)

\(\cdot62,87+35,14+4,13+8,35+4,86+5,65\)

\(=\left(62,87+4,13\right)+\left(35,14+4,86\right)+\left(8,35+5,65\right)\)

\(=67+40+14\)

\(=121\)

Ta có \(\frac{11}{4}\)> \(\frac{x}{20}\)> \(\frac{5}{11}\)

=>    \(\frac{605}{220}\)> \(\frac{11x}{220}\)> \(\frac{100}{220}\)

=> 605 > 11x > 100

=>  56 > x > 9

\(\frac{11}{4}>\frac{x}{20}>\frac{5}{11}\)

\(\Leftrightarrow\frac{605}{220}>\frac{11x}{220}>\frac{100}{220}\)

\(\Leftrightarrow\frac{55}{20}>\frac{x}{20}>\frac{9,\left(09\right)}{20}\)

\(\Rightarrow x\in\left\{11;22;33;44\right\}\)

Ta quy đồng mẫu số:

\(\frac{11}{4}\)>\(\frac{x}{20}\)>\(\frac{5}{11}\)

-> \(\frac{11.20.11}{4.20.11}\) >  \(\frac{x.11.4}{4.20.11}\)>   \(\frac{5.4.20}{4.20.11}\)

-> \(\frac{2420}{880}\)>   \(\frac{44x}{880}\)>    \(\frac{400}{880}\)

-> 2420  >  44x  >  400

-> 44x thuộc {401;...;2419}

-> x thuộc {10;11;12;13;...;51;52;53;54}