E = x^(4)*y^(4)+x^(5)*y^(5)+x^(6)*y^(6)+x^(7)*y^(7)+x^(8)*y^(8)+x^(9)*y^(9)+x^(10)*y^(10) tại x=-1, y=1 nha

a) x,y nguyên => x+4; y-8 nguyên

=> x+4; y-8\(\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

ta có bảng

Vậy (x;y)={(-10;7);(-7;6);(-6;5);(-5;2);(-3;14);(-2;11);(-1;10);(2;9)}

b) 2x+xy+3y+6=10

<=> x(2+y)+3(y+2)=10

<=> (y+2)(x+3)=10

x,y nguyên => y+2; x+3 nguyên 

=> y+2; x+3\(\in\)Ư(10)={-10;-5;-2;-1;1;2;5;10}

ta có bảng

a)y=5/6

b)y=11/4

\(a\left(\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right).y=\frac{1}{3}\)
\(\left(\frac{1}{2}-\frac{1}{10}\right).y=\frac{1}{3}\)

\(\frac{2}{5}.y=\frac{1}{3}\)

      \(y=\frac{1}{3}:\frac{2}{5}\)

     \(y=\frac{5}{6}\)

\(b,\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{11}\right).y=\frac{2}{3}\)

     \(\left(\frac{1}{1}-\frac{1}{11}\right).y=\frac{2}{3}\)

      \(\frac{10}{11}.y=\frac{2}{3}\)

              \(y=\frac{2}{3}:\frac{10}{11}\)

               \(y=\frac{22}{30}\)

a: \(\dfrac{-4}{8}=\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}\)

=>\(\dfrac{x}{-10}=\dfrac{-7}{y}=\dfrac{z}{-24}=\dfrac{-1}{2}\)

=>\(\left\{{}\begin{matrix}x=\left(-10\right)\cdot\dfrac{\left(-1\right)}{2}=5\\y=\dfrac{-7\cdot2}{-1}=14\\z=\dfrac{-24\cdot\left(-1\right)}{2}=\dfrac{24}{2}=12\end{matrix}\right.\)

b: \(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

=>\(\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{z}{-24}=\dfrac{-1}{2}\)

=>\(\dfrac{x}{2}=\dfrac{18}{y}=\dfrac{z}{24}=\dfrac{1}{2}\)

=>\(x=2\cdot\dfrac{1}{2}=1;y=18\cdot\dfrac{2}{1}=36;z=\dfrac{24}{2}=12\)