a+1 là bội của a-1

=>\(a+1⋮a-1\)

=>\(a-1+2⋮a-1\)

=>\(2⋮a-1\)

=>\(a-1\in\left\{1;-1;2;-2\right\}\)

=>\(a\in\left\{2;0;3;-1\right\}\)

mà a là số tự nhiên

nên \(a\in\left\{2;0;3\right\}\)

Đặt \(A=2^2+2^3+...+2^{2021}\)

=>\(2A=2^3+2^4+...+2^{2022}\)

=>\(2A-A=2^3+2^4+...+2^{2022}-2^2-2^3-...-2^{2021}\)

=>\(A=2^{2022}-4\)

\(\left(x-2\right)^6=4+2^2+2^3+...+2^{2021}\)

=>\(\left(x-2\right)^6=4+2^{2022}-4=2^{2022}\)

=>\(\left(x-2\right)^6=\left(2^{337}\right)^6\)

=>\(\left[{}\begin{matrix}x-2=2^{337}\\x-2=-2^{337}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2+2^{337}\\x=-2^{337}+2\end{matrix}\right.\)

Ta có: \(28-\left(2x-1\right)^3=2^2\cdot5\)

=>\(28-\left(2x-1\right)^3=20\)

=>\(\left(2x-1\right)^3=8=2^3\)

=>2x-1=2

=>2x=3

=>\(x=\dfrac{3}{2}\)

`28 - (2x-1)^3 = 2^2 xx 5`

`=> 28 - (2x -1)^3 = 20`

`=>        (2x -1)^3 = 28 - 20`

`=> (2x-1)^3 = 8`

`=> (2x - 1)^3 = 2^3`

`=> 2x - 1= 2`

`=> 2x =2+1`

`=> 2x = 3`

`=> x = 3/2`

Vậy `x =3/2`

\(\dfrac{x+1}{x-2}\) = \(\dfrac{3}{5}\)

(\(x+1\)).5= (\(x-2\)).3

5\(x+5\) = 3\(x\) - 6

5\(x-3x\) = - 6 - 5

    2\(x\) = -11

       \(x=-\dfrac{11}{2}\)

Vậy \(x=-\dfrac{11}{2}\) 

Ta có: \(\dfrac{x+1}{x-2}=\dfrac{3}{5}\)

=>5(x+1)=3(x-2)

=>5x+5=3x-6

=>5x-3x=-6-5

=>2x=-11

=>\(x=-\dfrac{11}{2}\)

    -|2,68 - 2\(x\)| - 5,9

Vì |2,68 - 2\(x\)| ≥ 0 ⇒ -|2,68 - 2\(x\)| ≤ 0 ⇒ - |2,68 - 2\(x\)| - 5,9 ≤ -5,9

Dấu bằng xảy ra khi:

2,68 - 2\(x\)  = 0 ⇒ 2\(x\) = 2,68 ⇒ \(x\) = 2,68 : 2 ⇒ \(x=1,34\)

Vậy giá trị lớn nhất của biểu thức:

- |2,68 - 2\(x\)| - 5,9 là -5,9 xảy ra khi \(x=1,34\)

x+2/5=7/10

x=7/10-2/5

x=7/20-4/10

x=3/10

vậy x=3/10

3/10

Ta có: \(\left|\dfrac{7}{5}x+\dfrac{2}{3}\right|=\left|\dfrac{4}{3}x-\dfrac{1}{4}\right|\)

=>\(\left[{}\begin{matrix}\dfrac{7}{5}x+\dfrac{2}{3}=\dfrac{4}{3}x-\dfrac{1}{4}\\\dfrac{7}{5}x+\dfrac{2}{3}=-\dfrac{4}{3}x+\dfrac{1}{4}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\dfrac{7}{5}x-\dfrac{4}{3}x=-\dfrac{1}{4}-\dfrac{2}{3}\\\dfrac{7}{5}x+\dfrac{4}{3}x=\dfrac{1}{4}-\dfrac{2}{3}\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}\dfrac{1}{15}x=\dfrac{-11}{12}\\\dfrac{41}{15}x=\dfrac{-5}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{12}:\dfrac{1}{15}=-\dfrac{11}{12}\cdot15=-11\cdot\dfrac{5}{4}=-\dfrac{55}{4}\\x=-\dfrac{5}{12}:\dfrac{41}{15}=-\dfrac{5}{12}\cdot\dfrac{15}{41}=\dfrac{-5\cdot5}{4\cdot41}=\dfrac{-25}{164}\end{matrix}\right.\)

              Chưa đủ dữ liệu để xác định x;y;z em nhé.

a: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{6xy}=\dfrac{1}{6}\)

=>\(\dfrac{6y+6x+1}{6xy}=\dfrac{xy}{6xy}\)

=>6x+6y+1-xy=0

=>6x-xy+6y+1=0

=>x(6-y)+6y-36+37=0

=>-x(y-6)+6(y-6)=-37

=>(y-6)(-x+6)=-37

=>(x-6)(y-6)=37

=>\(\left(x-6;y-6\right)\in\left\{\left(1;37\right);\left(37;1\right);\left(-1;-37\right);\left(-37;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(7;43\right);\left(43;7\right);\left(5;-31\right);\left(-31;5\right)\right\}\)

b: 

\(\dfrac{x}{4}-\dfrac{5}{y}=\dfrac{3}{2}\)

=>\(\dfrac{xy-20}{4y}=\dfrac{3}{2}\)

=>\(\dfrac{xy-20}{4y}=\dfrac{6y}{4y}\)

=>xy-20=6y

=>xy-6y=20

=>y(x-6)=20

=>(x-6;y)\(\in\){(1;20);(20;1);(-1;-20);(-20;-1);(2;10);(10;2);(-2;-10);(-10;-2);(4;5);(5;4);(-4;-5);(-5;-4)}

=>(x;y)\(\in\){(7;20);(26;1);(5;-20);(-14;-1);(8;10);(16;2);(4;-10);(-4;-2);(10;5);(11;4);(2;-5);(1;-4)}

\(x;y\) nguyên mới đúng chứ em ơi?