Ta có:  \(2\left(x-\frac{1}{2}\right)+3\left(-1+\frac{x}{3}\right)=x\left(\frac{2}{x}-1\right)\)

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

           \(2x-1+-3+x=2-x\)

           \(\left(2x+x\right)+\left(-3\right)-1=2-x\)

           \(3x+\left(-4\right)=2-x\)

           \(3x+x=2-\left(-4\right)\)  

           \(4x=6\)

           \(x=6:4\)

           \(x=\frac{6}{4}=\frac{3}{2}\)

mik nha

a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)

b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)

\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)

c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)

\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)

\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)

\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)

\(\Rightarrow x=-2\)

d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)

\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)

\(\Rightarrow x=\dfrac{25}{9}\)

e) \(\dfrac{1}{2}x+650\%x-x=-6\)

\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)

\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)

\(\Rightarrow6x=-6\)

\(\Rightarrow x=\dfrac{-6}{6}=-1\)

g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)

\(\Rightarrow2x-1-3+x=2-x\)

\(\Rightarrow3x-4=2-x\)

\(\Rightarrow3x+x=2+4\)

\(\Rightarrow4x=6\)

\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)

Ta có công thức:

\(1^3+2^3+.....+n^3=\left(1+2+3+....+n\right)^2\)

\(\Rightarrow1^3+2^3+3^3+.....+10^3=\left(1+2+3+....+10\right)^2=\left(x+1\right)^2\)

\(\Rightarrow x+1=55\Rightarrow x=54\)

1) Do x ∈ Z và 0 < x < 3

⇒ x ∈ {1; 2}

2) Do x ∈ Z và 0 < x ≤ 3

⇒ x ∈ {1; 2; 3}

3) Do x ∈ Z và -1 < x ≤ 4

⇒ x ∈ {0; 1; 2; 3; 4}

a: \(\Leftrightarrow3x-9+13⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{4;2;16;-10\right\}\)

b: \(\Leftrightarrow x\left(x+3\right)-13⋮x+3\)

\(\Leftrightarrow x+3\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{-2;-4;10;-16\right\}\)

c: \(\Leftrightarrow x^2-1+4⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{2;0;3;-1;5;-3\right\}\)

đó là ngũ đúng ko, chưa học nên ko biết j hết trơn

^ là mũ