\(\dfrac{93}{17}:x+\left[\dfrac{-4}{17}\right]:x+\dfrac{22}{7}:\dfrac{52}{3}=\dfrac{4}{11}\)

\(=>\left(\dfrac{93}{17}+\dfrac{-4}{17}\right):x+\dfrac{11}{26}=\dfrac{4}{11}\)

\(=>\dfrac{89}{17}:x+\dfrac{11}{26}=\dfrac{4}{11}\)

\(=>\dfrac{89}{17}:x=\dfrac{4}{11}-\dfrac{11}{26}\)

\(=>\dfrac{89}{17}:x=\dfrac{-17}{286}\)

\(=>x=\dfrac{89}{17}:\dfrac{-17}{286}\)

\(=>x=\dfrac{-25454}{289}=-88\dfrac{22}{289}\)

\(\dfrac{93}{17}:x-\dfrac{4}{17}:x+\dfrac{22}{7}:\dfrac{52}{3}=\dfrac{4}{11}\)

\(\Rightarrow\left(\dfrac{93}{17}-\dfrac{4}{17}\right):x+\dfrac{33}{182}=\dfrac{4}{11}\)

\(\Rightarrow\dfrac{89}{17}:x=\dfrac{365}{2002}\)

\(\Rightarrow x=\dfrac{178178}{6205}\)

\(\frac{36x22x51}{11x17x72}\)=\(\frac{2x3}{2}\)=3

giảm ước 36 và 72, 22 và 11, 51 và 17

49439808/11

\(=\dfrac{36\cdot22\cdot51}{11\cdot17\cdot72}=\dfrac{1}{2}\cdot3\cdot2=3\)

(93/17+-4/17) : x + 33/182=4/11

89/17 : x = 4/11 + 33/182

89/17 : x = Sia đề

a: 17-3x=11

=>3x=17-11=6

=>\(x=\frac63=2\)

b: Sửa đề: \(120-2\left(x+3\right)=2^2\cdot5^2\)

=>\(120-2\left(x+3\right)=4\cdot25=100\)

=>2(x+3)=120-100=20

=>x+3=10

=>x=10-3=7

c: 11⋮x-2

=>x-2∈{1;-1;11;-11}

=>x∈{3;1;13;-9}

mà x>0

nên x∈{1;3;13}

d: Ta có: \(12=2^2\cdot3;18=2\cdot3^2\)

Do đó: ƯCLN(12;18)=2*3=6

12⋮x; 18⋮x

=>x∈ƯC(12;18)

=>x∈Ư(6)

=>x∈{1;2;3;6}