Đề vẫn giải được nhưng số khá xấu, không phù hợp với lớp 7. Bạn xem lại xem đã đúng đề chưa vậy>

\(\dfrac{-19}{2}\) < a < \(\dfrac{-20}{3}\)

- \(\dfrac{57}{6}\) < a < \(\dfrac{-40}{6}\)

a = - 56/6; -55/6; -54/6; - 53/6; - 52/6; - 51/6; ......; -42/6; -41/6

b, \(\dfrac{1}{3}\) < \(\dfrac{4}{b}\) < \(\dfrac{1}{2}\) 

\(\dfrac{4}{12}\) < \(\dfrac{4}{b}\)< \(\dfrac{4}{8}\)

b = 11; 10; 9 

2(x-3)+3(x+1)=4x-1

=>2x-6+3x+3=4x-1

=> 2x+3x-4x=-1+6-3

=> x(2+3-4)=2

=>x=2

Vậy x=2

\(2\times\left(x-3\right)+3\times\left(x+1\right)=4x-1\)

\(2x-6+3x+3=4x-1\)

\(\Rightarrow2x-6+3x+3-\left(4x-1\right)=0\)

\(2x-6+3x+3-4x+1=0\)

\(2x+3x-4x-6+3+1=0\)

\(\left(2+3-4\right)x-\left(6-3-1\right)=0\)

\(x-2=0\)

\(x=0+2\)

\(x=2\)

2(x-3)+3(x+1)=4x-1

=>2x-6+3x+3=4x-1

=> 2x+3x-4x=-1+6-3

=> x(2+3-4)=2

=>x=2

Vậy x=2

- \(\dfrac{5}{6}\) + 3\(x\) = \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)\(x\)

3\(x\) + \(\dfrac{1}{2}\)\(x\) = \(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)

\(\dfrac{7x}{2}\) = \(\dfrac{3}{2}\)

\(x\) = \(\dfrac{3}{2}\) : \(\dfrac{7}{2}\)

\(x\)  = \(\dfrac{3}{7}\)

:)) anh Hải pro là ai vậy bn :))

Ta có \(\overline{a50}+\overline{b1}+\overline{1c}=999\)

⇒ \(\overline{a00}+50+\overline{b0}+1+10+c=999\)

⇒ \(\left(\overline{a00}+\overline{b0}+c\right)+\left(50+1+10\right)=999\)

⇒ \(\overline{abc}+61=999\)

⇒ \(\overline{abc}=999-61\)

⇒ \(\overline{abc}=938\)

Vậy số cần tìm là 938

Vậy a = 9; b = 3; c = 8

\(\dfrac{3x+5}{x-2}\) ϵ Z, đk x # 2

⇔ 3 + \(\dfrac{11}{x-2}\) ϵ Z 

⇔ \(\dfrac{11}{x-2}\) ϵ Z 

x - 2 ϵ Ư(11) = { -11; -1; 1; 11}

x ϵ {-9; 1; 3; 13}

\(\dfrac{3x+5}{x-2}\in Z\)

\(\Leftrightarrow\dfrac{3x-6+11}{x-2}\in Z\)

\(\Leftrightarrow\dfrac{3\left(x-2\right)+11}{x-2}\in Z\)

\(\Leftrightarrow\dfrac{3\left(x-2\right)}{x-2}+\dfrac{11}{x-2}\in Z\)

\(\Leftrightarrow3+\dfrac{11}{x-2}\in Z\)

Vì 3 thuộc Z nên \(\dfrac{11}{x-2}\)phải thuộc Z

\(\Rightarrow\dfrac{11}{x-2}\in Z\)

\(\Leftrightarrow11⋮x-2\)

\(\Rightarrow\) x - 2 là ước của 11

\(\Rightarrow x-2\in\left\{-11;-1;1;11\right\}\)

Ta có bảng sau:

\(\begin{matrix}x-2&-11&-1&1&11\\x&-9&1&3&13\end{matrix}\)

Vậy: \(x\in\left\{-9;1;3;13\right\}\)