a; \(\dfrac{x}{8}\) = \(\dfrac{3}{4}\) + \(\dfrac{-5}{8}\)

     \(\dfrac{x}{8}\) =  \(\dfrac{1}{8}\)

     \(x\) =  \(\dfrac{1}{8}\) \(\times\) 8 

      \(x\) = 1

b;  \(\dfrac{x}{12}\) = \(\dfrac{3}{4}\) + \(\dfrac{-2}{3}\)

     \(\dfrac{x}{12}\) =  \(\dfrac{1}{12}\)

       \(x\) = \(\dfrac{1}{12}\) \(\times\) 12

       \(x\) = 1

c; 1 + \(\dfrac{11}{3}\) = \(\dfrac{24}{x}\)

    \(\dfrac{14}{3}\)      =  \(\dfrac{24}{x}\)

     \(x\)        = 24 :  \(\dfrac{14}{3}\) 

      \(x\)       = \(\dfrac{36}{7}\)

d; \(\dfrac{x}{6}\) - \(\dfrac{3}{4}\) = \(\dfrac{1}{12}\)

     \(\dfrac{x}{6}\)      = \(\dfrac{1}{12}\) + \(\dfrac{3}{4}\)

      \(\dfrac{x}{6}\)     = \(\dfrac{5}{6}\)

      \(x\)    = \(\dfrac{5}{6}\) \(\times\) 6

       \(x\)   = 5

\(A=1+2-3-4+5+6-7-8+...-2020\)

\(A=\left(1+2-3-4\right)+\left(5+6-7-8\right)+...+\left(2017+2018-2019-2020\right)\)

\(A=\left(-4\right)+\left(-4\right)+...+\left(-4\right)\)

\(A=\left(-4\right)\cdot\dfrac{2020}{4}\)

\(A=-2020\)

A=1+2-3-4+5+6-7-8+9+..+2018-2019-2020

A= (1+2-3-4) + (5+6-7-8) +...+ (2017+2018 - 2019 - 2020)

A= -4 + (-4) +... + (-4) (505 thừa số -4)

A= -4 x 505 = -2020

đk (\(x\); y \(\in\) Z; y ≠ -1)

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

\(\dfrac{xy+x-3}{3.\left(y+1\right)}\)  = \(\dfrac{1}{6}\)

\(\dfrac{xy+x-3}{y+1}\)    = \(\dfrac{1}{6}\) \(\times\) 3 

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

2.(\(xy+x-3\))    = y + 1

2\(xy\) + 2\(x\) - 6       = y + 1

2\(xy\) - y  + 2\(x\) - 1 = 5 + 1

y.(2\(x\) - 1) + (2\(x\) - 1)  = 6

(2\(x\) - 1).(y + 1) = 6

6 = 6; Ư(6) = {-6; -3; -2; -1; 1; 2; 3; 6}

Lập bảng ta có:

Theo bảng trên ta có các cặp (\(x;y\)) nguyên thỏa mãn đề bài là:

(\(x\); y) = (0; -7); (-1; -3); (2; 1); (1; 5)

a; \(\dfrac{x-1}{12}\) = \(\dfrac{5}{3}\)

      \(x-1\) = \(\dfrac{5}{3}\) \(\times\) 12

      \(x\)  - 1  = 20

      \(x\)        = 20 + 1

      \(x\)        = 21

b;   \(\dfrac{-x}{8}\) = \(\dfrac{-50}{x}\)

     -\(x\).\(x\)  = -50.8

       -\(x^2\)  = -400

        \(x^2\) = 400

        \(\left[{}\begin{matrix}x=-20\\x=20\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-20; 20}

c; \(\dfrac{x}{3}\) = \(\dfrac{14}{x+1}\)

     \(x\).(\(x\)+1) =  14.3

     \(x^2\) + \(x\)  = 42

      \(x^2\) + \(x\) - 42 = 0

      \(x^2\) - 6\(x\) + 7\(x\) - 42  = 0

      \(x\).(\(x\) - 6) + 7.(\(x\) - 6) = 0

        (\(x\) - 6).(\(x\) + 7) = 0

         \(\left[{}\begin{matrix}x-6=0\\x+7=0\end{matrix}\right.\)

         \(\left[{}\begin{matrix}x=6\\x=-7\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-7; 6}

d; \(x-\dfrac{2}{9}\) = \(\dfrac{1}{6}\)

   \(x\)          = \(\dfrac{1}{6}\) + \(\dfrac{2}{9}\)

   \(x\)          = \(\dfrac{7}{18}\)

Vậy \(x\) = \(\dfrac{7}{18}\)

\(\dfrac{-5}{18}+\dfrac{5}{12}=\dfrac{-10}{36}+\dfrac{15}{36}=\dfrac{5}{36}\)

\(\dfrac{-5}{18}+\dfrac{5}{12}=\dfrac{5}{36}\)

Đây

\(\left|2x-1\right|=16\)

\(\Rightarrow\) Ta có: \(\left\{{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\)

               \(\left\{{}\begin{matrix}2x=16+1\\2x=-16+1\end{matrix}\right.\)

               \(\left\{{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\)

               \(\left\{{}\begin{matrix}x=17:2\\x=-15:2\end{matrix}\right.\)

          \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{17}{2}\\x=\dfrac{-15}{2}\end{matrix}\right.\)

   \(\Rightarrow x\in\left\{{}\begin{matrix}\dfrac{17}{2}\\\dfrac{-15}{2}\end{matrix}\right.\)

|2\(x\) - 1| = 16

\(\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\)

\(\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=\dfrac{17}{2}\\x=-\dfrac{15}{2}\end{matrix}\right.\)

Vậy \(x\) \(\in\) {-\(\dfrac{15}{2}\); \(\dfrac{17}{2}\)}

Nếu \(m-15>1\Rightarrow7\left(m-15\right)\) luôn có ít nhất 3 ước dương nên \(7\left(m-15\right)\) là hợp số (ktm)

 Vơi \(m-15=1\)

\(\Rightarrow m=16\Rightarrow7\left(m-15\right)=7\) là SNT (thỏa mãn)

Vậy \(m=16\)

`#3107.101107`

`a)`

`(x + 3)(x^2 + 1) = 0`

TH1: `x + 3 = 0 \Rightarrow x = -3`

TH2: `x^2 + 1 = 0 \Rightarrow x^2 = -1` (vô lý)

Vậy, `x = -3`

`b)`

`(x^2 + 2)(x - 4) = 0`

TH1: `x^2 + 2 = 0 \Rightarrow x^2 = -2` (vô lý)

TH2: `x - 4 = 0 \Rightarrow x = 4`

Vậy, `x = 4.`

A; (\(x\) + 3).(\(x^2\) + 1) = 0

     vì  \(x^2\) ≥ 0 ∀ \(x\) ⇒ \(x^2\) + 1 ≥ 0 ∀ \(x\)

     (\(x\)+ 3).(\(x^2\) + 1) = 0 ⇔ \(x\) + 3 = 0 ⇒ \(x\) = -3

Vậy \(x\) = - 3

b; (\(x^2\) + 2).(\(x\) - 4) = 0

    Vì \(x^2\) ≥ 0 ∀ \(x\); 

   (\(x^2\) + 2).(\(x\) - 4) = 0 ⇔ \(x\) - 4 = 0 ⇒ \(x\) = 4 

Vậy \(x\) = 4

bài giải

số học sinh nam là:

105 : ( 3 + 4 ) x 3 = 45 ( hs )

số học sinh nữ là:

105 - 45 = 60 ( hs )

đáp số: hs nam: 45

             hs nữ: 60