C

 C.- 5; - 4; - 3; - 2; - 1; 0; 1

x * 6 = 5/7 - 1/14

x * 6 = 10/14 - 1/14

x * 6 = 9/14

x = 9/14 : 6

x = 9/14 * 1/6

x = 9/84

3/28 

\(-\dfrac{9}{7}+1+\dfrac{12}{-7}< x< -\dfrac{5}{6}+\dfrac{1}{3}+\dfrac{5}{2}\\ \Rightarrow-2< x< 2\)

Mà \(x\) nguyên \(\Rightarrow x\in\left\{-1;0;1\right\}\)

Vậy có 3 giá trị thỏa mãn

a, 3x ( y+1) + y + 1 = 7

(y+1)(3x +1) =7

th1 : \(\left\{{}\begin{matrix}y+1=1\\3x+1=7\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=2\end{matrix}\right.\)

th2: \(\left\{{}\begin{matrix}y+1=-1\\3x+1=-7\end{matrix}\right.\)=> x = -8/3 (loại)

th3: \(\left\{{}\begin{matrix}y+1=7\\3x+1=1\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}y=6\\x=0\end{matrix}\right.\)

th 4 : \(\left\{{}\begin{matrix}y+1=-7\\3x+1=-1\end{matrix}\right.\)=> x=-2/3 (loại)

Vậy (x,y)= (2 ;0);  (0; 6)

b, xy - x + 3y - 3 = 5

   (x( y-1) + 3( y-1) = 5

          (y-1)(x+3) = 5

 th1: \(\left\{{}\begin{matrix}y-1=1\\x+3=5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=2\\x=8\end{matrix}\right.\)

th2: \(\left\{{}\begin{matrix}y-1=-1\\x+3=-5\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=0\\x=-8\end{matrix}\right.\)

th3: \(\left\{{}\begin{matrix}y-1=5\\x+3=1\end{matrix}\right.\) => \(\left\{{}\begin{matrix}y=6\\x=-2\end{matrix}\right.\)

th4: \(\left\{{}\begin{matrix}y-1=-5\\x+3=-1\end{matrix}\right.\) =>  \(\left\{{}\begin{matrix}y=-4\\x=-4\end{matrix}\right.\)

vậy (x, y) = ( 8; 2); ( -8; 0);  (-2; 6); (-4; -4)

c, 2xy + x + y = 7 => y = \(\dfrac{7-x}{2x+1}\) ; y ϵ Z ⇔ 7-x ⋮ 2x+1

⇔ 14 - 2x ⋮ 2x + 1 ⇔ 15 - 2x - 1  ⋮ 2x + 1

th1 : 2x + 1 = -1=> x = -1; y = \(\dfrac{7-(-1)}{-1.2+1}\) = -8

th2: 2x+ 1 = 1=> x =0; y = 7

th3: 2x+1 = -3 => x =  x=-2  => y = \(\dfrac{7-(-2)}{-2.2+1}\) = -3 

th4: 2x+ 1 = 3 => x = 1 => y = \(\dfrac{7+1}{2.1+1}\) = 2

th5: 2x + 1 = -5 => x = -3=> y = \(\dfrac{7-(-3)}{-3.2+1}\) = -2

th6: 2x + 1 = 5 => x = 2; ; y = \(\dfrac{7-2}{2.2+1}\) =1

th7 : 2x + 1 = -15 => x = -8; y = \(\dfrac{7-(-8)}{-8.2+1}\) = -1

th8 : 2x+1 = 15 => x = 7; y = \(\dfrac{7-7}{2.7+1}\) = 0

kết luận

(x,y) = (-1; -8); (0 ;7); ( -2; -3) ; ( 1; 2); ( -3; -2); (2;1); (-8;-1);(7;0)

3xy−2x+5y=293xy−2x+5y=29

9xy−6x+15y=879xy−6x+15y=87

(9xy−6x)+(15y−10)=77(9xy−6x)+(15y−10)=77

3x(3y−2)+5(3y−2)=773x(3y−2)+5(3y−2)=77

(3y−2)(3x+5)=77(3y−2)(3x+5)=77

⇒(3y−2)⇒(3y−2) và (3x+5)(3x+5) là Ư(77)=±1,±7,±11,±77Ư(77)=±1,±7,±11,±77

Ta có bảng giá trị sau:

Do x,y∈Zx,y∈Z nên (x,y)∈{(−4;−3),(−2;−25),(2;3),(24;1)}

Answer:

a. \(-5< x< 5\)

\(\Rightarrow x\in\left\{\pm4;\pm3;\pm2;\pm1;0\right\}\)

Tổng các số nguyên x thoả mãn:

\((-4) + (-3) + (-2) + (-1) + 0 + 1 + 2 + 3 + 4\)

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

\(=0\)

Chọn C

|\(x-5\)| = 7

\(\left[{}\begin{matrix}x-5=-7\\x-5=7\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-2\\x=12\end{matrix}\right.\)

\(x\in\) {-2; 12}

Có hai giá trị \(x\) thỏa mãn. Vậy chọn C.2 

Theo c) \(f\left(\frac{5}{7}\right)=f\left(\frac{2}{7}+\frac{3}{7}\right)=f\left(\frac{2}{7}\right)+f\left(\frac{3}{7}\right)\)

      \(f\left(\frac{2}{7}\right)=f\left(\frac{1}{7}+\frac{1}{7}\right)=f\left(\frac{1}{7}\right)+f\left(\frac{1}{7}\right)=2.f\left(\frac{1}{7}\right)\) 

  \(f\left(\frac{3}{7}\right)=f\left(\frac{1}{7}+\frac{2}{7}\right)=f\left(\frac{1}{7}\right)+f\left(\frac{2}{7}\right)=f\left(\frac{1}{7}\right)+2f\left(\frac{1}{7}\right)=3.f\left(\frac{1}{7}\right)\)

       \(\implies\)\(f\left(\frac{5}{7}\right)=5.f\left(\frac{1}{7}\right)\)          (1)

 Theo b) \(f\left(\frac{1}{7}\right)=\frac{1}{7^2}.f\left(7\right)\)          (2)

Theo c) \(f\left(7\right)=f\left(3+4\right)=f\left(3\right)+f\left(4\right)\)

                                                 \(=2.f\left(3\right)+f\left(1\right)\) 

                                                 \(=6.f\left(1\right)+f\left(1\right)\) 

                                                 \(=7.f\left(1\right)\)

Theo a)\(f\left(1\right)=1\)\(\implies\)\(f\left(7\right)=7\)      (3)

    Từ (1);(2);(3)

       \(\implies\)       \(f\left(\frac{5}{7}\right)=\frac{5}{7}\)

︵✰He❤lloღ