\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;7;-7;14;-14\right\}\)

hay \(x\in\left\{3;1;4;0;9;-5;16;-12\right\}\)

\(A\left(x\right)\) chia hết cho \(B\left(x\right)\) khi \(A\left(2\right)=0\)

\(\Rightarrow3.2^2+5.2+m=0\)

\(\Rightarrow m=-22\)

a ) x - 5 \(\in\)B ( 6 )

\(\Rightarrow\)x - 5 \(\in\){ 0 ; 6 ; 12 ; 18 ; 24 ; 30 ; 36 ; 42 ; 48 ; ..... }

\(\Rightarrow\)x \(\in\){ 5 ; 11 ; 17 ; 23 ; 29 ; 35 ; 41 ; 47 ; 53 ; .... }

b ) x - 1 \(⋮\)4

\(\Rightarrow\)x - 1 \(\in\)B ( 4 )

\(\Rightarrow\)x - 1 \(\in\){ 0 ; 4 ; 8 ; 12 ; 16 ; 20 ; 24 ; 28 ; 32 ; 36 ; 40 ; ..... }

\(\Rightarrow\)x \(\in\){ 1 ; 5 ; 9 ; 13 ; 17 ; 21 ; 25 ; 29 ; 33 ; 37 ; 41 ; .... }

Thanks bạn nhìu nhé ~!

Lời giải:

$2x+xy-2y=7$

$x(2+y)-2y=7$

$x(2+y)-2(y+2)=3$

$(x-2)(y+2)=3$

Do $x,y$ là số nguyên nên $x-2, y+2$ cũng là số nguyên. Do đó ta có bảng sau:

\(2x+xy-2y=7\)

\(\Rightarrow x\left(2+y\right)-2y-4+4=7\)

\(\Rightarrow x\left(2+y\right)-2\left(y+2\right)=3\)

\(\Rightarrow\left(x-2\right)\left(y+2\right)=3\)

\(\Rightarrow\left(x-2\right);\left(y+2\right)\in\left\{-1;1;-3;3\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(1;-5\right);\left(3;1\right);\left(-1;-3\right);\left(5;-1\right)\right\}\left(x;y\inℤ\right)\)

a) x : 1/4 = 7/5
    x         = 7/5 x 1/4
    x         = 7/20
b) 9/2 - x = 3/5
            x = 9/2 - 3/5
            x = 39/10
c) 12/5 : x = 7/2
              x = 12/5 : 7/2
              x = 24/35

a) \(x\div\dfrac{1}{4}=\dfrac{7}{5}\)

    \(x=\dfrac{7}{5}\times\dfrac{1}{4}\)

    \(x=\dfrac{7}{20}\)

b) \(\dfrac{9}{2}-x=\dfrac{3}{5}\)

            \(x=\dfrac{9}{2}-\dfrac{3}{5}\)

            \(x=\dfrac{39}{10}\)

c) \(\dfrac{12}{5}\div x=\dfrac{7}{2}\)

             \(x=\dfrac{12}{5}\div\dfrac{7}{2}\)

             \(x=\dfrac{12}{5}\times\dfrac{2}{7}\)

             \(x=\dfrac{24}{35}\)

a)\(x=\dfrac{1}{4}\)x\(\dfrac{7}{5}=\dfrac{7}{20}\)

b)\(x=\dfrac{9}{2}-\dfrac{3}{5}=\dfrac{39}{10}\)

c)\(x=\dfrac{12}{5}:\dfrac{7}{2}=\dfrac{24}{35}\)

\(2x+1\inƯ\left(28\right)\left(EZ\right)\)

\(x+15⋮x+3\)

\(\Leftrightarrow\left(x+3\right)+12⋮x+3\)

\(\Rightarrow12⋮x+3\left(EZ\right)\)

1) \(B\left(24\right)=\left\{24;48;72;96\right\}\)

\(B\left(39\right)=\left\{39;78\right\}\)

2) a) \(x+20⋮x+2\)

\(\Rightarrow x+20-\left(x+2\right)⋮x+2\)

\(\Rightarrow x+20-x-2⋮x+2\)

\(\Rightarrow18⋮x+2\)

\(\Rightarrow x+2\in\left\{1;2;3;6;9;18\right\}\)

\(\Rightarrow x\in\left\{-1;0;1;4;7;16\right\}\)

\(\Rightarrow x\in\left\{0;1;4;7;16\right\}\left(x\in N\right)\)

b) \(x+5⋮4x+69\)

\(\Rightarrow4\left(x+5\right)-\left(4x+69\right)⋮4x+69\)

\(\Rightarrow4x+20-4x-69⋮4x+69\)

\(\Rightarrow-49⋮4x+69\)

\(\Rightarrow4x+69\in\left\{1;7;49\right\}\)

\(\Rightarrow x\in\left\{-17;-\dfrac{31}{2};-20\right\}\)

\(\Rightarrow x\in\varnothing\left(x\in N\right)\)

c) \(10x+23⋮2x+1\)

\(\Rightarrow10x+23-5\left(2x+1\right)⋮2x+1\)

\(\Rightarrow10x+23-10x-5⋮2x+1\)

\(\Rightarrow18⋮2x+1\)

\(\Rightarrow2x+1\in\left\{1;2;3;6;9;18\right\}\)

\(\Rightarrow x\in\left\{0;\dfrac{1}{2};1;\dfrac{5}{2};4;\dfrac{17}{2}\right\}\)

\(\Rightarrow x\in\left\{0;1;4\right\}\left(x\in N\right)\)

Đính chính câu 1

Không có số có 2 chữ số thỏa đề bài

\(a,\left|x-\dfrac{1}{2}\right|+\dfrac{1}{3}=\dfrac{2}{3}\\ \Rightarrow\left|x-\dfrac{1}{2}\right|=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{3}\\x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\\x=-\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{1}{6}\end{matrix}\right.\\ b,\dfrac{x}{-2}=\dfrac{y}{5}=\dfrac{x-y}{-2-5}=\dfrac{14}{-7}=-2\\ \Rightarrow x=-2.\left(-2\right)=4;y=-2.5=-10\)

x-y=-14 nhé

a)

\(\left|x-\dfrac{1}{2}\right|+\dfrac{1}{3}=\dfrac{2}{3}\\ \left|x-\dfrac{1}{2}\right|=\dfrac{1}{3}\\ \left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{1}{3}\\x-\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{1}{3}+\dfrac{1}{2}\\x=-\dfrac{1}{3}+\dfrac{1}{2}\end{matrix}\right.\\ \left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{1}{6}\end{matrix}\right.\)

b)

\(\dfrac{x}{-2}=\dfrac{y}{5}\)

mà `x-y=14` nên áp dụng tính chất dãy tỉ số bằng nhau ta có

\(\dfrac{x}{-2}=\dfrac{y}{5}=\dfrac{x-y}{-2-5}=\dfrac{14}{-7}=-2\\ =>\left\{{}\begin{matrix}x=-2\cdot\left(-2\right)=4\\y=-2\cdot5=-10\end{matrix}\right.\)