a) Giả sử \(C=\frac{2x+3}{7}=t\left(t\in Z\right)\)

\(\Rightarrow x=\frac{7t-3}{2}\). Để \(x\in Z\) thì t phải lẻ. Nói cách khác \(t=2k+1\left(k\in Z\right)\)

Suy ra  \(x=\frac{7\left(2k+1\right)-3}{2}=14k+2\)

Vậy để \(\frac{2x+3}{7}\in Z\) thì \(x=14k+2\left(k\in Z\right)\)

b) Ta thấy \(C=\frac{6x-1}{3x+2}=\frac{\left(6x+4\right)-5}{3x+2}=2-\frac{5}{3x+2}\)

Do x nguyên nên C đạt GTNN khi \(\frac{5}{3x+2}\) lớn nhất. Điều này xảy ra khi 3x + 2 = 2 hay x = 0.

Vậy \(minC=-\frac{1}{2}\) khi x = 0.

a) ta có: \(\frac{x+6}{x+1}=\frac{x+1+5}{x+1}=1+\frac{5}{x+1}\)

Để \(\frac{x+6}{x+1}\in Z\) 

=> 5/x+1 thuộc Z

=> 5 chia hết cho x + 1

=> x + 1 thuộc Ư(5)={1;-1;5;-5}

...

rùi bn tự lập bảng xét giá trị hộ mk nha! câu b lm tương tự

c) ta có: \(\frac{2x+1}{x-3}=\frac{2x-6+7}{x-3}=\frac{2.\left(x-3\right)+7}{x-3}=2+\frac{7}{x-3}\)

...

\(a,\frac{x+6}{x+1}\inℤ\Leftrightarrow x+6⋮x+1\)

\(\Rightarrow x+1+5⋮x+1\)

      \(x+1⋮x+1\)

\(\Rightarrow5⋮x+1\)

\(\Rightarrow x+1\inƯ\left(5\right)\)

\(\Rightarrow x+1\in\left\{-1;1;-5;5\right\}\)

\(\Rightarrow x\in\left\{-2;0;-6;4\right\}\)

vậy_

\(c,\frac{2x+1}{x-3}\inℤ\Leftrightarrow2x+1⋮x-3\)

\(\Rightarrow2x-6+7⋮x-3\)

\(\Rightarrow2\left(x-3\right)+7⋮x-3\)

      \(2\left(x-3\right)⋮x-3\)

\(\Rightarrow7⋮x-3\)

\(\Rightarrow x-3\inƯ\left(7\right)\)

\(\Rightarrow x-3\in\left\{-1;1;-7;7\right\}\)

\(\Rightarrow x\in\left\{2;4;-4;10\right\}\)

vậy_

phần b thì làm tương tự phần a

Bài làm:

c) \(-\frac{2}{5}+\frac{5}{3}\left(\frac{3}{2}-\frac{4}{15}x\right)=-\frac{7}{6}\)

\(\Leftrightarrow-\frac{2}{5}+\frac{5}{2}-\frac{4}{9}x=-\frac{7}{6}\)

\(\Leftrightarrow\frac{4}{9}x=-\frac{2}{5}+\frac{5}{2}+\frac{7}{6}\)

\(\Leftrightarrow\frac{4}{9}x=\frac{49}{15}\)

\(\Leftrightarrow x=\frac{49}{15}\div\frac{4}{9}\)

\(\Rightarrow x=\frac{147}{20}\)

Vậy \(x=\frac{147}{20}\)

Bài 2:

a) Ta có: \(F=\frac{3x-2}{x+3}=\frac{\left(3x+9\right)-11}{x+3}=3-\frac{11}{x+3}\)

Để F nguyên \(\Rightarrow\frac{11}{x+3}\inℤ\Leftrightarrow x+3\inƯ\left(11\right)=\left\{-11;-1;1;11\right\}\)

\(\Rightarrow x\in\left\{-14;-4;-2;8\right\}\)

Vậy \(x\in\left\{-14;-4;-2;8\right\}\)thì F nguyên

2b) Tách

\(G=\frac{x^2-2x+4}{x+1}=\frac{x^2+x-3x-3+7}{x+1}=\frac{x\left(x+1\right)-3\left(x+1\right)+7}{x+1}\)

\(=\frac{x\left(x+1\right)}{x+1}-\frac{3\left(x+1\right)}{x+1}+\frac{7}{x+1}=x-3+\frac{7}{x+1}\)

G là số nguyên <=> \(\frac{7}{x+1}\)là số nguyên <=> \(7⋮x+1\)<=> \(x+1\inƯ\left(7\right)=\left\{1;-1;7;-7\right\}\)

<=> \(x\in\left\{0;-2;6;-8\right\}\)

đè hinh như là 6\(\sqrt{x}\) nhi bạn

\(a,\frac{-24}{x}+\frac{18}{x}=\frac{-24+18}{x}=\frac{-6}{x}\)

\(\Leftrightarrow x\inƯ(-6)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

\(b,\frac{2x-5}{x+1}=\frac{2x+2-7}{x+1}=\frac{2(x+1)-7}{x+1}=2-\frac{7}{x+1}\)

\(\Leftrightarrow7⋮x+1\Leftrightarrow x+1\inƯ(7)=\left\{\pm1;\pm7\right\}\)

Xét các trường hợp rồi tìm được x thôi :>

\(c,\frac{3x+2}{x-1}-\frac{x-5}{x-1}=\frac{3x+2-x-5}{x-1}=\frac{2x+7}{x-1}=\frac{2x-2+9}{x-1}=\frac{2(x-1)+9}{x-1}=2+\frac{9}{x-1}\)

\(\Leftrightarrow9⋮x-1\Leftrightarrow x-1\inƯ(9)=\left\{\pm1;\pm3;\pm9\right\}\)

\(\Leftrightarrow x\in\left\{2;0;4;-2;10;-8\right\}\)

d, TT

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