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\(y=\frac{2x-4+3+4}{x-2}=2+\frac{7}{x-2}\)
x-2=U(7)={-7,-1,1,7}
x={-5,1,3,9}
\(\frac{2x+3}{x-2}=\frac{2x-4+7}{x-2}=\frac{2\left(x-2\right)+7}{x-2}=2+\frac{7}{x-2}\)
Để 2 + \(\frac{7}{x-2}\)là ssoos nguyên <=> \(\frac{7}{x-2}\) là số nguyên
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a) Để \(f\left(x\right)=3\)
\(\Leftrightarrow\frac{2x+1}{2x+3}=3\)
\(\Leftrightarrow3.\left(2x+3\right)=2x+1\)
\(\Leftrightarrow6x+9=2x+1\)
\(\Leftrightarrow6x-2x=1-9\)
\(\Leftrightarrow4x=-8\)
\(\Leftrightarrow x=-2\)
Để f(x) nguyên
\(\Leftrightarrow2x+1⋮2x+3\)
\(\Leftrightarrow2x+3-2⋮2x+3\)
mà \(2x+3⋮2x+3\)
\(\Rightarrow2⋮2x+3\)
\(\Rightarrow2x+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
Lập bảng rồi tìm x nguyên nhé