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a) Đặt \(A=\frac{x}{x+3}=\frac{x+3-3}{x+3}=\frac{x+3}{x+3}-\frac{3}{x+3}=1-\frac{3}{x+3}\)
Để A nguyên thì \(\frac{3}{x+3}\) nguyên => \(3⋮x+3\)
=> \(x+3\in\left\{1;-1;3;-3\right\}\)
=> \(x\in\left\{-2;-4;0;-6\right\}\)
Vậy \(x\in\left\{-2;-4;0;-6\right\}\)
b) Đặt \(B=\frac{x-1}{2x+1}\)
Để B nguyên thì 2B nguyên
Ta có:
\(2B=\frac{2.\left(x-1\right)}{2x+1}=\frac{2x-2}{2x+1}=\frac{2x+1-3}{2x+1}=\frac{2x+1}{2x+1}-\frac{3}{2x+1}=1-\frac{3}{2x+1}\)
Để 2B nguyên thì \(\frac{3}{2x+1}\) nguyên => \(3⋮2x+1\)
=> \(2x+1\in\left\{1;-1;3;-3\right\}\)
=> \(2x\in\left\{0;-2;2;-4\right\}\)
=> \(x\in\left\{0;-1;1;-2\right\}\)
Vậy \(x\in\left\{0;-1;1;-2\right\}\)
a) \(\frac{x+3}{x-2}=\frac{x-2+5}{x-2}=\frac{x-2}{x-2}+\frac{5}{x-2}=1+\frac{5}{x-2}\)
Vì \(1\in Z\Rightarrow\frac{5}{x-2}\in Z\Rightarrow5⋮x-2\)
\(\Rightarrow x-2\inƯ\left(5\right)\)
\(\Rightarrow x-2\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x\in\left\{-3;1;3;7\right\}\)
b) \(\frac{1-2x}{x+3}=\frac{-2x+1}{x+3}=\frac{\left(-2x-6\right)+7}{x+3}\)
\(=-\frac{2.\left(x+3\right)+7}{x+3}\)
\(=\frac{-2.\left(x+3\right)}{x+3}+\frac{7}{x+3}=-2+\frac{7}{x+3}\)
Vì \(-2\in Z\Rightarrow\frac{7}{x+3}\in Z\Rightarrow7⋮x+3\)
\(\Rightarrow x+3\inƯ\left(7\right)\)
\(\Rightarrow x+3\in\left\{-7;-1;1;7\right\}\)
\(\Rightarrow x\in\left\{-10;-4;-2;4\right\}\)
=> (2*x^3+2*x+1)/x
=> 2*x^3/(x+2)+4*x^2/(x+2)+1/(x+2)
=> 2*(x^2+1)
1. Ta có \(\frac{n^2-2n+3}{n-2}=\frac{n\left(n-2\right)+3}{n-2}=n+\frac{3}{n-2}\)
Để \(\frac{n^2-2n+3}{n-2}\in Z\) thì \(\frac{3}{n-2}\in Z\Rightarrow n-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(\Rightarrow n\in\left\{-1;1;3;5\right\}\)
2. \(\frac{x}{4}=\frac{10}{x+3}\)
ĐK: \(x\ne-3\)
\(\frac{x}{4}=\frac{10}{x+3}\)
\(\Leftrightarrow\frac{x}{4}-\frac{10}{x+3}=0\)
\(\Leftrightarrow\frac{x^2+3x-40}{4\left(x+3\right)}=0\)
\(\Leftrightarrow x^2+3x-40=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-8\end{cases}}\left(tmđk\right)\)
b) \(\frac{x+2}{7}=\frac{-49}{\left(x+2\right)^2}\)
ĐK: \(x\ne-2\)
\(\frac{x+2}{7}=\frac{-49}{\left(x+2\right)^2}\)
\(\Leftrightarrow\left(x+2\right)^3=-49.7\)
\(\Leftrightarrow\left(x+2\right)^3=-343\)
\(\Leftrightarrow x+2=-7\)
\(\Leftrightarrow x=-9\left(tmđk\right)\)
bn Huyền ơi ở câu 1 bn chép sai đầu bài của bạn Thảo rùi
\(a,\frac{x+22}{x+1}\inℤ\Leftrightarrow x+22⋮x+1\)
\(\Rightarrow x+1+21⋮x+1\)
\(x+1⋮x+1\)
\(\Rightarrow21⋮x+1\)
\(\Rightarrow x+1\inƯ\left(21\right)\)
\(\Rightarrow x+1\in\left\{-1;1;-3;3;-7;7;-21;21\right\}\)
\(\Rightarrow x\in\left\{-2;0;-4;2;-8;6;-22;20\right\}\)
vậy___
\(b,\frac{3x+1}{2x+1}\inℤ\Leftrightarrow3x+1⋮2x+1\)
\(\Rightarrow2\left(3x+1\right)⋮2x+1\)
\(\Rightarrow6x+2⋮2x+1\)
\(\Rightarrow6x+2+1-1⋮2x+1\)
\(\Rightarrow6x+3-1⋮2x+1\)
\(\Rightarrow3\left(2x+1\right)-1⋮2x+1\)
\(3\left(2x+1\right)⋮2x+1\)
\(\Rightarrow1⋮2x+1\)
\(\Rightarrow2x+1\inƯ\left(1\right)\)
đến đây lm như phần a
\(c,\frac{2x+1}{6-n}\inℤ\Leftrightarrow2x+1⋮6-n\)
\(\Rightarrow2x+1+11-11⋮6-n\)
\(\Rightarrow2x+12-11⋮6-n\)
\(\Rightarrow2\left(x+6\right)-11⋮6-n\)
\(2\left(x+6\right)⋮6-n\)
\(\Rightarrow11⋮6-n\)
tự lm tp
phần c thì k chắc lắm
\(\frac{2x^2+1}{x+2}=\frac{2x^2+4x-4x-8+9}{x+2}=\frac{2x\left(x+2\right)-4\left(x+2\right)+9}{x+2}=2x-4+\frac{9}{x+2}\)
\(\Rightarrow x+2\inƯ\left(9\right)\Rightarrow x+2\in\left\{-9;-3;-1;1;3;9\right\}\Rightarrow x\in\left\{-11;-5;-3;-1;1;7\right\}\)
Cách 2:
\(\frac{2x^2+1}{x+2}=\frac{2\left(x^2-2^2\right)+9}{x+2}=\frac{2\left(x-2\right)\left(x+2\right)+9}{x+2}=2\left(x-1\right)+\frac{9}{x+2}\)
\(\Rightarrow x+2\inƯ\left(9\right)\Rightarrow x+2\in\left\{-9;-3;-1;1;3;9\right\}\Rightarrow x\in\left\{-11;-5;-3;-1;1;7\right\}\)