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\(a,\)\(A=\frac{a^2+4a+4}{a^3+2a^2-4a-8}\)
\(=\frac{\left(a+2\right)^2}{a^2\left(a+2\right)-4\left(a+2\right)}\)
\(=\frac{\left(a+2\right)^2}{\left(a+2\right)\left(a^2-4\right)}\)
\(=\frac{\left(a+2\right)^2}{\left(a+2\right)\left(a+2\right)\left(a-2\right)}\)
\(=\frac{1}{a-2}\)
\(a,A=\frac{\left(a+2\right)^2}{\left(a+2\right)\left(a^2-4\right)}=\frac{a+2}{\left(a-2\right)\left(a+2\right)}=\frac{1}{a-2}\)
b, Để A có giá trị là một số nguyên thì \(1⋮a-2\)
=> \(\orbr{\begin{cases}a-2=1\\a-2=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}a=3\\a=1\end{cases}}}\)
a) \(ĐKXĐ:\hept{\begin{cases}a\ne\pm2\\a\ne1\\a\ne0\end{cases}}\)
\(A=\left(\frac{4a}{2+a}+\frac{8a^2}{4-a^2}\right):\left(\frac{a-3}{a^2-2a}-\frac{2}{a}\right)\)
\(\Leftrightarrow A=\frac{8a-4a^2+8a^2}{\left(2-a\right)\left(2+a\right)}:\frac{a-3-2a+4}{a\left(a-2\right)}\)
\(\Leftrightarrow A=\frac{4a^2+8a}{\left(2-a\right)\left(2+a\right)}:\frac{-a+1}{a\left(a-2\right)}\)
\(\Leftrightarrow A=\frac{4a}{2-a}:\frac{-a+1}{a\left(a-2\right)}\)
\(\Leftrightarrow A=\frac{4a^2\left(a-2\right)}{\left(a-2\right)\left(a-1\right)}\)
\(\Leftrightarrow A=\frac{4a^2}{a-1}\)
b) Để A nhận giá trị nguyên
\(\Leftrightarrow\frac{4a^2}{a-1}\inℤ\)
\(\Leftrightarrow4a^2⋮a-1\)
\(\Leftrightarrow4\left(a^2-1\right)+4⋮a-1\)
\(\Leftrightarrow4\left(a-1\right)\left(a+1\right)+4⋮a-1\)
\(\Leftrightarrow4⋮a-1\)
\(\Leftrightarrow a-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow a\in\left\{0;2;-1;3;-3;5\right\}\)
Ta sẽ loại các giá trị ở đkxđ
Vậy để \(A\inℤ\Leftrightarrow a\in\left\{2;-1;3;-3;5\right\}\)
a) \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-5x}{x^2-1}\right)\cdot\frac{x-3}{x}\left(x\ne\pm1;x\ne0\right)\)
\(\Leftrightarrow A=\left[\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}+\frac{x^2-5x}{\left(x-1\right)\left(x+1\right)}\right]\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-5x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\frac{x^2-x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\frac{x\left(x-1\right)\left(x-3\right)}{\left(x-1\right)\left(x+1\right)x}=\frac{x-3}{x+1}\)
Vậy \(A=\frac{x-3}{x+1}\left(x\ne\pm1;x\ne0\right)\)
b) \(A=\frac{x-3}{x+1}\left(x\ne\pm1;x\ne0\right)\)
Để A nhận giá trị nguyên thì x-3 chia hết chi x+1
=> (x+1)-4 chia hết chi x+1
=> 4 chia hết cho x+1
x nguyên => x+1 nguyên => x+1 thuộc Ư (4)={-4;-2;-1;1;2;4}
Ta có bảng
| x+1 | -4 | -2 | -1 | 1 | 2 | 4 |
| x | -5 | -3 | -2 | 0 | 1 | 3 |
| ĐCĐK | tm | tm | tm | ktm | ktm | tm |
Vậy x={-5;-3;-2;3} thì A đạt giá trị nguyên
c) I3x-1I=5
\(\Rightarrow\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=6\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{-4}{3}\end{cases}}}\)
Đên đây thay vào rồi tính nhé
a) \(ĐKXĐ:\hept{\begin{cases}x\ne\pm1\\x\ne0\end{cases}}\)
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-5x}{x^2-1}\right)\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\frac{\left(x+1\right)^2-\left(x-1\right)^2+x^2-5x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\frac{x^2+2x+1-x^2+2x-1+x^2-5x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-3}{x}\)
\(\Leftrightarrow A=\frac{\left(x^2-x\right)\left(x-3\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow A=\frac{x-3}{x+1}\)
b) Để \(A\inℤ\)
\(\Leftrightarrow x-3⋮x+1\)
\(\Leftrightarrow x+1-4⋮x+1\)
\(\Leftrightarrow4⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
\(\Leftrightarrow x\in\left\{0;-2;-3;1;3;-5\right\}\)
Mà \(x\ne0;x\ne1\)
\(\Leftrightarrow x\in\left\{-2;-3;3;-5\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{-2;-3;3;-5\right\}\)
c) Khi \(\left|3x-1\right|=5\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=5\\3x-1=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=6\\3x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{4}{3}\end{cases}}\)
Vì khi x = 2 hoặc x = -4/3 thì x không thuộc tập hợp các giá trị làm cho A nguyên
Vậy khi |3x - 1| = 5 thì để cho A nguyên \(\Leftrightarrow x\in\varnothing\)
ĐKXĐ: \(a\ne\pm1;2;4\)
\(P=\frac{a^3-5a^2+4a+a^2-5a+4}{a^3-5a^2+4a-2a^2+10a-8}=\frac{a\left(a^2-5a+4\right)+\left(a^2-5a+4\right)}{a\left(a^2-5a+4\right)-2\left(a^2-5a+4\right)}\)
\(P=\frac{\left(a+1\right)\left(a^2-5a+4\right)}{\left(a-2\right)\left(a^2-5a+4\right)}=\frac{a+1}{a-2}\)
b/ \(P=\frac{a+1}{a-2}=1+\frac{3}{a-2}\)
\(P\) nguyên khi \(a-2=Ư\left(3\right)=\left\{-3;-1;1;3\right\}\)
\(a-2=-3\Rightarrow a=-1\left(l\right)\)
\(a-2=-1\Rightarrow a=1\left(l\right)\)
\(a-2=1\Rightarrow a=3\)
\(a-2=3\Rightarrow a=5\)
Vậy \(\left[{}\begin{matrix}a=3\\a=5\end{matrix}\right.\) thì P nguyên
\(P=\frac{a^3-4a^2-a+4}{a^3-7a^2+14a-8}=\frac{\left(a-4\right)\left(a+1\right)\left(a-1\right)}{\left(a-1\right)\left(a-2\right)\left(a-4\right)}=\frac{a+1}{a-2}\)
b \(P=\frac{a-2+3}{a-2}=1+\frac{3}{a-2}\)
Để P nhận giá trị nguyên \(\left(a-2\right)\inƯ\left(3\right)=\left\{1;-1;-3;3\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}a-2=1\\a-2=-1\\a-2=3\\a-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=3\\a=1\\a=5\\a=-1\end{matrix}\right.\)