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\(a,ĐKXĐ\hept{\begin{cases}x-3\ne0\\x+3\ne0\end{cases}\Leftrightarrow x\ne\pm3}\)
Ta có: \(M=\frac{3}{x-3}-\frac{6x}{9-x^2}+\frac{x}{x+3}\)
\(=\frac{3}{x-3}+\frac{6x}{x^2-9}+\frac{x}{x+3}\)
\(=\frac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x+3}{x-3}\)
\(b,x=\frac{1}{2}\Rightarrow M=\frac{\frac{1}{2}+3}{\frac{1}{2}-3}=-\frac{7}{5}\)
a) Phân thức xác định \(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x+1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}}\)
b) \(M=\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{x+3}{x^2-1}\)
\(M=\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{\left(x-1\right)\left(x+1\right)}\)
\(M=\frac{x^2+2x+1-x^2+2x-1-x-3}{\left(x-1\right)\left(x+1\right)}\)
\(M=\frac{3x-3}{\left(x-1\right)\left(x+1\right)}\)
\(M=\frac{3\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(M=\frac{3}{x+1}\)
Để M nguyên thì :
\(3⋮x+1\)
\(\Rightarrow x+1\inƯ\left(3\right)=\left\{1;3;-1;-3\right\}\)
\(\Rightarrow x\in\left\{0;2;-2;-4\right\}\)( thỏa mãn ĐKXĐ )
Vậy.......
ĐKXĐ \(x\ne1;x\ne-1\)
\(M=\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{x+3}{x^2-1}\)
\(M=\frac{\left(x+1\right)^2}{x^2-1}-\frac{\left(x-1\right)^2}{x^2-1}-\frac{x+3}{x^2-1}\)
\(M=\frac{x^2+2x+1-x^2+2x-1-x-3}{\left(x-1\right)\left(x+1\right)}\)
\(M=\frac{3x-3}{\left(x-1\right)\left(x+1\right)}\)
\(M=\frac{3}{x+1}\)
Để M nguyên \(\Leftrightarrow\text{ }3\text{ }⋮\text{ }x+1\text{ }hay\text{ }x+1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
=>\(x\in\left\{-4;-2;0;2\right\}\)thì M nguyên
A=x3/x2--4.x+2/x-x-4xx-4/xx-2
Điều kiện x \(\ne\)+-2
Ý b c tự làm
a) ĐKXĐ của A : \(\hept{\begin{cases}2x-3\ne0\\2x+3\ne0\\9-4x^2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}2x\ne3\\2x\ne-3\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne\frac{3}{2}\\x\ne-\frac{3}{2}\end{cases}}}\)
=> Giá trị của biểu thức A được xác định khi x khác 3/2 và x khác -3/2
\(A=\frac{5}{2x-3}+\frac{2}{2x+3}-\frac{2x+5}{9-4x^2}\)
\(=\frac{5}{2x-3}+\frac{2}{2x+3}+\frac{2x+5}{\left(2x+3\right)\left(2x-3\right)}\)
\(=\frac{5.\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{2.\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{2x+5}{\left(2x+3\right)\left(2x-3\right)}\)
\(=\frac{10x+15+4x-6+2x+5}{\left(2x+3\right)\left(2x-3\right)}\)
..... chắc tôi làm sai oy !
\(x\ne+-3\)
\(3\left(x-3\right)+1\left(x+3\right)+18\)
3x-9+x+3+18
4x+15
x=-15/4
a) ĐKXĐ:
\(\hept{\begin{cases}x-3\ne0,9-x^2\ne0,x+3\ne0\\1-\frac{x+1}{x+3}\ne0\end{cases}}\Leftrightarrow x\ne\pm3\).
\(M=\left(\frac{3+x}{x-3}+\frac{18}{9-x^2}+\frac{x-3}{x+3}\right)\div\left(1-\frac{x+1}{x+3}\right)\)
\(M=\frac{\left(3+x\right)\left(x+3\right)-18+\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\div\frac{2}{x+3}\)
\(M=\frac{x^2+6x+9-18+x^2-6x+9}{\left(x-3\right)\left(x+3\right)}\times\frac{x+3}{2}\)
\(M=\frac{2x^2\left(x+3\right)}{2\left(x-3\right)\left(x+3\right)}\)
\(M=\frac{x^2}{x-3}\)
b) \(M=\frac{x^2}{x-3}=\frac{x^2-3x+3x-9+9}{x-3}=x+3+\frac{9}{x-3}\inℤ\Leftrightarrow\frac{9}{x-3}\inℤ\)
mà \(x\inℤ\)nên \(x-3\inƯ\left(9\right)=\left\{-9,-3,-1,1,3,9\right\}\Leftrightarrow x\in\left\{-6,0,2,4,6,12\right\}\).
a) ĐKXĐ : x ≠ ±3
\(=\left[\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\frac{18}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}\right]\div\left(\frac{x+3}{x+3}-\frac{x+1}{x+3}\right)\)
\(=\left[\frac{x^2+6x+9-18+x^2-6x+9}{\left(x-3\right)\left(x+3\right)}\right]\div\left(\frac{x+3-x-1}{x+3}\right)\)
\(=\frac{2x^2}{\left(x-3\right)\left(x+3\right)}\div\frac{2}{x+3}=\frac{2x^2}{\left(x-3\right)\left(x+3\right)}\times\frac{x+3}{2}=\frac{x^2}{x-3}\)
b) \(M=\frac{x^2}{x-3}=\frac{x^2-3x+3x-9+9}{x-3}=\frac{x\left(x-3\right)+3\left(x-3\right)+9}{x-3}=x+3+\frac{9}{x-3}\)
Vì x nguyên nên x + 3 nguyên
nên để M nguyên thì 9/x-3 nguyên
hay x - 3 ∈ Ư(9) [ bạn tự xét tiếp :)) ]