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a)Đa thức B có nghĩa\(\Leftrightarrow x+1\ne0\)và\(x-1\ne0\)và\(x\ne0\Leftrightarrow x\ne-1\)và\(x\ne1\)và\(x\ne0\)
b)Ta có:\(B=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(\frac{x^2+1}{x+1}-\frac{x+1}{x+1}\right)\left(\frac{4.x}{\left(x-1\right).x}-\frac{2.\left(x-1\right)}{x.\left(x-1\right)}\right)\)
\(=\frac{x^2+1-x-1}{x+1}\left(\frac{4x}{x\left(x-1\right)}-\frac{2x-2}{x\left(x-1\right)}\right)=\frac{x^2-x}{x+1}.\frac{4x-2x+2}{x\left(x-1\right)}=\frac{x\left(x-1\right)}{x+1}.\frac{2x+2}{x\left(x-1\right)}\)
\(=\frac{2x+2}{x+1}=\frac{2\left(x+1\right)}{x+1}=2\)
a) B có nghĩa \(\Leftrightarrow\hept{\begin{cases}x+1\ne0\\x-1\ne0\\x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne-1\\x\ne1\\x\ne0\end{cases}}\)
b) \(B=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(=\frac{\left(x^2+1\right)-\left(x+1\right)}{x+1}.\frac{4x-\left(2x-2\right)}{x\left(x-1\right)}\)
\(=\frac{x^2+1-x-1}{x+1}.\frac{4x-2x+2}{x\left(x-1\right)}\)
\(=\frac{x^2-x}{x+1}.\frac{2x+2}{x\left(x-1\right)}=\frac{x\left(x-1\right)}{x+1}.\frac{2\left(x+1\right)}{x\left(x-1\right)}=2\)
a) ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ne0\\x-1\ne0\\x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne1\\x\ne0\end{matrix}\right.\)
b) Rút gọn :
GT \(\Leftrightarrow\frac{x^2+1-x-1}{x+1}.\frac{4x-2x+2}{x\left(x-1\right)}\)
\(\Leftrightarrow\frac{x\left(x-1\right)}{x+1}.\frac{2\left(x+1\right)}{x\left(x-1\right)}=2\)
a) A có nghĩa \(\Leftrightarrow\left(x+1\right)^2-3x\ne0\), \(x^3+1\ne0\),\(x+1\ne0\),\(3x^2+6x\ne0\) và \(x^2-4\ne0\)
+) \(\left(x+1\right)^2-3x\ne0\Leftrightarrow x^2+2x+1-3x\ne0\)
\(\Leftrightarrow x^2-x+1\ne0\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ne0\)(luôn đúng)
+) \(x^3+1\ne0\Leftrightarrow x^3\ne-1\Leftrightarrow x\ne-1\)
+) \(x+1\ne0\Leftrightarrow x\ne-1\)
+) \(3x^2+6x\ne0\Leftrightarrow3x\left(x+2\right)\ne0\)
\(\Leftrightarrow x\ne0;x\ne-2\)
+) \(x^2-4\ne0\Leftrightarrow x^2\ne4\Leftrightarrow x\ne\pm2\)
Vậy ĐKXĐ của A là \(x\ne-1;x\ne0;x\ne\pm2\)
a, \(Đkxđ:\hept{\begin{cases}x\ne-1\\x\ne0\\x\ne-2\end{cases}}\)
\(A=\left[\frac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\frac{2x^2+4x-1}{x^3+1}-\frac{1}{x+1}\right]:\frac{x^2-4}{3x^2+6x}\)
\(=\left[\frac{x^2+2x+1}{x^2-x+1}-\frac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{1}{x+1}\right].\frac{3x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{\left(x^2+2x+1\right)\left(x+1\right)-2x^2-4x+1-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)
\(=\frac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}.\frac{3x}{x-2}\)
\(=\frac{3x}{x-2}=3+\frac{6}{x-2}\)
b, Để A nguyên thì \(\Leftrightarrow6\)chia hết cho \(x-2\)
Hay \(\left(x-2\right)\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| x-2 | -6 | -3 | -2 | -1 | 1 | 2 | 3 | 6 |
| x | -4 | -1 | 0 | 1 | 3 | 4 | 5 | 8 |
Vậy ............................
Hình như đề sai.Sửa đề luôn nha !
\(ĐKXĐ:x\ne\pm2\)
\(A=\left(\frac{x}{x^2-4}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right):\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)
\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{-6}=\frac{1}{x-2}\)
b
Để \(A< 0\Rightarrow\frac{1}{x-2}< 0\Rightarrow x-2< 0\Rightarrow x< 2\)
c
Để A nguyên thì \(\frac{1}{x-2}\) nguyên
\(\Rightarrow1⋮x-2\)
\(\Rightarrow x-2\in\left\{1;-1\right\}\Rightarrow x\in\left\{3;1\right\}\)
\(a,\text{để a xác định thì }\hept{\begin{cases}x-2\ne0\\2-x\ne0\end{cases}\Rightarrow x\ne2}\)
\(b,\left[\left(\frac{x+1}{x-2}+\frac{3}{2-x}-3x\right):\frac{1-3x}{x-2}\right]-\frac{x^2+4}{x-2}\)
\(=\left[\left(\frac{x+1}{x-2}-\frac{3}{x-2}-3x\right):\frac{1-3x}{x-2}\right]-\frac{x^2+4}{x-2}\)
\(=\left(1-3x\right)\cdot\frac{\left(x-2\right)}{1-3x}-\frac{x^2+4}{x-2}=\frac{\left(x-2\right)^2}{x-2}-\frac{x^2+4}{x-2}=\frac{-4x}{x-2}\)
Vậy với \(x=\frac{1}{2}\text{ }\Rightarrow A=\frac{-\frac{4.1}{2}}{\frac{1}{2}-2}=\frac{4}{3}\)
a/ ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\\x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\\x\ne0\end{matrix}\right.\)
Vậy ....
b/ Ta có :
\(B=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(=\left(\frac{x^2+1}{x+1}-\frac{x+1}{x+1}\right)\left(\frac{4x}{x\left(x-1\right)}-\frac{2\left(x-1\right)}{x\left(x-1\right)}\right)\)
\(=\frac{x^2+1-x-1}{x+1}.\frac{4x-2x+2}{x\left(x-1\right)}\)
\(=\frac{x\left(x-1\right)}{x+1}.\frac{2\left(x+1\right)}{x\left(x-1\right)}\)
\(=2\)
Vậy...