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a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)
\(A=\frac{2x+1-3x-1+x^2}{3x}\)
\(A=\frac{x^2-x}{3x}\)
\(A=\frac{x\left(x-1\right)}{3x}\)
\(A=\frac{x-1}{3}\)
b) Thay x = 4 ta có :
\(A=\frac{4-1}{3}=\frac{3}{3}=1\)
c) Để A thuộc Z thì \(x-1⋮3\)
\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)
\(\Rightarrow x\in\left\{1;4;7;...\right\}\)
Vậy.....
a) Điều kiện: \(x\ne\left\{0;\pm2\right\}\)
\(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=[\frac{x^2}{x.\left(x-2\right).\left(x+2\right)}-\frac{6}{3.\left(x-2\right)}+\frac{1}{x+2}]:\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{x-2.\left(x+2\right)+x-2}{\left(x-2\right).\left(x+2\right)}.\frac{x+2}{6}\)
\(=\frac{6}{\left(x-2\right).\left(x+2\right)}.\frac{x+2}{6}\)
\(=-\frac{1}{x-2}\)
b) \(A\) \(Max\)
\(\Rightarrow-\frac{1}{x-2}Max\)
\(\Rightarrow\frac{1}{x-2}Min\)
\(\Rightarrow\left(x-2\right)\) \(Max\)
\(\Rightarrow x\) \(Max\)
\(\Rightarrow x\in\varnothing\)
a) \(ĐKXĐ:\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\left[\frac{\left(x+1\right)\left(x+2\right)}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x\left(x+1\right)}{3x\left(x+1\right)}\right]:\frac{2\left(1-2x\right)}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\frac{\left(x+1\right)\left(x+2\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{2-8x^2}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{1+2x-3x-1+x^2}{3x}\)
\(=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
b)\(\text{Với }x\ne0,x\ne-1,x\ne\frac{1}{2}\text{ ta có:}\)
\(\text{Để A< 0\Leftrightarrow}\frac{x-1}{3}< 0\Rightarrow x-1< 0\Leftrightarrow x< 1\)
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 ............................
a) Đk: x > 0 và x khác +-1
Ta có: A = \(\left(\frac{x+1}{x}-\frac{1}{1-x}-\frac{x^2-2}{x^2-x}\right):\frac{x^2+x}{x^2-2x+1}\)
A = \(\left[\frac{\left(x-1\right)\left(x+1\right)+x-x^2+2}{x\left(x-1\right)}\right]:\frac{x\left(x+1\right)}{\left(x-1\right)^2}\)
A = \(\frac{x^2-1+x-x^2+2}{x\left(x-1\right)}\cdot\frac{\left(x-1\right)^2}{x\left(x+1\right)}\)
A = \(\frac{x+1}{x}\cdot\frac{x-1}{x\left(x+1\right)}=\frac{x-1}{x^2}\)
b) Ta có: A = \(\frac{x-1}{x^2}=\frac{1}{x}-\frac{1}{x^2}=-\left(\frac{1}{x^2}-\frac{1}{x}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\frac{1}{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> 1/x - 1/2 = 0 <=> x = 2 (tm)
Vậy MaxA = 1/4 <=> x = 2
