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a) Ta có :A = \(\left(\frac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\frac{1-2x^2+4x}{x^3-1}+\frac{1}{x-1}\right):\frac{x^2+x}{x^3+x}\)
ĐK: \(\hept{\begin{cases}x\ne0\\x\ne1\end{cases}}\)
A = \(\left(\frac{\left(x-1\right)^2}{x^2+x+1}-\frac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{1}{x-1}\right):\frac{x\left(x+1\right)}{x\left(x^2+1\right)}\)
= \(\frac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-3x^2+3x-1+3x^2-3x}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}\)
= \(\frac{x^3-1}{\left(x-1\right)\left(x^2+x+1\right)}.\frac{x^2+1}{x+1}=1.\frac{x^2+1}{x+1}=\frac{x^2+1}{x+1}\)
b) Để A > - 1 <=> \(\frac{x^2+1}{x+1}>-1\)
<=> \(\frac{x^2+1}{x+1}+1>0\)
<=> \(\frac{x^2+x+2}{x+1}>0\)
Vì x2 + x + 2 >0 \(\forall x\)
=> A > 0 <=> x + 1 > 0 <=> 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 ............................
Ta có \(A=[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)]:\frac{x-1}{x^3}\)
\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)
\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)
Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)
Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)
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)\(\text{Với }x\ne0,x\ne-1,x\ne\frac{1}{2}\text{ ta có:}\)
\(\text{Để A< 0\Leftrightarrow}\frac{x-1}{3}< 0\Rightarrow x-1< 0\Leftrightarrow x< 1\)
\(ĐKXĐ:x\ne\pm1\)
a) \(A=\left(\frac{1}{1-x}+\frac{2}{1+x}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\left(\frac{\left(1+x\right)}{\left(1+x\right)\left(1-x\right)}+\frac{2\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)
\(=\frac{1+x+2-2x-5+x}{1-x^2}:\frac{2x-1}{1-x^2}\)
\(=\frac{8}{1-x^2}.\frac{1-x^2}{2x-1}=\frac{8}{2x-1}\)
b) Để A nguyên thì \(\frac{8}{2x-1}\inℤ\)
\(\Leftrightarrow8⋮2x-1\Rightarrow2x-1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)
Mà dễ thấy 2x - 1 lẻ nên\(2x-1\in\left\{\pm1\right\}\)
+) \(2x-1=1\Rightarrow x=1\left(ktmđkxđ\right)\)
+) \(2x-1=-1\Rightarrow x=0\left(tmđkxđ\right)\)
Vậy x nguyên bằng 0 thì A nguyên
c) \(\left|A\right|=A\Leftrightarrow A\ge0\)
\(\Rightarrow\frac{8}{2x-1}\ge0\Rightarrow2x-1>0\Leftrightarrow x>\frac{1}{2}\)
Vậy \(x>\frac{1}{2}\)thì |A| = A
a, \(A=\left(\frac{1}{1-x}+\frac{2}{1+x}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\frac{1}{2};x\ne\pm1\right)\)
\(\Leftrightarrow A=\left(\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{2-2x}{\left(1-x\right)\left(1+x\right)}-\frac{5-x}{\left(1-x\right)\left(1+x\right)}\right):\frac{\left(x+1\right)\left(x-1\right)}{2x-1}\)
\(\Leftrightarrow A=\frac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{2x-1}\)
\(\Leftrightarrow A=\frac{-2\left(1-x^2\right)}{\left(1-x^2\right)\left(2x-1\right)}=\frac{2}{2x-1}\)
Vậy \(A=\frac{2}{2x-1}\left(x\ne\frac{1}{2};x\ne\pm1\right)\)
b) \(A=\frac{2}{2x-1}\left(x\ne\frac{1}{2};x\ne\pm1\right)\)
Để A nhận giá trị nguyên thì 2 chia hết cho 2x-1
Mà x nguyên => 2x-1 nguyên
=> 2x-1 thuộc Ư (2)={-2;-1;1;2}
Ta có bảng
| 2x-1 | -2 | -1 | 1 | 2 |
| 2x | -1 | 0 | 2 | 3 |
| x | -1/2 | 0 | 1 | 3/2 |
Đối chiếu điều kiện
=> x=0
\(A=\left(\frac{1}{1-x}-1\right):\left(x+1-\frac{1-2x}{1-x}\right)\) \(\left(ĐK:x\ne1;x\ne2\right)\)
\(=\frac{1-1+x}{1-x}:\frac{\left(1-x\right)\left(x+1\right)-\left(1-2x\right)}{1-x}\)
\(=\frac{x}{1-x}\cdot\frac{1-x}{1-x^2-1+2x}\)
\(=\frac{x}{-x^2+2x}\)
\(=\frac{x}{-x\left(x-2\right)}=-\frac{1}{x-2}=\frac{1}{2-x}\)
b) Để A=\(\frac{1}{2}\) \(\Leftrightarrow\)\(\frac{1}{2-x}=\frac{1}{2}\)
\(\Leftrightarrow2-x=2\)
\(\Leftrightarrow-x=0\Leftrightarrow x=0\)
c) Để A>1 \(\Leftrightarrow\)\(\frac{1}{2-x}>1\)
\(\Leftrightarrow\)\(\frac{1}{2-x}-1>0\)
\(\Leftrightarrow\)\(\frac{1-2+x}{2-x}>0\)
\(\Leftrightarrow\)\(\frac{x-1}{2-x}>0\)
\(\Leftrightarrow\begin{cases}x-1>0\\2-x>0\end{cases}\) hoặc \(\begin{cases}x-1< 0\\2-x< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>1\\x< 2\end{cases}\) hoặc \(\begin{cases}x< 1\\x>2\end{cases}\)(vô nghiệm)
\(\Leftrightarrow1< x< 2\)
Vậy \(1< x< 2\) thì A<1
a) ĐK : \(a\ne\pm1\); \(a\ne\frac{-1}{2}\)
\(P=[\frac{\left(x-1\right)\left(1-x\right)}{1-x^2}+\frac{x\left(1+x\right)}{1-x^2}-\frac{3x+1}{1-x^2}]:\frac{2x+1}{x^2-1}\)
\(=\left(\frac{-x^2+2x-1+x^2+x-3x-1}{1-x^2}\right):\frac{2x+1}{x^2+1}\)
\(=\left(\frac{-2}{1-x^2}\right):\frac{-2x-1}{1-x^2}\)
\(=\frac{2}{2x+1}\)
b)
\(\frac{2}{2x+1}=\frac{3}{x-1}\)
\(\Leftrightarrow2\left(x-1\right)=3\left(2x+1\right)\)
<=> x=-5/4 (nhận)
c) P>1
\(\Leftrightarrow\frac{2}{2x+1}>1\)
\(\Leftrightarrow2x+1>0\)
Khi đó : 2 > 2x+1
<=> x < 1/2
mà x thuộc Z nên
\(P>1\Leftrightarrow x\hept{\begin{cases}x\in Z\\x\ne-1\\x\le0\end{cases}}\)
a/ \(P=\left(\frac{x-1}{x+1}-\frac{x}{x-1}-\frac{3x+1}{1-x^2}\right):\frac{2x+1}{x^2-1}\)
\(P=\left(\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{3x+1}{x^2-1}\right):\frac{2x+1}{x^2-1}\)
\(P=\left(\frac{x^2-2x+1}{x^2-1}-\frac{x^2+x}{x^2-1}+\frac{3x+1}{x^2-1}\right).\frac{x^2-1}{2x+1}\)
\(P=\frac{x^2-2x+1-x^2-x+3x+1}{x^2-1}.\frac{x^2-1}{2x+1}\)
\(P=\frac{2}{2x+1}\)
b/ để \(P=\frac{3}{x-1}\)
<=> \(\frac{2}{2x+1}=\frac{3}{x-1}\)
=> \(2x-2=6x+3\)
<=> \(2x-6x=3+2\)
<=> \(-4x=5\)
<=> \(x=\frac{-5}{4}\)
c/ để \(P>1\)
<=> \(\frac{2}{2x+1}\)\(>1\)
<=> \(\frac{2}{2x+1}-1>0\)
<=> \(\frac{2}{2x+1}-\frac{2x+1}{2x+1}>0\)
<=> \(\frac{3-2x}{2x+1}>0\)
<=> \(\hept{\begin{cases}3-2x>0\\2x+1>0\end{cases}}\)hoặc \(\hept{\begin{cases}3-2x< 0\\2x+1< 0\end{cases}}\)
<=> \(\hept{\begin{cases}x< \frac{3}{2}\\x>\frac{-1}{2}\end{cases}}\)hoặc \(\hept{\begin{cases}x>\frac{3}{2}\\x< \frac{-1}{2}\end{cases}}\)
<=> \(\frac{-1}{2}< x< \frac{3}{2}\)hoặc \(x\in\varnothing\)
vậy \(\frac{-1}{2}< x< \frac{3}{2}\)thì \(P< 1\)
học tốt