P = \(\left(\frac{x-1}{x+1}-\frac{x}{x-1}-\frac{3x+1}{1-x^2}\right):\frac{2x+1}{x^2-1}\)
a) Rút gọn P
b) tìm g.trị của x để P=\(\frac{3}{x-1}\)
c) tìm g.trị nguyên của x để A>1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
21 tháng 4 2020
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
KN
24 tháng 1 2020
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 \(x\ne-1;x\ne0;x\ne\pm2\)
24 tháng 1 2020
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 ............................
NM
Nguyễn Minh Quang
Giáo viên
9 tháng 2 2021
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 \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)
27 tháng 1 2018
a) \(ĐKXĐ:\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}\)
\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\left[\frac{\left(x+1\right)\left(x+2\right)}{3x\left(x+1\right)}+\frac{6x}{3x\left(x+1\right)}-\frac{9x\left(x+1\right)}{3x\left(x+1\right)}\right]:\frac{2\left(1-2x\right)}{x+1}-\frac{3x+1-x^2}{3x}\)
\(=\frac{\left(x+1\right)\left(x+2\right)+6x-9x\left(x+1\right)}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{2-8x^2}{3x\left(x+1\right)}.\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)
\(=\frac{1+2x-3x-1+x^2}{3x}\)
\(=\frac{x\left(x-1\right)}{3x}=\frac{x-1}{3}\)
27 tháng 1 2018
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) 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