cho biểu thức A = \(\left(\frac{1-x^3}{1-x}-x\right):\frac{1-x^2}{1-x-x^2+x^3}\) vs x \(\ne\pm1\)
a, Rút gọn A
b, Tìm g.trị của biểu thức A tại x = \(-1\frac{2}{3}\) .
c, Tìm g.trị của x để A<0.
K
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T
28 tháng 7 2018
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
\(P>1\Leftrightarrow x\hept{\begin{cases}x\in Z\\x\ne-1\\x\le0\end{cases}}\)
TN
28 tháng 7 2018
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
H
10 tháng 4 2019
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{8}{x^2-1}\right):\left(\frac{1}{x-1}-\frac{7x+3}{1-x^2}\right)\)
\(A=\left[\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}-\frac{x^2-2x+1}{\left(x+1\right)\left(x-1\right)}+\frac{8}{\left(x+1\right)\left(x-1\right)}\right]:\left[\frac{x+1}{\left(x+1\right)\left(x-1\right)}-\frac{3-7x}{\left(x+1\right)\left(x-1\right)}\right]\)
\(A=\left[\frac{x^2+2x+1-x^2+2x-1+8}{\left(x+1\right)\left(x-1\right)}\right]:\frac{x+1-3+7x}{\left(x+1\right)\left(x-1\right)}\)
\(A=\frac{4x+8}{\left(x+1\right)\left(x-1\right)}.\frac{\left(x+1\right)\left(x-1\right)}{8x-2}\)
......................
H
12 tháng 3 2019
\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{8x}{x^2-1}\right):\left(\frac{2x-2x^2-6}{x^2-1}-\frac{2}{x-1}\right)\)
\(A=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{8x}{\left(x+1\right)\left(x-1\right)}\right):\left(\frac{2x-2x^2-6}{\left(x-1\right)\left(x+1\right)}-\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}\right)\)
\(A=\left(\frac{x^2+2x+1-x^2+2x-1-8x}{\left(x-1\right)\left(x+1\right)}\right):\left(\frac{2x-2x^2-6-2x-2}{\left(x+1\right)\left(x-1\right)}\right)\)
\(A=\left(\frac{4x-8x}{\left(x-1\right)\left(x+1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{-2x^2-8}\)
..........
NN
12 tháng 3 2019
\(\frac{x+32}{2008}+\frac{x+31}{2009}+\frac{x+29}{2011}+\frac{x+28}{2012}+\frac{x+2056}{4}=0\) \(=0\)
\(\Leftrightarrow\)\(\frac{x+32}{2008}+1+\frac{x+31}{2009}+1+\frac{x+29}{2011}+1\)\(+\frac{x+28}{2012}+1+\frac{x+2056}{4}-4\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+32}{2008}+\frac{2008}{2008}+\frac{x+31}{2009}+\frac{2009}{2009}+\)\(\frac{x+29}{2011}+\frac{2011}{2011}+\frac{x+28}{2012}+\frac{2012}{2012}+\)\(\frac{x+2056}{4}-\frac{16}{4}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+32+2008}{2008}+\frac{x+31+2009}{2009}\)\(+\frac{x+29+2011}{2011}+\frac{x+28+2012}{2012}\)\(+\frac{x+2056-16}{4}\)\(=0\)
\(\Leftrightarrow\)\(\frac{x+2040}{2008}+\frac{x+2040}{2009}+\frac{x+2040}{2011}\)\(+\frac{x+2040}{2012}+\frac{x+2040}{4}=0\)
\(\Leftrightarrow\)\(\left(x+2040\right).\left(\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+2040=0\\\frac{1}{2008}+\frac{1}{2009}+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{4}=0\end{cases}}\)(vô lí)
\(\Leftrightarrow\)\(x=-2040\)
Vậy phương trình có nghiệm là : x = -2040
13 tháng 12 2020
a, \(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right):\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x+1}{\left(x+1\right)^2}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}=\frac{x+1}{x-1}\)
b, Thay x = -2 ta được :
\(\frac{x+1}{x-1}=\frac{-2+1}{-2-1}=\frac{1}{3}\)
Vậy A nhận giá trị 1/3
13 tháng 12 2020
\(A=\left(\frac{1}{x-1}+\frac{x}{x^2-1}\right)\div\frac{2x+1}{x^2+2x+1}\)
\(=\left(\frac{1}{x-1}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\div\frac{2x+1}{\left(x+1\right)^2}\)
\(=\left(\frac{x+1}{\left(x-1\right)\left(x+1\right)}+\frac{x}{\left(x-1\right)\left(x+1\right)}\right)\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{2x+1}{\left(x-1\right)\left(x+1\right)}\times\frac{\left(x+1\right)^2}{2x+1}\)
\(=\frac{x+1}{x-1}\)
Với x = -2 (tmđk) => \(A=\frac{-2+1}{-2-1}=\frac{-1}{-3}=\frac{1}{3}\)
M
25 tháng 3 2018
d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
25 tháng 3 2018
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
SN
2 tháng 8 2015
a)\(\left(\frac{1-x^3+1-x-x}{1-x}\right):\frac{-\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)^2}=\left(\frac{-x^3-2x+2}{1-x}\right)\cdot\left(1-x\right)=-x^3-2x+2\)
b) \(-\left(-1\frac{2}{3}\right)-2\cdot\left(-1\frac{2}{3}\right)+2=\frac{5}{3}+\frac{10}{3}+2=7\)