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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
\(A=\left(\frac{1}{1-x}-1\right):\left(x+1-\frac{1-2x}{1-x}\right)\) \(\left(ĐK:x\ne1;x\ne2\right)\)
\(=\frac{1-1+x}{1-x}:\frac{\left(1-x\right)\left(x+1\right)-\left(1-2x\right)}{1-x}\)
\(=\frac{x}{1-x}\cdot\frac{1-x}{1-x^2-1+2x}\)
\(=\frac{x}{-x^2+2x}\)
\(=\frac{x}{-x\left(x-2\right)}=-\frac{1}{x-2}=\frac{1}{2-x}\)
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 \(x\ne\pm1\), sau khi rút gọn ta được: (bạn tư làm)
\(P=\frac{x}{x+1}\)
b) Khi \(\left|x-\frac{2}{3}\right|=\frac{1}{3}\) thì hoặc \(x-\frac{2}{3}=\frac{1}{3}\) hoặc \(x-\frac{2}{3}=-\frac{1}{3}\)
Hay là \(x=1\) hoặc \(x=\frac{1}{3}\)
Do để P có nghĩa thì \(x\ne\pm1\) nên \(x=\frac{1}{3}\), khi đó:
\(P=\frac{\frac{1}{3}}{\frac{1}{3}+1}=\frac{1}{4}\)
c) P > 1 khi \(\frac{x}{x+1}>1\)
\(\Leftrightarrow1-\frac{1}{x+1}>1\)
\(\Leftrightarrow\frac{1}{x+1}< 0\)
\(\Leftrightarrow x< -1\)
e) Đề không rõ ràng
I don't now
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d) \(A>0\Leftrightarrow\frac{-1}{x-2}>0\)
\(\Leftrightarrow x-2< 0\) ( vì \(-1< 0\))
\(\Leftrightarrow x< 2\)
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)
\(A=\frac{-1}{x-2}\)
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