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a) điều kiện : x-1\(\ne0\)
\(\frac{1}{x-1}>\frac{1}{2}\Rightarrow\frac{1\cdot2}{\left(x-1\right)\cdot2}>\frac{1\left(x-1\right)}{2\left(x-1\right)}\Leftrightarrow2>x-1\Leftrightarrow-x>-1-2\Leftrightarrow-x>-3\)
\(\Leftrightarrow x< 3\)
b) \(\frac{2x+3}{-2}< \frac{3}{-2}\Leftrightarrow2x+3>3\Leftrightarrow2x>3-3\Leftrightarrow2x>0\Leftrightarrow x>0\)
c) điều kiện :\(x\ne0\)
\(\frac{2x-1}{x}< \frac{1+x}{x}\Leftrightarrow2x-1< 1+x\Leftrightarrow2x-x< 1+1\Leftrightarrow x< 2\)
nhiều thế
a) \(\frac{5x-2}{2}\ge\frac{3-x}{3}\Leftrightarrow\frac{3\left(5x-2\right)}{6}\ge\frac{2\left(3-x\right)}{6}\Leftrightarrow15x-6\ge6-2x\Leftrightarrow x\ge\frac{12}{17}\)
0 [ 12/17
a) vì 5>0 rồi => \(\frac{x+2}{5}\ge0\Leftrightarrow x+2\ge0\Leftrightarrow x\ge-2\)
b) th1: x+2<0 và x-3>0 <=> x<-2 và x>3 => vô lí
th2: x+2>0 và x-3<0 <=> x>-2 và x<3 => -2<x<3
c) \(\Leftrightarrow\frac{x-1}{x-3}-1>0\Leftrightarrow\frac{x-1-x+3}{x-3}>0\Leftrightarrow\frac{2}{x-3}>0\) đến đây làm như câu a, nếu k làm đc liên hệ mình làm nốt
a)\(pt\Leftrightarrow-\frac{x}{2x^2-5}-\frac{25}{2x^2-50}+\frac{x}{x^2-5}+\frac{5}{x^2-5}=\frac{x}{2x^2+10x}-\frac{5}{2x^2+10x}\)
=>\(-\frac{x}{2x^2+10x}+\frac{5}{2x^2+10x}-\frac{x}{2x^2-50}-\frac{25}{2x^2-50}+\frac{x}{x^2-5}+\frac{5}{x^2-5}=0\)
\(\Leftrightarrow-\frac{5\left(x^2+8x-5\right)}{2\left(x-5\right)x\left(x^2-5\right)}=0\)
\(\Rightarrow\frac{1}{x-5}=0\Leftrightarrow\frac{1}{x}=0\Rightarrow\frac{1}{x^2-5}=0\)
=>x2+8x-5=0
=>82-(-4(1.5))=84
=>x1=(-8)+8:2=\(\sqrt{21}-4\)
=>x2=(-8)+8:2=\(-\sqrt{21}-4\)
=>x=±\(\sqrt{21}-4\)
b)\(\Leftrightarrow-\frac{x}{x+1}+\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}=\frac{16}{x^2-1}\)
\(\Rightarrow-\frac{16}{x^2-1}-\frac{x}{x+1}+\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}=0\)
\(\Rightarrow\frac{4\left(x-4\right)}{\left(x-1\right)\left(x+1\right)}=0\Leftrightarrow\frac{1}{x-1}=0\Rightarrow\frac{1}{x+1}=0\)
=>x=4
c)\(\Leftrightarrow-\frac{x^2}{x+1}-\frac{x}{x+1}+\frac{2}{x+1}+x+2=\frac{x}{x+1}-\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}\)
\(\Rightarrow-\frac{x^2}{x+1}-\frac{2x}{x+1}+\frac{3}{x+1}-\frac{x}{x-1}+x-\frac{1}{x-1}+2=0\)
\(\Rightarrow\frac{2\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}=0\Leftrightarrow\frac{1}{x-1}=0\Rightarrow\frac{1}{x+1}=0\)
=>x=3
a) mau thuc chung la 21
b)mau thu chung la 30
c)mau thuc chung la 12
d)khai trien hang dang thuc
b) \(\frac{x-1}{2x-1}\ge1\)( ĐKXĐ : x ≠ 1/2 )
<=> \(\frac{x-1}{2x-1}-1\ge0\)
<=> \(\frac{x-1}{2x-1}-\frac{2x-1}{2x-1}\ge0\)
<=> \(\frac{x-1-2x+1}{2x-1}\ge0\)
<=> \(\frac{-x}{2x-1}\ge0\)
Xét hai trường hợp :
1. \(\hept{\begin{cases}-x\ge0\\2x-1>0\end{cases}}\Rightarrow loai\)
2. \(\hept{\begin{cases}-x\le0\\2x-1< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge0\\x< \frac{1}{2}\end{cases}}\Rightarrow0\le x< \frac{1}{2}\)
Vậy với \(0\le x< \frac{1}{2}\)thì \(\frac{x-1}{2x-1}\ge1\)
\(\frac{1}{x-8}< \frac{2}{x-6}\)( ĐKXĐ : x ≠ 8 ; x ≠ 6 )
<=> \(\frac{1}{x-8}-\frac{2}{x-6}< 0\)
<=> \(\frac{x-6}{\left(x-8\right)\left(x-6\right)}-\frac{2\left(x-8\right)}{\left(x-8\right)\left(x-6\right)}< 0\)
<=> \(\frac{x-6-2x+16}{\left(x-8\right)\left(x-6\right)}< 0\)
<=> \(\frac{-x+10}{\left(x-8\right)\left(x-6\right)}< 0\)
Xét hai trường hợp :
1.\(\hept{\begin{cases}-x+10>0\\\left(x-8\right)\left(x-6\right)< 0\end{cases}}\Rightarrow6< x< 8\)
2. \(\hept{\begin{cases}-x+10< 0\\\left(x-8\right)\left(x-6\right)>0\end{cases}}\Rightarrow x>10\)
Vậy \(\orbr{\begin{cases}6< x< 8\\x>10\end{cases}}\)thì \(\frac{1}{x-8}< \frac{2}{x-6}\)