tìm x để: 1/ B = 0
2/ B < 0
B=\(\frac{x-2}{3x+2}\)
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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)\(\text{Với }x\ne0,x\ne-1,x\ne\frac{1}{2}\text{ ta có:}\)
\(\text{Để A< 0\Leftrightarrow}\frac{x-1}{3}< 0\Rightarrow x-1< 0\Leftrightarrow x< 1\)
5 tháng 11 2016
a) \(\left(x-\frac{1}{3}\right)\left(5x+2\right)>0\)
<=> \(\left[\begin{array}{nghiempt}x-\frac{1}{3}>0\\5x+3< 0\end{array}\right.\) hoặc \(\left[\begin{array}{nghiempt}x-\frac{1}{3}< 0\\5x+3>0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x>\frac{1}{3}\\5x< 3\end{array}\right.\) hoặc \(\left[\begin{array}{nghiempt}x< \frac{1}{3}\\5x>3\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x>\frac{1}{3}\\x< \frac{3}{5}\end{array}\right.\) hoặc \(\left[\begin{array}{nghiempt}x< \frac{1}{3}\\x>\frac{3}{5}\end{array}\right.\)
Vậy...
5 tháng 11 2016
a) \(\left(x-\frac{1}{3}\right)\left(5x+2\right)>0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{3}>0\\5x+2>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{3}< 0\\5x+2< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{1}{3}\\x>-\frac{2}{5}\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{3}\\x< -\frac{2}{5}\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{3}\\x< -\frac{2}{5}\end{array}\right.\)
b) \(\left(5x+3\right)\left(3x-2\right)< 0\)
\(\Leftrightarrow\begin{cases}5x+3>0\\3x-2< 0\end{cases}\) hoặc \(\begin{cases}5x+3< 0\\3x-2>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>-\frac{3}{5}\\x< \frac{2}{3}\end{cases}\) hoặc \(\begin{cases}x< -\frac{3}{5}\\x>\frac{2}{5}\end{cases}\) (loại)
\(\Leftrightarrow-\frac{3}{5}< x< \frac{2}{3}\)
3 tháng 4 2021
a, \(B=\left(\frac{9-3x}{x^2+4x-5}-\frac{x+5}{1-x}-\frac{x+1}{x+5}\right):\frac{7x-14}{x^2-1}\)
\(=\left(\frac{9-3x}{\left(x-1\right)\left(x+5\right)}+\frac{\left(x+5\right)^2}{\left(x-1\right)\left(x+5\right)}-\frac{\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+5\right)}\right):\frac{7\left(x-2\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}.\frac{\left(x-1\right)\left(x+1\right)}{7\left(x-2\right)}\)
\(=\frac{35+7x}{x+5}\frac{x+1}{7\left(x-2\right)}=\frac{7\left(x+5\right)\left(x+1\right)}{7\left(x+5\right)\left(x-2\right)}=\frac{x+1}{x-2}\)
b, Ta có : \(\left(x+5\right)^2-9x-45=0\)
\(\Leftrightarrow x^2+10x+25-9x-45=0\Leftrightarrow x^2+x-20=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)
TH1 : Thay x = 4 vào biểu thức ta được : \(\frac{4+1}{4-2}=\frac{5}{2}\)
TH2 : THay x = 5 vào biểu thức ta được : \(\frac{5+1}{5-2}=\frac{6}{3}=2\)
c, Để B nhận giá trị nguyên khi \(\frac{x+1}{x-2}\inℤ\Rightarrow x-2+3⋮x-2\)
\(\Leftrightarrow3⋮x-2\Rightarrow x-2\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
3 tháng 4 2021
d, Ta có : \(B=-\frac{3}{4}\Rightarrow\frac{x+1}{x-2}=-\frac{3}{4}\)ĐK : \(x\ne2\)
\(\Rightarrow4x+4=-3x+6\Leftrightarrow7x=2\Leftrightarrow x=\frac{2}{7}\)( tmđk )
e, Ta có B < 0 hay \(\frac{x+1}{x-2}< 0\)
TH1 : \(\hept{\begin{cases}x+1< 0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x< -1\\x>2\end{cases}}}\)( ktm )
TH2 : \(\hept{\begin{cases}x+1>0\\x-2< 0\end{cases}}\Rightarrow\hept{\begin{cases}x>-1\\x< 2\end{cases}\Rightarrow-1< x< 2}\)
LV
1
N
13 tháng 6 2017
a) Xét A = 0
\(\Leftrightarrow\frac{X-2}{3X+2}=0\)
\(\Leftrightarrow X-2=0\)
\(\Leftrightarrow X=2\)
b) Xét A < 0
\(\Leftrightarrow\frac{X-2}{3X+2}=0\)
\(\Leftrightarrow\orbr{\begin{cases}X-2< 0\\3X+2< 0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}X< 1\\X< -1\end{cases}}\)
11 tháng 3 2020
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??
S
20 tháng 12 2018
\(P=1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(P=1+\frac{x+3}{\left(x+3\right)\left(x+2\right)}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{4x^2.2}{4x^2\left(x-2\right)}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{1}{x+2}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{2}{x-2}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{x-2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(P=1+\frac{1}{x+2}:\left(\frac{2x+4-x-x+2}{\left(x+2\right)\left(x-2\right)}\right)\)
\(P=1+\frac{1}{x+2}:\frac{6}{\left(x+2\right)\left(x-2\right)}=1+\frac{\left(x+2\right)\left(x-2\right)}{6\left(x+2\right)}=1+\frac{x-2}{6}\)
\(=\frac{x+4}{6}.P=0\Leftrightarrow x=-4\)
\(P>0\Leftrightarrow x>-4\)
1/ Để B=0 thì \(\frac{x-2}{3x+2}\)=0
Vì \(3x+2\ne0\)
=>x-2=0
=>x=0+2
=>x=2
Vậy x=2 thì B=0
2/ Để B<0 thì \(\frac{x-2}{3x+2}\)<0
=>x-2<0 hoặc 3x+2<0
+)Nếu x-2<0
=>x-2+2<0+2
=>x<2
+)Nếu 3x+2<0
=>3x+2-2<0-2
=>3x<-2
=>3x:3<(-2):3
=>x<\(\frac{-2}{3}\)
Vậy x<2 hoặc x<\(\frac{-2}{3}\)thì B<0