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10 tháng 4 2021
\(a,\)Với \(x\ne-3,x\ne2\) ta có :
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)
\(=\dfrac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)
\(=\dfrac{x-4}{x-2}\)
\(b,\) \(A=-3\Leftrightarrow\dfrac{x-4}{x-2}=-3\)
\(\Leftrightarrow x-4=-3\left(x-2\right)\)
\(\Leftrightarrow x-4+3x-6=0\)
\(\Leftrightarrow4x=10\Rightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)
2 tháng 3 2022
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
b: \(C=\dfrac{x^3-x^2-2x-2x+4}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^3-x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2\left(x-1\right)-4\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}=x-1\)
Để C=0 thì x-1=0
hay x=1
c: Để C>0 thì x-1>0
hay x>1
Vậy: \(\left\{{}\begin{matrix}x\in Z\backslash\left\{1\right\}\\x\notin\left\{2;-2\right\}\end{matrix}\right.\)
3 tháng 3 2022
a, ĐKXĐ:\(\left\{{}\begin{matrix}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x^2+x-6\ne0\\x\ne2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)
b, \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+3\right)}-\dfrac{5}{\left(x-2\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x-4}{x-2}\)
\(c,A=\dfrac{-3}{4}\\ \Leftrightarrow\dfrac{x-4}{x-2}=\dfrac{-3}{4}\\ \Leftrightarrow4\left(x-4\right)=-3\left(x-2\right)\\ \Leftrightarrow4x-16x=-3x+6\\ \Leftrightarrow4x-16x+3x-6=0\\ \Leftrightarrow7x-22=0\\ \Leftrightarrow x=\dfrac{22}{7}\)
d, \(A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}=1-\dfrac{2}{x-2}\)
Để \(A\in Z\Rightarrow\dfrac{2}{x-2}\in Z\Rightarrow x-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Ta có bảng:
| x-2 | -2 | -1 | 1 | 2 |
| x | 0 | 1 | 3 | 4 |
Vậy \(x\in\left\{0;1;3;4\right\}\)
DX
26 tháng 2 2023
\(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(\Rightarrow A=\left(\dfrac{x-2\left(x+2\right)+1\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)
\(\Rightarrow A=\left(\dfrac{-6}{x^2-4}\right):\left(\dfrac{6}{x+2}\right)\)
\(\Rightarrow A=-\dfrac{6}{x^2-4}.\dfrac{x+2}{6}=-\dfrac{6\left(x+2\right)}{\left(x-2\right)\left(x+2\right)6}=-\dfrac{1}{x-2}\)
để A<0 thì :
\(\left\{{}\begin{matrix}x-2\ne0\\x-2\notin Z-\end{matrix}\right.\)\(\Leftrightarrow x\in\left\{3;4;5;6;7;8;9;....n\right\}\)
( Z- là tập hợp số nguyên âm )
Để A có giá trị nguyên thì :
\(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
8 tháng 12 2021
a: \(A=\dfrac{x^2-5x+6-x^2+x+2x^2-6}{x\left(x-3\right)}=\dfrac{2x^2-4x}{x\left(x-3\right)}=\dfrac{2x}{x-3}\)
8 tháng 12 2021
a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2}{x^2-4}\)
12 tháng 12 2021
a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-2x}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
8 tháng 12 2021
a) \(A=\dfrac{x+2+x^2-2x+1}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2-x+1}{\left(x-2\right)\left(x+2\right)}\)
8 tháng 12 2021
a: \(A=\dfrac{x+2+x^2-2x+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{x^2}{x^2-4}\)
ĐK: \(2+x\ne0\Leftrightarrow x\ne-2\)
A nguyên dương \(\Leftrightarrow-\dfrac{4}{2+x}>0\)
Mà -4<0 nên để A>0 thì 2+x<0 <=> x<-2
Vậy khi x<-2 thì A nguyên dương.
Để `A in N**`
`=>A>0,A in N`
`A>0`
`=>-4/(x+2)>0`
`=>x+2<0`
`=>x<-2`
`A in N`
`=>-4 vdots x+2`
`=>x+2 in Ư(4)={+-1,+-2,+-4}`
`=>x in {-3,-1,0,-4,-6,2}`
Mà `x<-1`
`=>x in {-3,-,4,-6}`