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Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
a) \(ĐKXĐ:x\ne\pm2\)
b)
\(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right).\dfrac{x+2}{2}\\ =\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right].\dfrac{x+2}{2}\\ =\left[\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{1\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right].\dfrac{x+2}{2}\\ =\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{2}\\ =\dfrac{-6}{\left(x-2\right)\left(x +2\right)}.\dfrac{x+2}{2}\\ =\dfrac{-3}{x-2}\)
c) Khi \(A=1\) ta có
\(1=\dfrac{-3}{x-2}\\ \Leftrightarrow x-2=\left(-3\right).1\\ \Leftrightarrow x-2=-3\\ \Leftrightarrow x=-3+2\\ \Leftrightarrow x=-1\)
Vậy \(A=1\Leftrightarrow x=-1\)
ta có
\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right).\frac{x+2}{2}\)
điều kiện xác định \(\hept{\begin{cases}x^2-4\ne0\\2-x\ne0\\x+2\ne0\end{cases}\Leftrightarrow x\ne\pm2}\)
b.\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right).\frac{x+2}{2}=\left(\frac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right)\frac{x+2}{2}\)
\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{2}=-\frac{3}{x-2}\)
c. khi \(x=1\Rightarrow A=-\frac{3}{x-2}=-\frac{3}{1-2}=3\)
\(a,A=\dfrac{x^2-2x+2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{x+6}\\ A=\dfrac{x^2-4}{x+6}\\ b,A>0\Leftrightarrow\dfrac{x^2-4}{x+6}>0\Leftrightarrow\dfrac{\left(x-2\right)\left(x+2\right)}{x+6}>0\\ TH_1:\left\{{}\begin{matrix}\left(x-2\right)\left(x+2\right)>0\\x+6>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x< -2\end{matrix}\right.\\x>-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>2\\-6< x< -2\end{matrix}\right.\\ TH_2:\left\{{}\begin{matrix}\left(x-2\right)\left(x+2\right)< 0\\x+6< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-2< x< 2\\x< -6\end{matrix}\right.\Leftrightarrow x< -6\)
Vậy xảy ra các TH: \(\left[{}\begin{matrix}x>2\\-6< x< -2\\x< -6\end{matrix}\right.\)
Bài 1 :
a, \(A=\frac{2x^2-4x+8}{x^3+8}=\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{2}{x+2}\)
b, Ta có : \(\left|x\right|=2\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
TH1 : Thay x = 2 vào biểu thức trên ta được :
\(\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}\)
TH2 : Thay x = -2 vào biểu thức trên ta được :
\(\frac{2}{-2+2}=\frac{2}{0}\)vô lí
c, ta có A = 2 hay \(\frac{2}{x+2}=2\)ĐK : \(x\ne-2\)
\(\Rightarrow2x+4=2\Leftrightarrow2x=-2\Leftrightarrow x=-1\)
Vậy với x = -1 thì A = 2
d, Ta có A < 0 hay \(\frac{2}{x+2}< 0\)
\(\Rightarrow x+2< 0\)do 2 > 0
\(\Leftrightarrow x< -2\)
Vậy với A < 0 thì x < -2
e, Để A nhận giá trị nguyên khi \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
| x + 2 | 1 | -1 | 2 | -2 |
| x | -1 | -3 | 0 | -4 |
2.
ĐKXĐ : \(x\ne\pm2\)
a. \(B=\frac{x^2-4x+4}{x^2-4}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)
b. | x - 1 | = 2 <=>\(\hept{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)<=>\(\hept{\begin{cases}x=3\\x=-1\end{cases}}\)
Với x = 3 thì \(B=\frac{3-2}{3+2}=\frac{1}{5}\)
Với x = - 1 thì \(B=\frac{-1-2}{-1+2}=-3\)
Vậy với | x - 1 | = 2 thì B đạt được 2 giá trị là B = 1/5 hoặc B = - 3
c. \(B=\frac{x-2}{x+2}=-1\)<=>\(-\left(x-2\right)=x+2\)
<=> \(-x+2=x+2\)<=>\(-x=x\)<=>\(x=0\)
d. \(B=\frac{x-2}{x+2}< 1\)<=>\(x-2< x+2\)luôn đúng \(\forall\)x\(\ne\pm2\)
e. \(B=\frac{x-2}{x+2}=\frac{x+2-4}{x+2}=1-\frac{4}{x+2}\)
Để B nguyên thì 4/x+2 nguyên => x + 2\(\in\){ - 4 ; - 2 ; - 1 ; 1 ; 2 ; 4 }
=> x \(\in\){ - 6 ; - 4 ; - 3 ; - 1 ; 0 ; 2 }
a: Khi x=3 thì \(A=\dfrac{3+2}{3-1}=\dfrac{5}{2}\)
b: \(B=\dfrac{x-1}{x}+\dfrac{2x+1}{x\left(x+1\right)}=\dfrac{x^2-1+2x+1}{x\left(x+1\right)}=\dfrac{x+2}{x+1}\)
\(P=A:B=\dfrac{x+2}{x-1}\cdot\dfrac{x+1}{x+2}=\dfrac{x+1}{x-1}\)
3: Để P>1/3 thì \(P-\dfrac{1}{3}>0\)
=>\(\Leftrightarrow3\left(x+1\right)-x+1>0\)
=>3x+3-x+1>0
=>2x+4>0
hay x>-2
a) \(A=\dfrac{x-2}{x+2}+\dfrac{x}{2-x}+\dfrac{8}{x^2-4}\left(dkxd:x\ne2;x\ne-2\right)\)
\(=\dfrac{x-2}{x+2}-\dfrac{x}{x-2}+\dfrac{8}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{8}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2-4x+4-x^2-2x+8}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-6x+12}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{-6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)\(=\dfrac{-6}{x+2}\)
Vậy với \(x\ne2;x\ne-2\) thì \(A=\dfrac{-6}{x+2}\).
b) Để \(A< 0\) thì: \(\dfrac{-6}{x+2}< 0\)
\(\Rightarrow x+2>0\) (vì \(-6< 0\))
\(\Leftrightarrow x>-2\)
Kết hợp với điều kiện xác định của x, ta được: \(x>-2;x\ne2\)
Vậy \(A< 0\) khi \(x>-2;x\ne2\).