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\(x^2-x+1=x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\)
\(-x^2+4x-5=-\left(x^2-2.x.2+2^2\right)-1=-\left(x-2\right)^2-1< 0\forall x\)
\(a\left(2a-3\right)-2a\left(a+1\right)=a\left(2a-3-2a-2\right)=-5a⋮5\forall a\inℤ\)
a,ĐK: \(\hept{\begin{cases}x\ne0\\x\ne\pm3\end{cases}}\)
b, \(A=\left(\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}:\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}=\frac{-3}{x-3}\)
c, Với x = 4 thỏa mãn ĐKXĐ thì
\(A=\frac{-3}{4-3}=-3\)
d, \(A\in Z\Rightarrow-3⋮\left(x-3\right)\)
\(\Rightarrow x-3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\Rightarrow x\in\left\{0;2;4;6\right\}\)
Mà \(x\ne0\Rightarrow x\in\left\{2;4;6\right\}\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ne-2\\x\ne3\\x\ne2\end{cases}}\)
\(A=\left(1-\frac{4}{x+2}\right):\left(1+\frac{1}{x-3}\right)\)
\(\Leftrightarrow A=\frac{x-2}{x+2}:\frac{x-2}{x-3}\)
\(\Leftrightarrow A=\frac{x-3}{x+2}\)
b) Để A nguyên
\(\Leftrightarrow x-3⋮x+2\)
\(\Leftrightarrow x+2-5⋮x+2\)
\(\Leftrightarrow5⋮x+2\)
\(\Leftrightarrow x+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Leftrightarrow x\in\left\{-3;-1;-7;3\right\}\)
Vậy để A nguyên \(\Leftrightarrow x\in\left\{-3;1;-7;3\right\}\)
c) Để A > 0
\(\Leftrightarrow\frac{x-3}{x+2}>0\)
\(\Leftrightarrow1-\frac{5}{x+2}>0\)
\(\Leftrightarrow\frac{5}{x+2}< 0\)
\(\Leftrightarrow x+2< 0\)(vì 5 > 0)
\(\Leftrightarrow x< -2\)
Vậy để A > 0 \(\Leftrightarrow x< -2\)
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\) ( Chữa đề nhé.)
a) \(ĐKXĐ:x\ne-3;x\ne2\)
\(\text{Với }x\ne-3;x\ne2,\text{ ta có: }A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\\ =\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\\ =\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{\left(x+3\right)\left(x-4\right)}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x-4}{x-2}\\ \text{Vậy }A=\dfrac{x-4}{x-2}\text{ với }x\ne-3;x\ne2\)
b) Lập bảng xét dấu:
x x-4 x-2 x-4 2 4 0 0 x-2 _ _ + _ + + 0 + _ +
\(\Rightarrow\left[{}\begin{matrix}x< 2\\x>4\end{matrix}\right.\)
Vậy để \(A>0\) thì \(x< 2\) hoặc \(x>4\)
c) \(\text{Với }x\ne-3;x\ne2\)
\(\text{Ta có : }A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}\\ =\dfrac{x-2}{x-2}-\dfrac{2}{x-2}=1-\dfrac{2}{x-2}\)
\(\Rightarrow\) Để A nhận giá trị nguyên
thì \(\Rightarrow\dfrac{2}{x-2}\in Z\)
\(\Rightarrow2⋮x-2\\ \Rightarrow x-2\inƯ_{\left(2\right)}\)
Mà \(Ư_{\left(2\right)}=\left\{\pm1;\pm2\right\}\)
Lập bảng giá trị:
| \(x-2\) | \(-2\) | \(-1\) | \(1\) | \(2\) |
| \(x\) | \(0\left(TM\right)\) | \(1\left(TM\right)\) | \(3\left(TM\right)\) | \(4\left(TM\right)\) |
\(\Rightarrow x\in\left\{-2;-1;1;2\right\}\)
Vậy với \(x\in\left\{-2;-1;1;2\right\}\)
thì \(A\in Z\)
Câu 2:
a) \(ĐKXĐ:x\ne\dfrac{3}{2};x\ne1\)
\(\text{Với }x\ne\dfrac{3}{2};x\ne1,\text{ ta có : }B=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\\ =\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right]:\left(\dfrac{3\left(1-x\right)}{1-x}+\dfrac{2}{1-x}\right)\\ =\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3-3x+2}{\left(1-x\right)}\\ =\dfrac{\left(-3x+5\right)\cdot\left(1-x\right)}{\left(2x-3\right)\left(x-1\right)\cdot\left(-3x+5\right)}\\ =-\dfrac{1}{2x-3}\)
Vậy \(B=-\dfrac{1}{2x-3}\) với \(x\ne\dfrac{3}{2};x\ne1\)
b) \(\text{Với }x\ne\dfrac{3}{2};x\ne1\)
Để \(B=\dfrac{1}{x^2}\)
\(\text{thì }\Rightarrow\dfrac{-1}{2x-3}=\dfrac{1}{x^2}\\ \Rightarrow2x-3=-x^2\\ \Leftrightarrow2x-3+x^2=0\\ \Leftrightarrow x^2-3x+x-3=0\\ \Leftrightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\left(TM\right)\)
Vậy với \(x=-1;x=3\) thì \(B=\dfrac{1}{x^2}\)
a) A=(\(\dfrac{x+2}{x+2}\)-\(\dfrac{4}{x+2}\)):(\(\dfrac{x-3}{x-3}\)+\(\dfrac{1}{x-3}\))(ĐKXĐ:x≠-2,x≠3)
⇔A=\(\dfrac{x-2}{x+2}\):\(\dfrac{x-2}{x+3}\)
⇔A=\(\dfrac{x-2}{x+2}\).\(\dfrac{x+3}{x-2}\)
⇔A=\(\dfrac{x+3}{x+2}\)
b)Để A>0
⇔\(\dfrac{x+3}{x+2}\)>0
⇔x+3>0 ,x+2>0 hoặc x+3<0,x+2<0(- với - thành +)
⇔x>-3,x>-2 hoặc x<-3,x<-2
⇔-2<x<-3
Vậy ......