Cho A = \(\dfrac{x}{x-1}+\dfrac{3-x}{x^2-2x-3}\)
Tìm x để A + \(\dfrac{1}{x^4-2x^2+1}\) > 0
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23 tháng 12 2021
\(A=\dfrac{x^2+x-2+x^2-x-2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x-3\right)}{2\left(x-2\right)\left(x+2\right)^2}=\dfrac{x-3}{x+2}\\ A\le0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2>0\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 3;x\ne0\left(ĐKXD\right)\)
29 tháng 1 2022
a: =>1/2x=7/2-2/3=21/6-4/6=17/6
=>x=17/3
b: =>2/3:x=-7-1/3=-22/3
=>x=2/3:(-22/3)=-1/11
c: =>1/3x+2/5x-2/5=0
=>11/15x=2/5
hay x=6/11
d: =>2x-3=0 hoặc 6-2x=0
=>x=3/2 hoặc x=3
19 tháng 12 2020
a) ĐKXĐ: \(x\notin\left\{0;-5\right\}\)
Ta có: \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x^2+10x}\)
\(=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+5x-x-5}{2\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)-\left(x+5\right)}{2\left(x+5\right)}\)
\(=\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}\)
\(=\dfrac{x-1}{2}\)
b) Để B=0 thì \(\dfrac{x-1}{2}=0\)
\(\Leftrightarrow x-1=0\)
hay x=1(nhận)
Vậy: Để B=0 thì x=1
Để \(B=\dfrac{1}{4}\) thì \(\dfrac{x-1}{2}=\dfrac{1}{4}\)
\(\Leftrightarrow4\left(x-1\right)=2\)
\(\Leftrightarrow4x-4=2\)
\(\Leftrightarrow4x=6\)
hay \(x=\dfrac{3}{2}\)(nhận)
Vậy: Để \(B=\dfrac{1}{4}\) thì \(x=\dfrac{3}{2}\)
c) Thay x=3 vào biểu thức \(B=\dfrac{x-1}{2}\), ta được:
\(B=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)
Vậy: Khi x=3 thì B=1
d) Để B<0 thì \(\dfrac{x-1}{2}< 0\)
\(\Leftrightarrow x-1< 0\)
\(\Leftrightarrow x< 1\)
Kết hợp ĐKXĐ, ta được:
\(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)
Vậy: Để B<0 thì \(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)
Để B>0 thì \(\dfrac{x-1}{2}>0\)
\(\Leftrightarrow x-1>0\)
hay x>1
Kết hợp ĐKXĐ, ta được: x>1
Vậy: Để B>0 thì x>1
24 tháng 6 2021
`đk:x ne +-3,x ne -2`
`B=(21/(x^2-9)-(x-4)/(3-x)-(x-1)/(3+x)):(1-1/(x+3))`
`=(21/(x^2-9)+(x-4)/(x-3)-(x-1)/(x+3)):((x+3-1)/(x+3))`
`=((21+x^2-x-12-x^2+4x-3)/((x-3)(x+3))):(x+2)/(x+3)`
`=(3x+6)/((x-3)(x+3))*(x+3)/(x+2)`
`=(3x+6)/((x-3)(x+2))`
`=3/(x-3)`
`b)|2x+1|=5`
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2(tm)\\x=-3(l)\end{array} \right.\)
`=>B=3/(2-3)=-3`
`c)B=-3/5`
`<=>3/(x-3)=3/(-5)`
`<=>x-3=-5`
`<=>x=-2(l)`
`d)B<0`
`<=>3/(x-3)<0`
Mà `3>0`
`=>x-3<0<=>x<3`
24 tháng 6 2021
a) đk: \(x\ne\pm3\)
\(B=\left[\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3}\right]:\left(\dfrac{x+3-1}{x+3}\right)\)
= \(\left[\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\dfrac{x+2}{x+3}\)
= \(\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}\)
= \(\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)
b) Để \(\left|2x+1\right|=5\)
<=> \(\left[{}\begin{matrix}2x+1=5< =>x=2\left(c\right)\\2x+1=-5< =>x=-3\left(l\right)\end{matrix}\right.\)
Thay x = 2, ta có;
B = \(\dfrac{3}{2-3}=-3\)
c) Để B = \(\dfrac{-3}{5}\)
<=> \(\dfrac{3}{x-3}=\dfrac{-3}{5}\)
<=> x - 3 = -5
<=> x = -2
d) Để B < 0
<=> \(\dfrac{3}{x-3}< 0\)
<=> x - 3 < 0
<=> x < 3
21 tháng 2 2021
a)\(x\in R\)
b)\(x\ne1\)
c) \(x\notin\left\{1;2\right\}\)
d) \(x\notin\left\{3;-3\right\}\)
e) \(x\ne1\)
f) \(x\notin\left\{2;3\right\}\)
24 tháng 6 2021
`a)P=(x/(x+2)-(x^3-8)/(x^3+8)*(x^2-2x+4)/(x^2-4)):4/(x+2)`
`đk:x ne 0,x ne -2`
`P=(x/(x+2)-((x-2)(x^2+2x+4))/((x+2)(x^2-2x+4))*(x^2-2x+4)/((x-2)(x+2)))*(x+2)/4`
`=(x/(x+2)-(x^2+2x+4)/(x+2)^2)*(x+2)/4`
`=(x^2+2x-x^2-2x-4)/(x+2)^2*(x+2)/4`
`=-4/(x+2)^2*(x+2)/4`
`=-1/(x+2)`
`b)P<0`
`<=>-1/(x+2)<0`
Vì `-1<0`
`<=>x+2>0`
`<=>x> -2`
`c)P=1/x+1(x ne 0)`
`<=>-1/(x+2)=1/x+1`
`<=>1/x+1+1/(x+2)=0``
`<=>x+2+x(x+2)+x=0`
`<=>x^2+4x+2=0`
`<=>` \(\left[ \begin{array}{l}x=\sqrt2-2\\x=-\sqrt2-2\end{array} \right.\)
`d)|2x-1|=3`
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2(l)\\x=-1(tm)\end{array} \right.\)
`x=-1=>P=-1/(-1+2)=-1`
`e)P=-1/(x+2)` thì nhỏ nhất cái gì nhỉ?
24 tháng 6 2021
a) đk: \(x\ne-2;2\)
\(P=\left[\dfrac{x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}.\dfrac{x^2-2x+4}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{4}{x+2}\)
= \(\left[\dfrac{x}{x+2}-\dfrac{x^2+2x+4}{\left(x+2\right)^2}\right].\dfrac{x+2}{4}\)
= \(\dfrac{x^2+2x-x^2-2x-4}{\left(x+2\right)^2}.\dfrac{x+2}{4}\) = \(\dfrac{-4}{4\left(x+2\right)}=\dfrac{-1}{x+2}\)
b) Để P < 0
<=> \(\dfrac{-1}{x+2}< 0\)
<=> x +2 > 0
<=> x > -2 ( x khác 2)
c) Để P= \(\dfrac{1}{x}+1\)
<=> \(\dfrac{-1}{x+2}=\dfrac{1}{x}+1\)
<=> \(\dfrac{1}{x}+\dfrac{1}{x+2}+1=0\)
<=> \(\dfrac{x+2+x+x\left(x+2\right)}{x\left(x+2\right)}=0\)
<=> x2 + 4x + 2 = 0
<=> (x+2)2 = 2
<=> \(\left[{}\begin{matrix}x=\sqrt{2}-2\left(c\right)\\x=-\sqrt{2}-2\left(c\right)\end{matrix}\right.\)
d) Để \(\left|2x-1\right|=3\)
<=> \(\left[{}\begin{matrix}2x-1=3< =>x=2\left(l\right)\\2x-1=-3< =>x=-1\left(c\right)\end{matrix}\right.\)
Thay x = -1, ta có:
P = \(\dfrac{-1}{-1+2}=-1\)