\(A=\dfrac{x^2+x-2+x^2-x-2-4}{x\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}=\dfrac{2\left(x-2\right)\left(x+2\right)\left(x-3\right)}{2\left(x-2\right)\left(x+2\right)^2}=\dfrac{x-3}{x+2}\\ A\le0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3\ge0\\x+2< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3\le0\\x+2>0\end{matrix}\right.\end{matrix}\right.\Rightarrow-2< x< 3;x\ne0\left(ĐKXD\right)\)

Bài 2:

a: a>b

=>2a>2b

=>2a+1>2b+1

mà 2b+1>2b-3

nên 2a+1>2b-3

b: 3x-1<=x+2

=>3x-x<=1+2

=>2x<=3

=>\(x\le\frac32\)

c: 2x+3>0

=>2x>-3

=>\(x>-\frac32\)

3x+1<x-4

=>3x-x<-4-1

=>2x<-5

=>\(x<-\frac52\)

2(x+1)+3>=3(5-x)

=>2x+2+3>=15-3x

=>2x+5>=15-3x

=>5x>=10

=>x>=2

\(\frac{x}{3}-\frac{x+1}{5}>1\)

=>\(\frac{5x-3\left(x+1\right)}{15}>1\)

=>5x-3(x+1)>15

=>5x-3x-3>15

=>2x>18

=>x>9

Bài 3:

Gọi độ dài quãng đường AB là x(km)

(Điều kiện: x>0)

Thời gian ô tô đi từ A đến B là \(\frac{x}{50}\) (giờ)

Thời gian ô tô đi từ B về A là \(\frac{x}{60}\) (giờ)

Tổng thời gian cả đi và về là 6h30p-1h=5h30p=5,5 giờ nên ta có:

\(\frac{x}{50}+\frac{x}{60}=5,5\)

=>\(\frac{6x}{300}+\frac{5x}{300}=5,5\)

=>\(x\cdot\frac{11}{300}=\frac{11}{2}\)

=>x=150(nhận)

Vậy: Độ dài quãng đường AB là 150km

Bài 1:

a: 2x-1=0

=>2x=1

=>\(x=\frac12\)

b: 3x-2=5+x

=>3x-x=2+5

=>2x=7

=>\(x=\frac72\)

c: 2(x-3)-4=3(1+x)-5x

=>2x-6-4=3+3x-5x

=>2x-10=3-2x

=>4x=13

=>\(x=\frac{13}{4}\)

d: \(\frac{x+1}{2}-\frac{2x}{3}=1\)

=>\(\frac{3\left(x+1\right)-4x}{6}=1\)

=>3(x+1)-4x=6

=>3x+3-4x=6

=>3-x=6

=>x=3-6=-3

e: x(x-2)+3(x-2)=0

=>(x-2)(x+3)=0

=>\(\left[\begin{array}{l}x-2=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-3\end{array}\right.\)

f: ĐKXĐ: x<>0; x<>1

\(\frac{x+1}{x-1}+\frac{3}{x}=\frac{x^2+2}{x^2-x}\)

=>\(\frac{x\left(x+1\right)+3\left(x-1\right)}{x\left(x-1\right)}=\frac{x^2+2}{x\left(x-1\right)}\)

=>\(x^2+x+3x-3=x^2+2\)

=>4x=5

=>x=5/4(nhận)

Chia nhỏ ra

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

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

`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`

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

ĐKXĐ: x>0; x<>1/2; x<>-1/2; x<>1/4

a: \(\frac{8x\cdot\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\cdot\sqrt{x}+1}{2x+\sqrt{x}}\)

\(=\frac{\left(2\sqrt{x}-1\right)\left(4x+2\sqrt{x}+1\right)}{\sqrt{x}\left(2\sqrt{x}-1\right)}-\frac{\left(2\sqrt{x}+1\right)\left(4x-2\sqrt{x}+1\right)}{\sqrt{x}\left(2\sqrt{x}+1\right)}\)

\(=\frac{4x+2\sqrt{x}+1}{\sqrt{x}}-\frac{4x-2\sqrt{x}+1}{\sqrt{x}}=\frac{4\sqrt{x}}{\sqrt{x}}=4\)

\(B=\left(\frac{8x\cdot\sqrt{x}-1}{2x-\sqrt{x}}-\frac{8x\cdot\sqrt{x}+1}{2x+\sqrt{x}}\right):\frac{2x+1}{2x-1}\)

\(=\frac{4\left(2x-1\right)}{2x+1}=\frac{8x-4}{2x+1}\)

b: A/B=1

=>A=B

=>\(\frac{x-2}{\sqrt{x}+2}=\frac{8x-4}{2x+1}\)

=>\(\left(8x-4\right)\left(\sqrt{x}+2\right)=\left(x-2\right)\left(2x+1\right)\)

=>\(8x\sqrt{x}+16x-4\sqrt{x}-8=2x^2+x-4x-2\)

=>\(2x^2-3x-2-8x\sqrt{x}-16x+4\sqrt{x}+8=0\)

=>\(2x^2-8x\cdot\sqrt{x}-19x+4\sqrt{x}+6=0\) (1)

Đặt \(t = \sqrt{x}\) (t>0; \(t<>\frac{1}{\sqrt2}\) )

(1) sẽ trở thành: \(2t^4-8t^3-19t^2+4t+6=0\)

=>t≃0,65 hoặc t≃5,37

t≃0,65

=>\(x=t^2\) ≃0,42(nhận)

t≃5,37

=>\(x=t^2\) ≃28,84(nhận)

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) \(x\notin\left\{2;3\right\}\)

a) 

b) 

c) 

d) 

e) 

f) 

`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?

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 = \(\dfrac{-1}{-1+2}=-1\)

ĐKXĐ: x<>2; x<>-2

a: \(A=\frac{1}{x-2}+\frac{x^2-2x}{x^2-4}+\frac{1}{x+2}\)

\(=\frac{x+2+x^2-2x+x-2}{\left(x-2\right)\left(x+2\right)}=\frac{x^2}{x^2-4}\)

\(2x^2+x=0\)

=>x(2x+1)=0

=>x=0(nhận) hoặc x=-1/2(nhận)

Thay x=0 vào A, ta được:

\(A=\frac{0^2}{0^2-4}=\frac{0}{-4}=0\)

Thay x=-1/2 vào A, ta được:

\(A=\left(-\frac12\right)^2:\left\lbrack\left(-\frac12\right)^2-4\right\rbrack=\frac14:\left(\frac14-4\right)=\frac14:\frac{-15}{4}=-\frac{1}{15}\)

b: \(A=-\frac13\)

=>\(\frac{x^2}{x^2-4}=-\frac13\)

=>\(-3x^2=x^2-4\)

=>\(-4x^2=-4\)

=>\(x^2=1\)

=>x=1(nhận) hoặc x=-1(nhận)

c: Để A nguyên thì \(x^2\) ⋮\(x^2-4\)

=>\(x^2-4+4\) ⋮\(x^2-4\)

=>4⋮\(x^2-4\)

=>\(x^2-4\in\left\lbrace1;-1;2;-2;4;-4\right\rbrace\)

=>\(x^2\in\left\lbrace5;3;6;2;8;0\right\rbrace\)

=>\(x^2=0\)

=>x=0