a: Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)

\(=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}\)

\(=\dfrac{3x}{x-3}\)

b: Ta có P=AB

nên \(P=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}=\dfrac{3x}{x+1}\)

Để \(P=\dfrac{9}{2}\) thì 9x+9=6x

\(\Leftrightarrow3x=-9\)

hay x=-3(loại)

a) \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\\ \Rightarrow A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-3+11x}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

    \(\Rightarrow A=\dfrac{3x}{x-3}\)

a. ĐKXĐ: \(x\ne\pm3\)

\(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)

\(=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{3-11x}{x^2-9}\)

\(=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-\left(3-11x\right)}{x^2-9}\)

\(=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{x^2-9}\)

\(=\dfrac{3x^2+9x}{x^2-9}=\dfrac{3x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x-3}\)

b. \(P=A.B\)

\(\Rightarrow P=\dfrac{3x}{x-3}.\dfrac{x-3}{x+1}=\dfrac{3x}{x+1}\) 

Ta có \(P=\dfrac{9}{2}\) 

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\dfrac{3x}{x+1}=\dfrac{9}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\6x=9x+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\-3x=9\end{matrix}\right.\) \(\Leftrightarrow x=-3\)

c. \(B< 1\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\dfrac{x-3}{x-1}< 1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\dfrac{x-3}{x-1}-1< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\dfrac{2}{1-x}< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\1-x< 0\end{matrix}\right.\) \(\Leftrightarrow x>1\)

Làm lại nha cái này đúng, kia sai nha=)

b)

Với \(\left\{{}\begin{matrix}x\ne3\\x\ne2\end{matrix}\right.\)

\(P=A-B=(\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)})+\dfrac{2x-1}{x-3}\\ =\left(\dfrac{2x-9-x^2-9}{\left(x-3\right)\left(x-2\right)}\right)+\dfrac{\left(2x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}+\dfrac{2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2+2x^2-4x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-3x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x^2-2x-x+2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(x-2\right)-\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{x-1}{x-3}\)

c)

Để P\(\ge1\) thì:

\(\dfrac{x-1}{x-3}\ge1\\ \Leftrightarrow x-3-x+1-1\ge0\\ \Leftrightarrow-3\ge0\left(vô.lý\right)\)

Vậy không tồn tại giá trị x để \(P\ge1\)

`HaNa☘D`

b)

\(P=A-B=\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(x+3\right)\left(x-3\right)}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9}{\left(x-3\right)\left(x-2\right)}-\dfrac{x^2-9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-9-x^2+9}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{2x-x^2}{\left(x-3\right)\left(x-2\right)}\\ =\dfrac{x\left(2-x\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}\\ =-\dfrac{x}{x-3}\)

c)

Để \(P\le1\) thì:

\(-\dfrac{x}{x-3}\le1\)

\(\Leftrightarrow\dfrac{x}{x-3}\ge1\\ \Leftrightarrow x-3-x\ge1\\ \Leftrightarrow-3\ge1\left(vô.lý\right)\)

Vậy không tồn tại giá trị x để \(P\le1\)

`HaNa♬D`

a: \(A=\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x^2+9x}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x}{x+3}\)

\(ĐK:x\ne\pm3\\ a,A=\dfrac{5x-15+2x+6-3x^2+2x+9}{\left(x+3\right)\left(x-3\right)}\\ A=\dfrac{-3x^2+9x-1}{\left(x-3\right)\left(x+3\right)}\\ b,\left|x-2\right|=1\Leftrightarrow x=1\left(x\ne3\right)\\ \Leftrightarrow A=\dfrac{-3+9-1}{\left(-2\right)\cdot4}=\dfrac{5}{-8}\)

\(a,A=\dfrac{x^2-6x+9-x^2+9}{x\left(x-3\right)}\cdot\dfrac{x}{2\left(x-1\right)}\\ A=\dfrac{-6x+18}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-6\left(x-3\right)}{2\left(x-3\right)\left(x-1\right)}=\dfrac{-3}{x-1}\\ b,A\in Z\Leftrightarrow x-1\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\)

a) Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)

\(=\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{11x-3}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{2x^2-6x+x^2+4x+3+11x-3}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x}{x-3}\)

b)

ĐKXĐ: \(x\notin\left\{3;-3;-1\right\}\)

Ta có: P=AB

\(=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}\)

\(=\dfrac{3x}{x+1}\)

Để \(P=\dfrac{9}{2}\) thì \(\dfrac{3x}{x+1}=\dfrac{9}{2}\)

\(\Leftrightarrow9\left(x+1\right)=6x\)

\(\Leftrightarrow9x-6x=-9\)

\(\Leftrightarrow3x=-9\)

hay x=-3(loại)

Vậy: Không có giá trị nào của x để \(P=\dfrac{9}{2}\)

a: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}=\dfrac{x+1}{x-3}\)

a: \(P=\dfrac{2x-9-x^2+9+2x^2-4x+x-2}{\left(x-2\right)\left(x-3\right)}\)

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

`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

a)\(B=\left(\dfrac{21}{x^2-9}-\dfrac{x-4}{3-x}-\dfrac{x-1}{3+x}\right):\left(1-\dfrac{1}{x+3}\right)\\ =\left(\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x+2}{x+3}\)

\(=\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)

b)\(\left|2x+1\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\left(loại\right)\end{matrix}\right.\)

với x=2 gt của B là

\(B=\dfrac{3}{2-3}=-3\)

c)\(B=\dfrac{3}{x-3}=-\dfrac{3}{5}\Leftrightarrow x-3=-5\Leftrightarrow x=-2\)

d) \(B=\dfrac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)

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