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. \(B< 1\)

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

1,

a) 3,2x +(-12)x +2,7=-4,9 b)-5,6x+2,9x-3,86=-9,8

=>3,2x+(-12)x=-4,9-2,7 =>-5,6x+2,9x=-9,8+3,86

=>3,2x+(-12)x=-7,6 =>-5,6x+2,9x=-5,94

=>x[3,2+(-12)]=-7,6 =>x[(-5,6)+2,9]=-5,94

=>2x=-7,6=>x=-7,6:2=-3,8 =>-2,7x=-5,94=>x=(-5,94):(-2,7)=2,2

Nhầm môn r bạn ơi

a) \(\Leftrightarrow\dfrac{3}{2}:x=\dfrac{1}{2}\\ \Leftrightarrow x=\dfrac{3}{2}:\dfrac{1}{2}\\ \Leftrightarrow x=3\)

b) \(\Leftrightarrow x=\dfrac{7}{9}-\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)

c) \(\Leftrightarrow x=\dfrac{8}{7}:\dfrac{6}{7}\\ \Leftrightarrow x=\dfrac{4}{3}\)

d) \(\Leftrightarrow x=\dfrac{9}{5}-\dfrac{3}{7}\\ \Leftrightarrow x=\dfrac{48}{35}\)

a) x = 3

b) x = \(\dfrac{1}{9}\)

c) x = \(\dfrac{4}{3}\)

d)\(\dfrac{48}{35}\)

1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\notin\left\{4;9\right\}\end{matrix}\right.\)

Ta có: \(A=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

\(1,A=\dfrac{2\sqrt{x}-9-x+9+2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ A=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\left(x\ge0;x\ne4;x\ne9\right)\\ 2,A< 1\Leftrightarrow\dfrac{\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}-3}< 0\\ \Leftrightarrow\dfrac{4}{\sqrt{x}-3}< 0\Leftrightarrow\sqrt{x}-3< 0\Leftrightarrow0\le x< 9\)

a: x=1-2/9=7/9

b: x=6/7-2/3=18/21-14/21=4/21

c: x=1/4+1/2=3/4

a) \(\dfrac{2}{9}+x=1\)

\(x=1-\dfrac{2}{9}\)

\(x=\dfrac{9}{9}-\dfrac{2}{9}=\dfrac{7}{9}\)

b) \(\dfrac{6}{7}-x=\dfrac{2}{3}\)

\(x=\dfrac{6}{7}-\dfrac{2}{3}\)

\(x=\dfrac{4}{21}\)

c) \(x-\dfrac{1}{2}=\dfrac{1}{4}\)

\(x=\dfrac{1}{4}+\dfrac{1}{2}\)

\(x=\dfrac{3}{4}\)

a)\(x=1-\dfrac{2}{9}=\dfrac{9}{9}-\dfrac{2}{9}=\dfrac{7}{9}\)

b)\(x=\dfrac{6}{7}-\dfrac{2}{3}=\dfrac{18}{21}-\dfrac{14}{21}=\dfrac{4}{21}\)

c)\(x=\dfrac{1}{4}+\dfrac{1}{2}=\dfrac{1}{4}+\dfrac{2}{4}=\dfrac{3}{4}\)

a) Bổ sung cho đầy đủ đề

b) (3x - 1)/4 = (2x - 5)/3

3(3x - 1) = 4(2x - 5)

9x - 3 = 8x - 20

9x - 8x = -20 + 3

x = -17

c) Điều kiện: x ≠ -1/3

3/(-2) = (x - 3)/(3x + 1)

3.(3x + 1) = -2(x - 3)

9x + 3 = -2x + 6

9x + 2x = 6 - 3

11x = 3

x = 3/11 (nhận)

Vậy x = 3/11

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 \(P=\dfrac{9}{2}\)

`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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