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. \(B< 1\)

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

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 \(P=\dfrac{9}{2}\)

Bạn ghi lại đề đi bạn. Khó hiểu quá!

a) \(C=\left(\dfrac{x}{x^2-x-6}-\dfrac{x-1}{3x^2-4x-15}\right):\dfrac{x^4-2x^2+1}{3x^2+11x+10}\cdot\left(x^2-2x+1\right)\) (ĐK: \(x\ne-\dfrac{5}{3};x\ne3;x\ne-2;x\ne1\))

\(C=\left[\dfrac{x}{\left(x-3\right)\left(x+2\right)}-\dfrac{x-1}{\left(x-3\right)\left(3x+5\right)}\right]:\dfrac{\left(x^2-1\right)^2}{\left(3x+5\right)\left(x+2\right)}\cdot\left(x-1\right)^2\)

\(C=\left[\dfrac{x\left(3x+5\right)}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-3\right)\left(3x+5\right)\left(x+2\right)}\right]\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{3x^2+5x-x^2-2x+x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x^2-1\right)^2\left(x-1\right)^2}\)

\(C=\dfrac{2x^2+4x+2}{\left(3x+5\right)\left(x+2\right)\left(x-3\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2\left(x+1\right)^2}{\left(3x+5\right)\left(x-3\right)\left(x+2\right)}\cdot\dfrac{\left(3x+5\right)\left(x+2\right)}{\left(x+1\right)^2\left(x-1\right)^4}\)

\(C=\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}\)

b) Thay x = 2003 ta có: 

\(C=\dfrac{2}{\left(2003-1\right)^4\left(2003-3\right)}=\dfrac{2}{2002^4\cdot2000}=\dfrac{1}{2002^4\cdot1000}\)

c) \(C>0\) khi: 

\(\dfrac{2}{\left(x-1\right)^4\left(x-3\right)}>0\) mà: \(\left\{{}\begin{matrix}2>0\\\left(x-1\right)^4>0\end{matrix}\right.\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\) (đpcm) 

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

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

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

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

\(=\dfrac{x-9\sqrt{x}+6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

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