$P=\dfrac{2}{x+3}+\dfrac{1}{x-3}+\dfrac{9-x}{9-x^2}$

$=\dfrac{2(x-3)+(x+3)}{(x+3)(x-3)}+\dfrac{9-x}{(3-x)(3+x)}$

$=\dfrac{3x-3}{x^2-9}+\dfrac{9-x}{9-x^2}$

$=\dfrac{3x-3}{x^2-9}-\dfrac{9-x}{x^2-9}$

$=\dfrac{3x-3-9+x}{x^2-9}$

$=\dfrac{4x-12}{x^2-9}$

$=\dfrac{4(x-3)}{(x-3)(x+3)}$

$=\dfrac{4}{x+3}$

Vậy $P=\dfrac{4}{x+3}$.

$(2x^4-3x^3-3x^2+6x-1):(x^2-2)$

$=2x^2-3x+1+\dfrac{4x+1}{x^2-2}$

Vậy thương là $2x^2-3x+1$, dư $4x+1$.

$(15x^4y^6-12x^3y^4-18x^2y^3):(-6x^2y^2)$

$=\dfrac{15x^4y^6}{-6x^2y^2}+\dfrac{-12x^3y^4}{-6x^2y^2}+\dfrac{-18x^2y^3}{-6x^2y^2}$

$=-\dfrac52x^2y^4+2xy^2+3y$

Vậy kết quả là $-\dfrac52x^2y^4+2xy^2+3y$.

1: \(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)

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

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

\(=\frac{x^2-6x+9-x^2-6x-9+x\left(x^2-9\right)}{\left(x-3\right)^2\cdot\left(x+3\right)^2}=\frac{x^3-9x-12x}{\left(x-3\right)^2\cdot\left(x+3\right)^2}\)

\(=\frac{x^3-21x}{\left(x-3\right)^2\cdot\left(x+3\right)^2}\)

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

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

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

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

3: \(\frac{x-3}{x+1}-\frac{x+2}{x-1}+\frac{8x}{x^2-1}\)

\(=\frac{\left(x-3\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)+8x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\frac{x+1}{\left(x-1\right)\left(x+1\right)}=\frac{1}{x-1}\)

1: \(\frac{1}{x^2+6x+9}+\frac{1}{6x-x^2-9}+\frac{x}{x^2-9}\)

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

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

\(=\frac{x^2-6x+9-x^2-6x-9+x\left(x^2-9\right)}{\left(x-3\right)^2\cdot\left(x+3\right)^2}=\frac{x^3-9x-12x}{\left(x-3\right)^2\cdot\left(x+3\right)^2}\)

\(=\frac{x^3-21x}{\left(x-3\right)^2\cdot\left(x+3\right)^2}\)

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

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

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

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

3: \(\frac{x-3}{x+1}-\frac{x+2}{x-1}+\frac{8x}{x^2-1}\)

\(=\frac{\left(x-3\right)\left(x-1\right)-\left(x+2\right)\left(x+1\right)+8x}{\left(x-1\right)\left(x+1\right)}\)

\(=\frac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\frac{x+1}{\left(x-1\right)\left(x+1\right)}=\frac{1}{x-1}\)

câu d

\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)

$\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}$

ĐKXĐ: $x\ne3,\ x\ne-3$

$\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}$

$=\dfrac{3(x+3)+6x+x(x-3)}{(x-3)(x+3)}$

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

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

$=\dfrac{(x+3)^2}{(x-3)(x+3)}$

$=\dfrac{x+3}{x-3}$

a: Sửa đề: \(M=\frac{x}{x+3}+\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{x^2-9}\)

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

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

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

b: \(A=\left(\frac{3x}{1-3x}-\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}\)

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

\(=\frac{-9x^2-3x-6x^2+2x}{3x+1}\cdot\frac{3x-1}{2x\left(3x+5\right)}=\frac{-15x^2-x}{3x+1}\cdot\frac{3x-1}{2x\left(3x+5\right)}\)

\(=\frac{-x\left(15x+1\right)}{3x+1}\cdot\frac{3x-1}{2x\left(3x+5\right)}=\frac{-\left(15x+1\right)\left(3x-1\right)}{2\left(3x+1\right)\left(3x+5\right)}\)

Phân thức `(7x^2 - x)/(x^2 - 9)` được xác đinh khi

`x^2-9≠0`

`(x-3).(x+3)≠0`

`=> x≠3; x ≠ -3`

`=> C`

a) Ta có: \(\dfrac{x}{x-3}-\dfrac{6}{x}-\dfrac{9}{x^2-3x}\)

\(=\dfrac{x^2}{x\left(x-3\right)}-\dfrac{6\left(x-3\right)}{x\left(x-3\right)}-\dfrac{9}{x\left(x-3\right)}\)

\(=\dfrac{x^2-6x+18-9}{x\left(x-3\right)}\)

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

b) Ta có: \(\dfrac{7}{x}-\dfrac{x}{x+6}+\dfrac{36}{x^2+6x}\)

\(=\dfrac{7\left(x+6\right)-x^2+36}{x\left(x+6\right)}\)

\(=\dfrac{7x+42-x^2+36}{x\left(x+6\right)}\)

\(=\dfrac{-\left(x^2-7x-78\right)}{x\left(x+6\right)}\)

\(=\dfrac{-\left(x^2-13x+6x-78\right)}{x\left(x+6\right)}\)

\(=\dfrac{-\left[x\left(x-13\right)+6\left(x-13\right)\right]}{x\left(x+6\right)}\)

\(=\dfrac{13-x}{x}\)

c) Ta có: \(\dfrac{6}{x-3}-\dfrac{2x-6}{x^2-9}-\dfrac{4}{x+3}\)

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

\(=\dfrac{6x+18-2x+6-4x+12}{\left(x-3\right)\left(x+3\right)}=\dfrac{36}{\left(x-3\right)\left(x+3\right)}\)

\(a,VP=\dfrac{x\left(x+3\right)}{x\left(2x-5\right)}=\dfrac{x+3}{2x-5}=VT\\ b,VP=\dfrac{\left(3-x\right)^2}{\left(3-x\right)\left(3+x\right)}=\dfrac{3-x}{x+3}=VT\\ c,VP=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{\left(3-x\right)\left(x^2-4x+16\right)}=\dfrac{x+4}{3-x}=VP\left(bạn.sửa.lại.đề.đi\right)\\ d,VT=\dfrac{x^3-2x^2+8x^2-16x+15x-30}{x^3-5x^2+8x^2-40x+15x-75}\\ =\dfrac{\left(x-2\right)\left(x^2+8x+15\right)}{\left(x-5\right)\left(x^2+8x+15\right)}=\dfrac{x-2}{x-5}=VP\)

a) Ta có: \(B=\left(\dfrac{x}{3x-9}+\dfrac{2x-3}{3x-x^2}\right)\cdot\dfrac{3x^2-9x}{x^2+6x+9}\)

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

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

\(=\dfrac{x^2-6x+9}{3x\left(x-3\right)}\cdot\dfrac{3x\left(x-3\right)}{\left(x+3\right)^2}\)

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

b) Ta có: \(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\dfrac{1}{x+2}\)

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

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

\(=\left(\dfrac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{1}{x+2}\)

\(=\dfrac{-6}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{1}\)

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