\(=\dfrac{\left(x+3\right)\left(x-1\right)}{\left(x+5\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x-7\right)}{\left(x+3\right)\left(x-2\right)}:\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x+2\right)\left(x+5\right)}\)

\(=\dfrac{\left(x-1\right)\left(x-7\right)}{\left(x+5\right)\left(x-2\right)}\cdot\dfrac{\left(x+2\right)\left(x+5\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x-7\right)\left(x+2\right)}{\left(x-2\right)\left(x-3\right)}\)

Chép đề đúng chưa bạn? 2 phân số đầu có ngoặc không vậy?

Nguyễn Công Tỉnh đúng r bạn, mình sửa lại r

\(ĐKXĐ:x\ne\pm3\)

\(P=\left(\frac{x^2-3x}{x^3+3x^2+9x+27}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)

\(\Leftrightarrow P=\left(\frac{x^2-3x}{\left(x+3\right)\left(x^2+9\right)}+\frac{3}{x^2+9}\right):\left(\frac{1}{x-3}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right)\)

\(\Leftrightarrow P=\frac{\left(x^2-3x\right)+3\left(x+3\right)}{\left(x+3\right)\left(x^2+9\right)}:\frac{x^2+9-6x}{\left(x-3\right)\left(x^2+9\right)}\)

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

\(\Leftrightarrow P=\frac{1}{x+3}:\frac{x-3}{x^2+9}\)

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

b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)

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

\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)

a) (2x - 1)(3x + 5) - 2(-4x + 1)² = 6x² + 10x - 3x - 5 - 2(16x² - 8x + 1) = 6x² - 3x - 5 - 32x² + 16x - 2 = -26x² + 13x - 7

b) \(\frac{x^2-16}{4x-x^2}=\frac{\left(x-4\right)\left(x+4\right)}{-x\left(x-4\right)}=-\frac{x+4}{x}\)

c) \(\frac{2x-9}{x^2-5x+6}+\frac{2x+1}{x-3}+\frac{x+3}{2-x}\)

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

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

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

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

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

d) (x - 1)³ - (x + 1)³ + 6(x + 1)(x - 1)

= (x - 1 - x - 1)[(x - 1)² + (x - 1)(x + 1) + (x + 1)²] + 6(x² - 1)

= -2(x² - 2x + 1  + x² - 1 + x² + 2x + 1) + 6x² - 6

= -2(3x² + 1) + 6x² - 6

= -6x² - 2 + 6x²  - 6

= -8

e) (2x + 7)² - (4x + 14)(2x - 8) + (8 - 2x)²

= (2x + 7)² - 2(2x + 7)(2x - 8) + (2x - 8)²

= (2x + 7 - 2x + 8)²

= 15² = 225

$\textbf{a)}$

\[\left(x+3-\frac1{x+3}\right)\cdot\frac{x+3}{x+4}\]

$\text{ĐKXĐ: }x\ne-3,\,-4.$

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

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

$=x+2.$

$\textbf{b)}$

$\left(2x-4-\frac{x-12}{3x+4}\right)\cdot\left(3x-2-\frac{10}{2x+1}\right)$

$\text{ĐKXĐ: }x\ne-\dfrac43,\,-\dfrac12.$

$=\dfrac{(2x-4)(3x+4)-(x-12)}{3x+4}\cdot\dfrac{(3x-2)(2x+1)-10}{2x+1}$

$=\dfrac{6x^2-5x-4}{3x+4}\cdot\dfrac{6x^2-x-12}{2x+1}.$