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a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

a: \(A=4x-3x^2+20-15x-9x^2-12x-4+\left(2x+1\right)^3-\left(8x^3-1\right)\)

\(=-12x^2-23x+16+8x^3+12x^2+6x+1-8x^3+1\)

\(=-17x+18\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

a) đk: x khác 1; \(\dfrac{3}{2}\)

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

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

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

b) Có \(\left|3x-2\right|+1=5\)

<=> \(\left|3x-2\right|=4\)

<=> \(\left[{}\begin{matrix}3x-2=4< =>x=2\left(Tm\right)\\3x-2=-4< =>x=\dfrac{-2}{3}\left(Tm\right)\end{matrix}\right.\)

TH1: Thay x = 2 vào P, ta có:

P = \(\dfrac{-1}{2.2-3}=-1\)

TH2: Thay x = \(\dfrac{-2}{3}\)vào P, ta có:

P = \(\dfrac{-1}{2.\dfrac{-2}{3}-3}=\dfrac{3}{13}\)

c) Để P > 0

<=> \(\dfrac{-1}{2x-3}>0\)

<=> 2x - 3 <0

<=> x < \(\dfrac{3}{2}\) ( x khác 1)

d) P = \(\dfrac{1}{6-x^2}\)

<=> \(\dfrac{-1}{2x-3}=\dfrac{1}{6-x^2}\)

<=> \(\dfrac{-1}{2x-3}=\dfrac{-1}{x^2-6}\)

<=> 2x - 3 = x² - 6

<=> x² - 2x - 3 = 0

<=> (x-3)(x+1) = 0

<=> \(\left[{}\begin{matrix}x=-1\left(Tm\right)\\x=3\left(Tm\right)\end{matrix}\right.\)

\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)

\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)

a: \(P=\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)

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

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