\(P=1+\frac{x+3}{x^2+5x+6}:\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)

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

\(P=1+\frac{1}{x+2}:\left(\frac{4x^2.2}{4x^2\left(x-2\right)}-\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{1}{x+2}\right)\)

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

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

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

\(=\frac{x+4}{6}.P=0\Leftrightarrow x=-4\)

\(P>0\Leftrightarrow x>-4\)

sai lớp :>>>

\(P=1+\dfrac{x+3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}+\dfrac{3x}{12-3x^2}-\dfrac{1}{2}\right)\)\(=1+\dfrac{x+3}{x^2+3x+2x+6}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}+\dfrac{3x}{3\left(4-x^2\right)}-\dfrac{1}{x+2}\right)\)\(=1+\dfrac{x+3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{2}{x-2}-\dfrac{3x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\)\(=1+\dfrac{1}{x+2}:\left(\dfrac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3x}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\)\(=1+\dfrac{1}{x+2}:\left(\dfrac{2x+4-3x-x+2}{\left(x-2\right)\left(x+2\right)}\right)\)

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

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

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

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

Lời giải:

ĐKXĐ: $x\neq \pm 2; x\neq -3$

Ta có:

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

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

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

$=1+\frac{1}{x+2}:\frac{-2(x^2-2x-2)}{x(x-2)(x+2)}$
$=1-\frac{x(x-2)}{2x^2-4x-4}=\frac{x^2-2x-4}{2x^2-4x-4}$

a: Ta có: \(\left(8x^3-4x^2\right):4x-\left(4x^2-5x\right):2x+\left(2x\right)^2\)

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

\(=6x^2-3x+\dfrac{5}{2}\)

b: Ta có: \(\left(3x^3-x^2y\right):x^2-\left(xy^2+x^2y\right):xy+2x\left(x-1\right)\)

\(=3x-y-y-x+2x^2-2x\)

\(=2x^2-2y\)

Để \(\sqrt{x}\) xác định

 \(\Leftrightarrow x\ge0\)

\(\Leftrightarrow-7x\le0\)

\(\Rightarrow\sqrt{-7x}\)không tồn tại 

\(\Leftrightarrow\frac{8x}{4x\sqrt{x-8x}}\)không tồn tại

=> A không tồn tại