a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)

b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)

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

= -2,164432983x10^-13 =)?

\(=x^6-6x^4+12x^2-8-x^3+x+6x^2-18x\\ =x^6-6x^4-x^3+18x^2-17x-8\)

Q = \(\frac{\left(x+2\right)^2}{x}\cdot\left(1-\frac{x^2}{x+2}\right)-\frac{x^2+6x+4}{x}\)

Q = \(\frac{\left(x+2\right)^2}{x}\cdot\frac{x+2-x^2}{x+2}-\frac{x^2+6x+4}{x}\)

Q = \(\frac{\left(x+2\right)\left(x+2-x^2\right)}{x}-\frac{x^2+6x+4}{x}\)

Q = \(\frac{x^2+2x-x^3+2x+4-2x^2-x^2-6x-4}{x}\)

Q = \(\frac{-x^3-2x^2-2x}{x}\)

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

Bạn ơi mik ra \(\dfrac{x^3+45x-54}{12\left(x-3\right)\left(x+3\right)}\) có đúng không bạn?

\(=\left(x-\dfrac{1}{3}\right)\left(\dfrac{4}{3}x+\dfrac{1}{9}-x+\dfrac{1}{3}\right)\\ =\left(x-\dfrac{1}{3}\right)\left(\dfrac{1}{3}x+\dfrac{4}{9}\right)\\ =\dfrac{1}{3}x^2+\dfrac{4}{9}x-\dfrac{1}{9}x-\dfrac{4}{27}\\ =\dfrac{1}{3}x^2+\dfrac{1}{3}x-\dfrac{4}{27}\)

\(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-1\right)\left(x+1\right).\)

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

\(=x^3-3x^2+3x-1-x^3+8+3x^2-3\)

\(=3x+4\)

\(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)^3-\left(x^3+8\right)+3\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1\right)^3-x^3-8+3\left(x-1\right)\left(x+1\right)\)

\(=\left(x-1-x\right)+3x\left(x-1\right)\left(x-1-x\right)-8+3\left(x-1\right)\left(x+1\right)\)(1)

\(=-1-3x\left(x-1\right)-8+3\left(x-1\right)\left(x+1\right)\)

\(=3\left(x-1\right)\left(-x+x+1\right)-9=3\left(x-1\right)-9=3\left(x-4\right)=3x-12\)

(1) là hằng đẳng thức \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)\)

\(a,M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}-\dfrac{2-x}{x\sqrt{x}+x}\right)\left(x>0;x\ne1\right)\\ M=\dfrac{x+\sqrt{x}+x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\\ M=\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(b,M=-\dfrac{1}{2}\Leftrightarrow\dfrac{2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=-\dfrac{1}{2}\\ \Leftrightarrow-4x=x+\sqrt{x}-2\\ \Leftrightarrow5x+\sqrt{x}-2=0\)

Đặt \(\sqrt{x}=t\)

\(\Leftrightarrow5t^2+t-2=0\\ \Delta=1^2-4\cdot5\left(-2\right)=41\\ \Leftrightarrow\left[{}\begin{matrix}t=\dfrac{-1-\sqrt{41}}{10}\\t=\dfrac{-1+\sqrt{41}}{10}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(1+\sqrt{41}\right)^2}{100}=\dfrac{-42-2\sqrt{41}}{100}\\x=\dfrac{\left(\sqrt{41}-1\right)^2}{100}=\dfrac{42-2\sqrt{41}}{100}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-21-\sqrt{41}}{50}\left(L\right)\\x=\dfrac{21-\sqrt{41}}{50}\left(N\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{21-\sqrt{41}}{50}\)

a: Ta có: \(M=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{2}{x}+\dfrac{x-2}{x\sqrt{x}+x}\right)\)

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

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

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