\(c,\)

\(\left(6x^3-19x^2+23x-12\right):\left(2x-3\right)\)

\(=\left(6x^3-10x^2-9x^2+8x+15x-12\right):\left(2x-3\right)\)

\(=\left[\left(6x^3-10x^2+8x\right)-\left(9x^2-15x+12\right)\right]:\left(2x-3\right)\)

\(=\left[2x\left(3x^2-5x+4\right)-3\left(3x^2-5x+4\right)\right]:\left(2x-3\right)\)

\(=\left[\left(3x^2-5x+4\right)\left(2x-3\right)\right]:\left(2x-3\right)\)

\(=3x^2-5x+4\)

\(e,\)

\(\left(6x^3-5x^2+4x-1\right):\left(2x^2-x+1\right)\)

\(=\left(6x^3-3x^2-2x^2+3x+x-1\right):\left(2x^2-x+1\right)\)

\(=\left[\left(6x^3-3x^2+3x\right)-\left(2x^2-x+1\right)\right]:\left(2x^2-x+1\right)\)

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

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

\(=3x-1\)

\(f,\)

\(\left(x^3-2x^2-5x+6\right):\left(x+2\right)\)

\(=\left(x^3-4x^2+2x^2+3x-8x+6\right):\left(x+2\right)\)

\(=\left[\left(x^3+2x^2\right)-\left(4x^2+8x\right)+\left(3x+6\right)\right]\)

\(=\left[x^2\left(x+2\right)-4x\left(x+2\right)+3\left(x+2\right)\right]\)

\(=\left[\left(x+2\right)\left(x^2-4x+3\right)\right]:\left(x+2\right)\)

\(=x^2-4x+3\)

\(g,\)

\(\left(x^3-2x^2-5x+6\right):\left(x+2\right)\)

\(=\left(x^3+2x^2-4x^2-8x+3x+6\right):\left(x+2\right)\)

\(=\left[\left(x^3+2x^2\right)-\left(4x^2+8x\right)+\left(3x+6\right)\right]:\left(x+2\right)\)

\(=\left[x^2\left(x+2\right)-4x\left(x+2\right)+3\left(x+2\right)\right]:\left(x+2\right)\)

\(=\left[\left(x+2\right)\left(x^2-4x+3\right)\right]:\left(x+2\right)\)

\(=x^2-4x+3\)

\(b,\)

\(\sqrt{10x^2-12x\sqrt{10}+36}\)

\(=\sqrt{\left(\sqrt{10}x\right)^2-2.x.\sqrt{10}.6+6^2}\)

\(=\sqrt{\left(\sqrt{10}x-6\right)^2}\)

\(=\left|\sqrt{10}x-6\right|\)

Với \(x=\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{2}{5}}\Rightarrow\left|\sqrt{10}.\left(\sqrt{\dfrac{5}{2}}+\sqrt{\dfrac{2}{5}}\right)-6\right|=1\)

\(e,\)

\(\sqrt{9x^4-24x^2+16}-\sqrt{x^4-8x^2+16}\)

\(=\sqrt{\left(3x^2\right)^2-2.3x^2.4+4^2}-\sqrt{\left(x^2\right)^2-2.x^2.4+4^2}\)

\(=\sqrt{\left(3x^2-4\right)^2}-\sqrt{\left(x^2-4\right)^2}\)

\(=\left|3x^2-4\right|-\left|x^2-4\right|\)

Với \(x=\sqrt{3}\Rightarrow\left|3.\left(\sqrt{3}\right)^2-4\right|-\left|\left(\sqrt{3}\right)^2-4\right|=4\) 

Ta có: \(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}}\)

\(\Rightarrow x^2=\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\sqrt{3}+2}\)

\(=\dfrac{1}{2\left(\sqrt{3}-1\right)}-\dfrac{3}{2\left(\sqrt{3}+1\right)}\)

\(=\dfrac{\sqrt{3}+1}{2\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\dfrac{3\left(\sqrt{3}-1\right)}{2\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

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

\(=\dfrac{-2\sqrt{3}+4}{4}\)

\(=\dfrac{\left(\sqrt{3}-1\right)^2}{4}\)

\(\Rightarrow x=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{4}}=\dfrac{\left|\sqrt{3}-1\right|}{2}=\dfrac{\sqrt{3}-1}{2}\) (do \(x\ge0\))

\(\Leftrightarrow2x=\sqrt{3}-1\)

\(\Leftrightarrow2x+1=\sqrt{3}\)

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

\(\Leftrightarrow4x^2+4x+1=3\)

\(\Leftrightarrow4x^2+4x-2=0\)

\(\Leftrightarrow2x^2+2x-1=0\)

Lại có: \(P=\dfrac{4\left(x+1\right)^{2024}-2x^{2023}+2x+3}{2x^2+3x}\)

\(=\dfrac{2x^{2023}\left[2x\left(x+1\right)-1\right]+2x+3}{2x^2+3x}\)

\(=\dfrac{2x^{2023}\left(2x^2+2x-1\right)+2x+3}{2x^2+3x}\)

\(=\dfrac{2x+3}{2x^2+3x}\left(1\right)\) (do \(2x^2+2x-1=0\))

Thay \(x=\dfrac{\sqrt{3}-1}{2}\) vào \(\left(1\right)\) ta được:

\(P=\dfrac{2.\dfrac{\sqrt{3}-1}{2}+3}{2.\left(\dfrac{\sqrt{3}-1}{2}\right)^2+3.\dfrac{\sqrt{3}-1}{2}}=1+\sqrt{3}\)

\(a,\)

\(B=\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}-\dfrac{4}{1-x^2}\) (Điều kiện xác định: \(x\ne\pm1\))

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

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

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

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

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

\(=-\dfrac{4}{x+1}\)

\(b,\)

\(x^2-x=0\)

\(\Leftrightarrow x\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Với \(x=0\Rightarrow B=-\dfrac{4}{0+1}=-4\)

Với \(x=1\Rightarrow B=-\dfrac{4}{1+1}=-2\)

\(c,\)

\(B=-3\Rightarrow-\dfrac{4}{x+1}=-3\)

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

\(\Leftrightarrow-3x-3+4=0\)

\(\Leftrightarrow-3x+1=0\)

\(\Leftrightarrow-3x=-1\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(d,\)

\(B< 0\Rightarrow-\dfrac{4}{x+1}< 0\)

\(\Leftrightarrow x+1>0\)

\(\Leftrightarrow x>-1\)

Kết hợp điều kiện \(x\ne\pm1\) 

\(\Rightarrow-1< x< 1\)