\(\text{Thay m=-4 vào phương trình (1)}\)

\(\left(1\right)x^2+x-4-2=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Delta=b^2-4ac=1^2-4.1.\left(-6\right)=25>0\)

\(\text{Vậy phương trình có hai nghiệm phân biệt:}\)

\(x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=-3\)

\(x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=2\)

\(D=\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}\)

\(D=\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9\left(x-1\right)}+24\dfrac{\sqrt{x-1}}{8}\)

\(D=\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+192\sqrt{x-1}\)

\(D=\sqrt{x-1}\left(\dfrac{1}{2}-\dfrac{9}{2}+3\right)\)

\(D=-\sqrt{x-1}\)

\(E=\sqrt{16x-32}+\sqrt{9x-18}-\sqrt{4x-8}\)

\(E=\sqrt{16\left(x-2\right)}+\sqrt{9\left(x-2\right)}-\sqrt{4\left(x-2\right)}\)

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

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

\(E=5\sqrt{x-2}\)

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

\(ĐKXĐ:x\ge0.x\ne1\)

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

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

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

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

\(A=\dfrac{-1}{\sqrt{x}-1}.\dfrac{\sqrt{x}+1}{1}+1\)

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

\(A=\dfrac{-\sqrt{x}-1}{\sqrt{x-1}}+1\)

\(A=\dfrac{-\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x-1}}\)

\(A=\dfrac{-2}{\sqrt{x}-1}\)

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

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

\(P=2\sqrt{x}+1-\sqrt{x}-1+1\)

\(P=\sqrt{x}+1\)

\(B=\left(\dfrac{3}{x+3\sqrt{x}}-\dfrac{1}{\sqrt{x}-3}\right).\dfrac{x-9}{\sqrt{x}}\)

\(ĐKXĐ:x>0,x\ne9\)

\(B=\left(\dfrac{3}{\sqrt{x}\left(\sqrt{x}+3\right)}-\dfrac{1}{\sqrt{x}-3}\right).\dfrac{x-9}{\sqrt{x}}\)

\(B=\left(\dfrac{3\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\dfrac{x-9}{\sqrt{x}}\)

\(B=\left(\dfrac{3\sqrt{x}-9-x-3\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\dfrac{x-9}{\sqrt{x}}\)

\(B=\dfrac{-x-9}{\sqrt{x}\left(x-9\right)}.\dfrac{x-9}{\sqrt{x}}\)

\(B=\dfrac{-x-9}{\sqrt{x}.\sqrt{x}}=\dfrac{-x-9}{x}\)

\(\text{Thay m=-4 vào phương trình:}\)

\(x^2+x+\left(-4\right)-2=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Delta=b^2-4ac=1^2-4.1.\left(-6\right)=25>0\)

\(\text{Vậy phương trình có hai nghiệm phân biệt:}\)

\(x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=-3\)

\(x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=2\)