có 4 hàng hàng số 2 mấy bạn kéo giùm mình cái phần dấu gạch chia ở dưới dài ra để kéo dài cả hai biểu thức luôn được ko dấu gạch dưới phần căn a-1 với căn a+1 đó ạ mình ko biết kéo dài ra rồi các bạn làm bình thường giúp mình nha mình đang rất cần làm ơn

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

     \(=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]\left[\dfrac{\sqrt{a}-1}{a-1}+\dfrac{2}{a-1}\right]\)

     \(=\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}.\dfrac{\sqrt{a}-1}{a-1}\)

     \(=\dfrac{1}{\sqrt{a}}\)

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

     \(=\left[\dfrac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]\left[\dfrac{\sqrt{a}-1}{a-1}+\dfrac{2}{a-1}\right]\)

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

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

\(a,M=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\\ M=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,M< \dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}< 0\\ \Leftrightarrow\dfrac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}< 0\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}< 0\\ \Leftrightarrow\sqrt{a}-4< 0\left(6\sqrt{a}>0\right)\\ \Leftrightarrow a< 16\\ \Leftrightarrow0< a< 16\left(kết.hợp.ĐKXĐ\right)\)

\(A=\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)=\dfrac{7\sqrt{a}}{a-9}-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{7\sqrt{a}}{a-9}-\dfrac{a+3\sqrt{a}-a+3\sqrt{a}+\sqrt{a}-3}{a-9}=\dfrac{3}{a-9}\)\(B=\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)=\dfrac{\sqrt{a}-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-3\right)}:\dfrac{a-9-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{3}{\sqrt{a}\left(\sqrt{a}-3\right)}.\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}{-5}=\dfrac{3\sqrt{a}-6}{-5\sqrt{a}}\)

\(C=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\left(\sqrt{a}-1\right)}\right).\dfrac{1-2a}{a}=\dfrac{a\sqrt{a}-a}{\sqrt{a}-1}.\dfrac{1-2a}{a}=\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}-1}.\dfrac{1-2a}{a}=1-2a\)\(D=\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a\sqrt{a}+1-\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1}=\dfrac{a\sqrt{a}+1-a\sqrt{a}+a+\sqrt{a}-1}{a-1}=\dfrac{a+\sqrt{a}}{a-1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-1}\)

\(E=\dfrac{a}{a-4}+\dfrac{1}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}=\dfrac{a+\sqrt{a}+2+\sqrt{a}-2}{a-4}=\dfrac{a+2\sqrt{a}}{a-4}=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-2}\)

1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn

2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)

\(a,Sửa:M=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\\ M=\dfrac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\\ b,M=\dfrac{\sqrt{a}-1}{\sqrt{a}}=1-\dfrac{1}{\sqrt{a}}< 1\left(\dfrac{1}{\sqrt{a}}>0\right)\\ c,M>0\Leftrightarrow\sqrt{a}-1>0\left(\sqrt{a}>0\right)\Leftrightarrow a>1\)

b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)

\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)

\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)

\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)

\(VT=0=VP\)