A= \(\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)\) ĐK:(a≥0, a≠9)
B= \(\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)\) ĐK:(a≥0, a≠9)
C= \(\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)\) ĐK:(a>0, a≠1)
D= \(\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}\) ĐK:(a≥0, a≠1)
E= \(\dfrac{a}{a-4}+\dfrac{1}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}\) ĐK:(a≥0, a≠4)
Giúp mìnk với nha !!!
\(A=\dfrac{7\sqrt{a}}{a-9}-\left(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{\sqrt{a}-1}{\sqrt{a}+3}\right)=\dfrac{7\sqrt{a}}{a-9}-\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-\left(\sqrt{a}-1\right)\left(\sqrt{a}-3\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}=\dfrac{7\sqrt{a}}{a-9}-\dfrac{a+3\sqrt{a}-a+3\sqrt{a}+\sqrt{a}-3}{a-9}=\dfrac{3}{a-9}\)\(B=\left(\dfrac{1}{\sqrt{a}-3}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+3}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-3}\right)=\dfrac{\sqrt{a}-\sqrt{a}+3}{\sqrt{a}\left(\sqrt{a}-3\right)}:\dfrac{a-9-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-3\right)}=\dfrac{3}{\sqrt{a}\left(\sqrt{a}-3\right)}.\dfrac{\left(\sqrt{a}-3\right)\left(\sqrt{a}-2\right)}{-5}=\dfrac{3\sqrt{a}-6}{-5\sqrt{a}}\)
\(C=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\sqrt{a}-a}\right).\left(\dfrac{1}{a}-2\right)=\left(\dfrac{a\sqrt{a}}{\sqrt{a}-1}-\dfrac{a^2}{a\left(\sqrt{a}-1\right)}\right).\dfrac{1-2a}{a}=\dfrac{a\sqrt{a}-a}{\sqrt{a}-1}.\dfrac{1-2a}{a}=\dfrac{a\left(\sqrt{a}-1\right)}{\sqrt{a}-1}.\dfrac{1-2a}{a}=1-2a\)\(D=\dfrac{a\sqrt{a}+1}{a-1}-\dfrac{a-1}{\sqrt{a}+1}=\dfrac{a\sqrt{a}+1-\left(a-1\right)\left(\sqrt{a}-1\right)}{a-1}=\dfrac{a\sqrt{a}+1-a\sqrt{a}+a+\sqrt{a}-1}{a-1}=\dfrac{a+\sqrt{a}}{a-1}=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-1}\)
\(E=\dfrac{a}{a-4}+\dfrac{1}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}=\dfrac{a+\sqrt{a}+2+\sqrt{a}-2}{a-4}=\dfrac{a+2\sqrt{a}}{a-4}=\dfrac{\sqrt{a}\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}=\dfrac{\sqrt{a}}{\sqrt{a}-2}\)
