Toán 9

T

thinhyamai

[TEX]\frac{1-a^2}{48}.\sqrt[0]{\frac{36}{(a-1)^2}}[/TEX]

[TEX]=-\frac{(a-1)(a+1)}{48}.\frac{6}{\left |a-1\right |}[/TEX]

[TEX]TH1:a>1[/TEX]

[TEX]\frac{-(a-1)(a+1).6}{48.(a-1)}=\frac{-(a+1)}{8}[/TEX]

[TEX]TH2:a<1[/TEX]

[TEX]\frac{-(a-1)(a+1).6}{-48.(a-1)}=\frac{a+1}{8}[/TEX]
:D
 
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