`a)` ` x= 1/4 \to \sqrt(x) = 1/2` , thay vào `A` ta có
` A = ( \sqrt(x))/(\sqrt(x) +6) = ( 1/2)/(1/2 +6) = 1/13`
`b)`
` B = 4/(x-1) + ( \sqrt(x) +3)/( \sqrt(x)+1) - 5/( 1 - \sqrt(x))` ĐKXĐ : ` x \ge 0 ; x \ne 1`
` = 4/(x-1) + ( (\sqrt(x) +3)(\sqrt(x)-1))/(x-1) + (5(\sqrt(x)+1))/(x-1)`
` = ( 4 + x - \sqrt(x) + 3\sqrt(x) - 3 + 5\sqrt(x) +5)/(x-1)`
` = ( x + 7\sqrt(x) +6)/( ( \sqrt(x) -1)( \sqrt(x) +1))`
` = ( ( \sqrt(x)+1)( \sqrt(x) +6))/( ( \sqrt(x) -1)( \sqrt(x) +1))`
` = ( \sqrt(x) +6)/( \sqrt(x) -1)`
`c)`
` P = A.B = ( \sqrt(x))/(\sqrt(x) +6) . ( \sqrt(x) +6)/( \sqrt(x) -1)`
` = ( \sqrt(x))/( \sqrt(x) -1)`
Để ` P < 0` thì ` \sqrt(x);\ \sqrt(x) -1` trái dấu
Mà ` \sqrt(x) > \sqrt(x) -1` nên
` \sqrt(x) > 0;\ \sqrt(x) -1 < 0`
+) ` \sqrt(x) > 0 \to x > 0`
+) ` \sqrt(x) -1 < 0 \to \sqrt(x) < 1 \to x < 1`
Vậy ` 0 < x < 1`
