Cho biểu thức$C=(\frac{1}{\sqrt{x} -3}- \frac{1}{\sqrt{x}}): (\frac{\sqrt{x} + 3}{\sqrt{x} -2}- \frac{\sqrt{x}+2}{\sqrt{x} -3})$ a) Rút gọn C b) Tính

`a)` ĐKXĐ : ` x \ne 9;\ x \ne 0;\ x \ne 4`

` C = ( 1/(\sqrt(x) -3) – 1/(\sqrt(x)) ) : ( ( \sqrt(x) +3)/(\sqrt(x)-2) – (\sqrt(x) +2)/(\sqrt(x)-3)) `

` = ( \sqrt(x)/(\sqrt(x). (\sqrt(x) -3)) – (\sqrt(x)-3)/(\sqrt(x). (\sqrt(x) -3))) : [( (\sqrt(x)+3)(\sqrt(x)-3))/((\sqrt(x)-2)(\sqrt(x)-3))- ((\sqrt(x)+2)(\sqrt(x)-2))/((\sqrt(x)-2)(\sqrt(x)-3))]`

` = 3/(\sqrt(x). (\sqrt(x) -3)) : ( (x-9)/((\sqrt(x)-2)(\sqrt(x)-3)) – (x-4)/((\sqrt(x)-2)(\sqrt(x)-3)))`

` = 3/(\sqrt(x). (\sqrt(x) -3)) : ( -5)/((\sqrt(x)-2)(\sqrt(x)-3))`

` = (-3(\sqrt(x)-2)(\sqrt(x)-3))/(5\sqrt(x).(\sqrt(x)-3)`

`  = ( 6 – 3\sqrt(x))/(5\sqrt(x))`

`b)`

`  \sqrt( 6 + 4\sqrt(2)) = \sqrt( 4 + 2.2. \sqrt(2) + 2) = \sqrt( (2+\sqrt(2))^2) = 2 + \sqrt(2)`

` \sqrt( 6 – 4\sqrt(2)) = \sqrt( 4  – 2.2 \sqrt(2)  + 2 ) = \sqrt ( ( 2 – \sqrt(2))^2) = 2 – \sqrt(2)`

 ` \to x =  \sqrt( 6 + 4\sqrt(2)) + \sqrt( 6 – 4\sqrt(2))  = 2 + \sqrt(2) + 2 – \sqrt(2) = 4`

`\to C = ( 6 – 3\sqrt(x))/(5\sqrt(x)) = ( 6 – 3\sqrt(4))/(5\sqrt(4)) = 0`

`c)`

` C < 1`

`\to ( 6 – 3\sqrt(x))/(5\sqrt(x)) < 1`

`\to 6 – 3\sqrt(x) < 5 \sqrt(x)`

`\to 5 \sqrt(x) + 3 \sqrt(x) -6 > 0`

`\to 8 \sqrt(x) > 6`

`\to \sqrt(x) > 3/4`

`\to x > 9/16`

Kết hợp với ĐKXĐ ta có ` x > 9/16;\ x \ne 4;\ x \ne 9`