`A)đk:x>=0,x ne 25`
`A=9=>A=(3+2)/(3-5)=-5/2`
`B)B=(3sqrtx-15+20-2sqrtx)/(x-25)`
`=(sqrtx+5)/(x-25)`
`=1/(sqrtx-5)`
`A=B.|x-4|`
`<=>A/B=|x-4|`
`<=>\sqrtx+2=|x-4|`
`<=>\sqrtx+2=(sqrtx+2)|sqrtx-2|`
`<=>|sqrtx-2|=1`
`+)sqrtx-2=1<=>x=9(tm)`
`+)sqrtx-2=-1<=>x=1(tm)`
Vậy `S={1,9}`
a, Thay x=9 vào biểu thức A ta có
\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}\)
\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-2,5\)
Vậy A =-2,5 khi x=9
a. A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)
=\(\dfrac{\sqrt{9}+2}{\sqrt{9}-5}=\dfrac{-5}{2}\)
b,\(B=\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}\)
\(B=\dfrac{3\left(\sqrt{x}-5\right)+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(B=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(B=\dfrac{1}{\sqrt{x}-5}\)( Điều cần chứng minh)
\(x=9\rightarrow\sqrt{x}=3\)
a) Thay \(\sqrt{x}=3\) vào A ta có :
\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-\dfrac{5}{2}\)
b) \(B=\dfrac{3}{\sqrt{x}+5}+\dfrac{20-2\sqrt{x}}{x-25}\)
\(B=\dfrac{3\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\dfrac{20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(B=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(B=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(B=\dfrac{1}{\sqrt{x}-5}\) (đpcm)
