Đáp án:
$a)B=\dfrac{\sqrt{x}}{x+1}\\b)x=\dfrac{1}{4}\quad\text{hoặc}\quad x=4\\B\le1/2\\x\in\varnothing$
Giải thích các bước giải:
`B=(\sqrt{x}(1-x)^2)/(x+1):[((x\sqrt{x}-1)/(\sqrt{x}-1)+\sqrt{x}).((x\sqrt{x}+1)/(\sqrt{x}+1)-\sqrt{x})](x>=0,x\ne1)`
`B=(\sqrt{x}(1-x)^2)/(x+1):[(((\sqrt{x}-1)(x+\sqrt{x}+1))/(\sqrt{x}-1)+\sqrt{x}).(((\sqrt{x}+1)(x-\sqrt{x}+1))/(\sqrt{x}+1)-\sqrt{x})]`
`B=(\sqrt{x}(1-x)^2)/(x+1):[(x+\sqrt{x}+1+\sqrt{x}).(x-\sqrt{x}+1-\sqrt{x})]`
`B=(\sqrt{x}(1-x)^2)/(x+1):[(x+2\sqrt{x}+1).(x-2\sqrt{x}+1)]`
`B=(\sqrt{x}(1-x)^2)/(x+1):[(\sqrt{x}+1)^2.(\sqrt{x}-1)^2]`
`B=(\sqrt{x}(1-x)^2)/(x+1):[(\sqrt{x}+1)(\sqrt{x}-1)]^2`
`B=(\sqrt{x}(1-x)^2)/(x+1):(x-1)^2`
`B=(\sqrt{x})/(x+1)`
`b)B=2/5`
`<=>(\sqrt{x})/(x+1)=2/5`
`<=>5\sqrt{x}=2x+2`
`<=>2x-5\sqrt{x}+2=0`
`<=>(2\sqrt{x}-1)(\sqrt{x}-2)=0`
`<=>[(2\sqrt{x}=1),(\sqrt{x}=2):}`
\(⇔\left[ \begin{array}{l}x=4(TMĐKXĐ)\\x=\dfrac{1}{4}(TMĐKXĐ)\end{array} \right.\)
Vậy với `x=1/4` hoặc `x=4` thì `B=2/5.`
`c)\text{Xét}\quadB-1/2`
`B-1/2=\sqrt{x}/(x+1)-1/2`
`B-1/2=(2\sqrt{x})/(2(x+1))-(x+1)/(2(x+1))`
`B-1/2=(2\sqrt{x}-x-1)/(2(x+1))`
`B-1/2=(-(x-2\sqrt{x}+1))/(2(x+1))`
`B-1/2=(-(\sqrt{x}-1)^2)/(2(x+1))<=0AAx>=0,x\ne1`
`=>B-1/2<=0`
`=>B<=1/2`
`d)B>3/\sqrt{x}(x>0)`
`<=>\sqrt{x}/(x+1)>3/\sqrt{x}`
`<=>\sqrt{x}/(x+1)-3/\sqrt{x}>0`
`<=>x/(\sqrt{x}(x+1))-(3(x+1))/(\sqrt{x}(x+1))>0`
`<=>(x-3(x+1))/(\sqrt{x}(x+1))>0`
`<=>(x-3x-3)/(\sqrt{x}(x+1))>0`
`<=>(-2x-3)/(\sqrt{x}(x+1))>0` vô lý
`\text{Vì}\quad x>0=>{(\sqrt{x}(x+1)>0),(-2x-3<-3<0):}=>(-2x-3)/(\sqrt{x}(x+1))<0`
`\text{Vậy không có giá trị nào của x để}\quadB>3/\sqrt{x}.`
