Question

[Ly Trần]

giúp tớ câu 3 í c vớii aa

Answer 1

[Nguyễn Lê Phước Thịnh]

Câu 3:

a: A=1/2

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}}=\dfrac{1}{2}\)

=>\(2\sqrt{x}-4=\sqrt{x}\)

=>\(2\sqrt{x}-\sqrt{x}=4\)

=>\(\sqrt{x}=4\)

=>x=16(nhận)

b: \(B=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{\sqrt{x}+3}{x-1}\)

\(=\dfrac{\sqrt{x}-3}{\sqrt{x}-1}+\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)+\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-2\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)

c: \(P=A\cdot B=\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)

\(\sqrt{P}< \dfrac{1}{2}\)

=>\(0< =P< \dfrac{1}{4}\)

=>P>=0 và P<1/4

=>P>=0 và P-1/4<0

=>\(\left\{{}\begin{matrix}\dfrac{\sqrt{x}-2}{\sqrt{x}+1}>=0\\\dfrac{\sqrt{x}-2}{\sqrt{x}+1}-\dfrac{1}{4}< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}-2>=0\\\dfrac{4\sqrt{x}-8-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{x}>=2\\3\sqrt{x}-9< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>=2\\\sqrt{x}< 3\end{matrix}\right.\)

=>\(2< =\sqrt{x}< 3\)

=>4<=x<9