C

c

1: Thay \(x=\dfrac{1}{25}\) vào C, ta được:

\(C=\left(\dfrac{1}{5}+2\right):\left(\dfrac{1}{5}-3\right)=\dfrac{11}{5}:\dfrac{-14}{5}=-\dfrac{11}{14}\)

2: Để C=-2 thì \(\sqrt{x}+2=-2\left(\sqrt{x}-3\right)\)

\(\Leftrightarrow\sqrt{x}+2+2\sqrt{x}-6=0\)

\(\Leftrightarrow3\sqrt{x}=4\)

hay \(x=\dfrac{16}{9}\)

Để \(C=\dfrac{7}{5}\) thì \(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}=\dfrac{7}{5}\)

\(\Leftrightarrow7\sqrt{x}-21=2\sqrt{x}+10\)

\(\Leftrightarrow5\sqrt{x}=31\)

hay \(x=\dfrac{961}{25}\)

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

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

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+1}\)

mình mới học lớp 5 hihi)

a: Ta có: \(A=\frac{-x}{4-x}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)

\(=\frac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

\(=\frac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}}{\sqrt{x}-2}\)

b: Thay x=36 vào A, ta được:

\(A=\frac{6}{6-2}=\frac64=\frac32\)

c: \(A=-\frac13\)

=>\(\frac{\sqrt{x}}{\sqrt{x}-2}=-\frac13\)

=>\(3\sqrt{x}=-\sqrt{x}+2\)

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

=>\(\sqrt{x}=\frac12\)

=>\(x=\frac14\left(nhận\right)\)

e: A:B=-2

=>\(\frac{\sqrt{x}}{\sqrt{x}-2}:\frac{\sqrt{x}+1}{\sqrt{x}-2}=-2\)

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

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

=>\(3\sqrt{x}=-2\) (vô lý)

=>x∈∅