A= \(\frac{\sqrt{x}+3}{\sqrt{x}-2}\)\(-\frac{\sqrt{x}+2}{\sqrt{x}-3}\)+\(\frac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}\)

=\(\frac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}\)

=\(\frac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}\)

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

Vay A= \(\frac{1}{\sqrt{x}-2}\)

O co the co kieu gen: AaBbDDEeGg khi giam phan se cho bao nhieu loai giao tu? Loai giao tu mang gen ABDEG chiem ti le la bao nhieu?

\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{156}\)
Ta có:
\(\frac{1}{2}=1-\frac{1}{2}\)  \(;\)   \(\frac{1}{4}=\frac{1}{2}-\frac{1}{4}\)\(;\)   \(\frac{1}{8}=\frac{1}{4}-\frac{1}{8}\)\(;...;\)   \(\frac{1}{256}=\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+...+\frac{1}{128}-\frac{1}{256}\)
\(\Rightarrow A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{256}{256}-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
Vậy \(A=\frac{255}{256}\)
◘_◘ Đúng 100%

Tim GTNN cua bieu thuc sau:

Q= \(\frac{-2\sqrt{3x}}{3+x}\) ( voi x≥0, x≠-3)