Đk:\(x>0;x\ne1\)
\(B=\left[\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
\(B=\dfrac{1}{2}\Leftrightarrow\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=2\)\(\Leftrightarrow x=9\) (tm)
Vậy..
a) \(B=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{x-1}\right):\dfrac{x\sqrt{x}-1}{x\sqrt{x}-\sqrt{x}}\)
\(B=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(B=\dfrac{1}{\sqrt{x}-1}\)
b) Với \(B=\dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}-1}=\dfrac{1}{2}\Leftrightarrow\sqrt{x}-1=2\)
\(\Leftrightarrow\sqrt{x}=3\)
\(\Leftrightarrow x=9\)
Vậy...
Chúc bạn học tốt
a, \(B=\left(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{1}{\sqrt{x}-1}\)
b, Thay B = 1/2 vào ta được :\(\dfrac{1}{2}=\dfrac{1}{\sqrt{x}-1}\)
\(\Leftrightarrow\sqrt{x}=3\)
\(\Leftrightarrow x=9\)
Vậy ...
a) B=√x+1+x√x(√x−1)(√x+1):(√x−1)(x+√x+1)√x(√x−1)(√x+1)B=x+1+xx(x−1)(x+1):(x−1)(x+x+1)x(x−1)(x+1)
B=1√x−1B=1x−1
b) Với ⇔1√x−1=12⇔√x−1=2⇔1x−1=12⇔x−1=2
⇔√x=3⇔x=3
⇔x=9⇔x=9
