Bài 1:Rút gọn B=2 / √x -1 – 1/ √x – 3 √x -5/ √x -x
Bài 2:Rút gọn P=x-2 √x +1/ √x-1.(x- √x -2/ √x+1 +3/1)
Bài 3:Cho A=( √x-1/x- √x – √x/x+ √x):(1-1/ √x)
a)Rút gọn A b)Tính A khi x=9
HUHU ,,help me
Bài 1:Rút gọn B=2 / √x -1 – 1/ √x – 3 √x -5/ √x -x
Bài 2:Rút gọn P=x-2 √x +1/ √x-1.(x- √x -2/ √x+1 +3/1)
Bài 3:Cho A=( √x-1/x- √x – √x/x+ √x):(1-1/ √x)
a)Rút gọn A b)Tính A khi x=9
HUHU ,,help me
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
B = \dfrac{2}{{\sqrt x – 1}} – \dfrac{1}{{\sqrt x }} – \dfrac{{3\sqrt x – 5}}{{\sqrt x – x}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.} \right)\\
= \dfrac{2}{{\sqrt x – 1}} – \dfrac{1}{{\sqrt x }} – \dfrac{{3\sqrt x – 5}}{{\sqrt x \left( {1 – \sqrt x } \right)}}\\
= \dfrac{2}{{\sqrt x – 1}} – \dfrac{1}{{\sqrt x }} + \dfrac{{3\sqrt x – 5}}{{\sqrt x \left( {\sqrt x – 1} \right)}}\\
= \dfrac{{2\sqrt x – 1.\left( {\sqrt x – 1} \right) + 3\sqrt x – 5}}{{\sqrt x \left( {\sqrt x – 1} \right)}}\\
= \dfrac{{2\sqrt x – \sqrt x + 1 + 3\sqrt x – 5}}{{\sqrt x \left( {\sqrt x – 1} \right)}}\\
= \dfrac{{4\sqrt x – 4}}{{\sqrt x \left( {\sqrt x – 1} \right)}}\\
= \dfrac{4}{{\sqrt x }}\\
3,\\
a,\\
A = \left( {\dfrac{{\sqrt x – 1}}{{x – \sqrt x }} – \dfrac{{\sqrt x }}{{x + \sqrt x }}} \right):\left( {1 – \dfrac{1}{{\sqrt x }}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\left( {\left\{ \begin{array}{l}
x > 0\\
x \ne 1
\end{array} \right.} \right)\\
= \left( {\dfrac{{\sqrt x – 1}}{{\sqrt x \left( {\sqrt x – 1} \right)}} – \dfrac{{\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}} \right):\dfrac{{\sqrt x – 1}}{{\sqrt x }}\\
= \left( {\dfrac{1}{{\sqrt x }} – \dfrac{1}{{\sqrt x + 1}}} \right):\dfrac{{\sqrt x – 1}}{{\sqrt x }}\\
= \dfrac{{\left( {\sqrt x + 1} \right) – \sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x }}{{\sqrt x – 1}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x + 1} \right)}}.\dfrac{{\sqrt x }}{{\sqrt x – 1}}\\
= \dfrac{1}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}\\
= \dfrac{1}{{x – 1}}\\
b,\\
x = 9 \Rightarrow A = \dfrac{1}{8}
\end{array}\)
Em viết lại đề của câu 2 nhé!