Cho P=( √x/1 – 1/ √x) 🙁 √x -1/ √x – √x -1 /x+ √x)
a)rút gon P
b)cho P=9/2 tính x
0 bình luận về “Cho P=( √x/1 – 1/ √x) :( √x -1/ √x – √x -1 /x+ √x)
a)rút gon P
b)cho P=9/2 tính x”
Giải thích các bước giải:
$\eqalign{ & a)\,P = (\sqrt x – \frac{1}{{\sqrt x }}):(\frac{{\sqrt x – 1}}{{\sqrt x }} – \frac{{\sqrt x – 1}}{{x + \sqrt x }}) \cr & = \frac{{x – 1}}{{\sqrt x }}:(\frac{{\sqrt x – 1}}{{\sqrt x }} – \frac{{\sqrt x – 1}}{{\sqrt x (\sqrt x + 1)}}) \cr & = \frac{{x – 1}}{{\sqrt x }}:\frac{{(\sqrt x – 1)(\sqrt x + 1) – (\sqrt x – 1)}}{{\sqrt x (\sqrt x + 1)}} \cr & = \frac{{x – 1}}{{\sqrt x }}:\frac{{x – 1 – \sqrt x + 1}}{{\sqrt x (\sqrt x + 1)}} \cr & = \frac{{x – 1}}{{\sqrt x }}.\frac{{\sqrt x (\sqrt x + 1)}}{{x – \sqrt x }} \cr & = \frac{{(\sqrt x – 1)(\sqrt x + 1).(\sqrt x + 1)}}{{\sqrt x (\sqrt x – 1)}} \cr & = \frac{{{{(\sqrt x + 1)}^2}}}{{\sqrt x }} \cr} $
b) ĐK: x>0
$\eqalign{ & P = \frac{9}{2} \cr & \Leftrightarrow \frac{{{{(\sqrt x + 1)}^2}}}{{\sqrt x }} = \frac{9}{2} \cr & \Leftrightarrow 2{(\sqrt x + 1)^2} = 9\sqrt x \cr & \Leftrightarrow 2(x + 2\sqrt x + 1) – 9\sqrt x = 0 \cr & \Leftrightarrow 2x – 5\sqrt x + 2 = 0 \cr & \Leftrightarrow (2\sqrt x – 1)(\sqrt x – 2) = 0 \cr & \Leftrightarrow 2\sqrt x – 1 = 0\,hoặc\,\sqrt x – 2 = 0 \cr & \Leftrightarrow \sqrt x = \frac{1}{2}\,hoặc\,\sqrt x = 2 \cr & \Leftrightarrow x = \frac{1}{4}\,hoặc\,x = 4(tm\,x > 0) \cr} $
Giải thích các bước giải:
$\eqalign{ & a)\,P = (\sqrt x – \frac{1}{{\sqrt x }}):(\frac{{\sqrt x – 1}}{{\sqrt x }} – \frac{{\sqrt x – 1}}{{x + \sqrt x }}) \cr & = \frac{{x – 1}}{{\sqrt x }}:(\frac{{\sqrt x – 1}}{{\sqrt x }} – \frac{{\sqrt x – 1}}{{\sqrt x (\sqrt x + 1)}}) \cr & = \frac{{x – 1}}{{\sqrt x }}:\frac{{(\sqrt x – 1)(\sqrt x + 1) – (\sqrt x – 1)}}{{\sqrt x (\sqrt x + 1)}} \cr & = \frac{{x – 1}}{{\sqrt x }}:\frac{{x – 1 – \sqrt x + 1}}{{\sqrt x (\sqrt x + 1)}} \cr & = \frac{{x – 1}}{{\sqrt x }}.\frac{{\sqrt x (\sqrt x + 1)}}{{x – \sqrt x }} \cr & = \frac{{(\sqrt x – 1)(\sqrt x + 1).(\sqrt x + 1)}}{{\sqrt x (\sqrt x – 1)}} \cr & = \frac{{{{(\sqrt x + 1)}^2}}}{{\sqrt x }} \cr} $
b) ĐK: x>0
$\eqalign{ & P = \frac{9}{2} \cr & \Leftrightarrow \frac{{{{(\sqrt x + 1)}^2}}}{{\sqrt x }} = \frac{9}{2} \cr & \Leftrightarrow 2{(\sqrt x + 1)^2} = 9\sqrt x \cr & \Leftrightarrow 2(x + 2\sqrt x + 1) – 9\sqrt x = 0 \cr & \Leftrightarrow 2x – 5\sqrt x + 2 = 0 \cr & \Leftrightarrow (2\sqrt x – 1)(\sqrt x – 2) = 0 \cr & \Leftrightarrow 2\sqrt x – 1 = 0\,hoặc\,\sqrt x – 2 = 0 \cr & \Leftrightarrow \sqrt x = \frac{1}{2}\,hoặc\,\sqrt x = 2 \cr & \Leftrightarrow x = \frac{1}{4}\,hoặc\,x = 4(tm\,x > 0) \cr} $