Question

E=\(\frac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{1}{1-\sqrt{x}}+\frac{2-x}{x-\sqrt{x}}\right)\)

a/ rút gọn

b/tìm x để E>1

c/ tìm GTNN của E vs x >1

d) Tìm x\(\in\)Z để E\(\in\)Z

e) Tìm x để E=\(\frac{9}{2}\)

Answer 1

a) e= \(\dfrac{x+\sqrt{x}}{x-2\sqrt{x}+1}:\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}-\dfrac{1}{1-\sqrt{x}}+\dfrac{2-x}{x-\sqrt{x}}\right)\)(xkhac0;1)

\(\Leftrightarrow e=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)\(\Leftrightarrow e=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\left(\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\right)\)

\(\Leftrightarrow e=\)\(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(\Leftrightarrow e=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}.\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)

\(\Leftrightarrow e=\dfrac{x}{\sqrt{x}-1}\)

vậy e=\(\dfrac{x}{\sqrt{x}-1}\)

​b )ta có e>1\(\Leftrightarrow\dfrac{x}{\sqrt{x}-1}>1\)

\(\Leftrightarrow x>\sqrt{x}-1\)

\(\Leftrightarrow x-\sqrt{x}+1>0\)

vì x-\(\sqrt{x}-1>0\) với mọi x khác 0 và khác 1

Answer 2

d: Để E là số nguyên thì \(x-1+1⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{1;-1\right\}\)

hay \(x=4\)

e: Để E=9/2 thì \(\dfrac{x}{\sqrt{x}-1}=\dfrac{9}{2}\)

=>\(2x-9\sqrt{x}+9=0\)

=>2x-3 căn x-6 căn x+9=0

=>2 căn x-3=0

hay x=9/4