bo như ngù

ĐKXĐ: x>=0; x<>1

a: \(A=\frac{x\cdot\sqrt{x}+x}{\sqrt{x}+1}-\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}\)

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

\(=x-\left(\sqrt{x}-1\right)=x-\sqrt{x}+1\)

b: A=4

=>\(x-\sqrt{x}+1=4\)

=>\(x-\sqrt{x}-3=0\)

=>\(x-\sqrt{x}+\frac14-\frac{13}{4}=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac{13}{4}\)

=>\(\sqrt{x}-\frac12=\frac{\sqrt{13}}{2}\)

=>\(\sqrt{x}=\frac{\sqrt{13}+1}{2}\)

=>\(x=\frac{\left(\sqrt{13}+1\right)^2}{4}=\frac{4+2\sqrt{13}}{4}=\frac{2+\sqrt{13}}{2}\)

c: \(A=x-\sqrt{x}+1\)

\(=x-\sqrt{x}+\frac14+\frac34=\left(\sqrt{x}-\frac12\right)^2+\frac34\ge\frac34>0\forall x\) thỏa mãn ĐKXĐ

=>x>=0 và x<>1

a) đk: \(x\ne0;4\); \(x>0\)

P = \(\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{1}{\sqrt{x}-2}\right]\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)

= \(\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\times\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)\)

= \(\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)}.\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)

b) Để P < \(\dfrac{1}{2}\)

<=> \(\dfrac{\sqrt{x}-1}{\sqrt{x}}< \dfrac{1}{2}\)

<=> \(1-\dfrac{1}{\sqrt{x}}< \dfrac{1}{2}\)

<=> \(\dfrac{1}{\sqrt{x}}>\dfrac{1}{2}\)

<=> \(\sqrt{x}< 2\)

<=> x < 4

<=> 0 < x < 4

c, Tìm gt nguyên của x để P có gt nguyên

\(A=\frac{15\sqrt{x}-11}{x-\sqrt{x}+3\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{45\sqrt{x}-11}{\left(\sqrt{x}+3\right)(\sqrt{x}-1)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{45\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{37\sqrt{x}-5x-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(a,E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(x>0;x\ne1\right)\\ E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\\ b,E>1\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\\ \Leftrightarrow\sqrt{x}-1>0\left[x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\right]\\ \Leftrightarrow x>1\left(tm\right)\)

\(c,E=\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\\ E=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\ge2\sqrt{\dfrac{\sqrt{x}-1}{\sqrt{x}-1}}+2=2+2=4\\ E_{min}=4\Leftrightarrow\sqrt{x}-1=1\Leftrightarrow x=4\)