b: Ta có: \(N=a^3+b^3+3ab\)

\(=\left(a+b\right)^3-3ab\left(a+b\right)+3ab\)

\(=1-3ab+3ab\)

=1

\(E=\frac{x^2}{x-2}.\left(\frac{x^2+4}{x}-4\right)+3\)( \(ĐK:x\ne2;x\ne0\))

\(=\frac{x^2}{x-2}.\frac{x^2-4x+4}{x}+3\)

\(=\frac{x^2}{x-2}.\frac{\left(x-2\right)^2}{x}+3=x\left(x-2\right)+3=x^2-2x+3\)

b, \(E=x^2-2x+3=\left(x-1\right)^2+2\ge2\forall x\)

Dấu "=" xảy ra khi \(x-1=0\Rightarrow x=1\)

Vậy GTNN của E là 2 khi x = 1

a: ĐKXĐ: x>0

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

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

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

b: E=2/5

=>\(\dfrac{\sqrt{x}}{x+\sqrt{x}+2}=\dfrac{2}{5}\)

=>\(5\sqrt{x}=2x+2\sqrt{x}+4\)

=>\(2x-3\sqrt{x}+4=0\)

=>\(x-\dfrac{3}{2}\cdot\sqrt{x}+2=0\)

=>\(x-2\cdot\sqrt{x}\cdot\dfrac{3}{4}+\dfrac{9}{16}+\dfrac{23}{16}=0\)

=>\(\left(\sqrt{x}-\dfrac{3}{4}\right)^2+\dfrac{23}{16}=0\)(vô lý)

Vậy: \(x\in\varnothing\)

chịu

\(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{\sqrt{x}+1}\left(đk:x\ne1,x\ge0\right)\)

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

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

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

ĐKXĐ: \(x\ne1,x\ge0\)

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

Với \(x\ge0\)

\(E=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{4\sqrt{x}}{3}\)

\(=\left(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\right).\dfrac{4\sqrt{x}}{3}\)

\(=\dfrac{4\sqrt{x}}{3\left(x+\sqrt{x}+1\right)}\)

a) Ta có: \(E=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)

\(=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)

\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)

\(=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b) Để \(E=\dfrac{8}{9}\) thì \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)

\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)

\(\Leftrightarrow24x-24\sqrt{x}-36\sqrt{x}+24=0\)

\(\Leftrightarrow24x-60\sqrt{x}+24=0\)

\(\Leftrightarrow24x-12\sqrt{x}-48\sqrt{x}+24=0\)

\(\Leftrightarrow12\sqrt{x}\left(2\sqrt{x}-1\right)-24\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(12\sqrt{x}-24\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}-1=0\\12\sqrt{x}-24=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}=1\\12\sqrt{x}=24\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=4\end{matrix}\right.\)