\(=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+16}=\dfrac{\left(x+16\right)\left(\sqrt{x}+2\right)}{\left(x-16\right)\left(\sqrt{x}+16\right)}\)

Lời giải:
\(A=\frac{(\sqrt{x}+2)(\sqrt{x}-2)}{(\sqrt{x}+3)(\sqrt{x}-2)}-\frac{5}{(\sqrt{x}+3)(\sqrt{x}-2)}-\frac{\sqrt{x}+3}{(\sqrt{x}-2)(\sqrt{x}+3)}\)

\(=\frac{x-4-5-\sqrt{x}-3}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{x-\sqrt{x}-12}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{(\sqrt{x}+3)(\sqrt{x}-4)}{(\sqrt{x}-2)(\sqrt{x}+3)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)

a: ĐKXĐ của A là: \(\begin{cases}x+2<>0\\ x^2-4<>0\\ x^2+3x+2<>0\end{cases}\)

=>\(\begin{cases}x<>-2\\ x^2<>4\\ \left(x+1\right)\left(x+2\right)<>0\end{cases}\)

=>x∉{-2;2;-1}

ĐKXĐ cua B là \(x^3-1<>0\)

=>\(x^3<>1\)

=>x<>1

b: \(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\)

\(=\frac{4x}{x+2}-\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x}{x+2}-\frac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}=\frac{4x\left(x+2\right)-4x^2-8x-16}{\left(x+2\right)^2}\)

\(=\frac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}=-\frac{16}{\left(x+2\right)^2}\)

\(A=\left(\frac{4x}{x+2}-\frac{x^3-8}{x^3+8}\cdot\frac{4x^2-8x+16}{x^2-4}\right):\frac{16}{x+2}\cdot\frac{x^2+3x+2}{x^2+x+1}\)

\(=\frac{-16}{\left.\left(x+2\right)^2\right.}\cdot\frac{x+2}{16}\cdot\frac{\left(x+1\right)\left(x+2\right)}{x^2+x+1}=\frac{-\left(x+1\right)}{x^2+x+1}\)

\(B=\frac{x^2+x-2}{x^3-1}\)

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

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

b: Đặt P=A+B

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

\(=\frac{1}{x^2+x+\frac14+\frac34}=\frac{1}{\left(x+\frac12\right)^2+\frac34}\le1:\frac34=\frac43\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x+1/2=0

=>x=-1/2

a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)

\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)

b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)

\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)

c: \(C=x-4+\left|x-4\right|\)

=x-4+x-4

=2x-8

\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)

\(\dfrac{\sqrt{x}}{\sqrt{x}-4}=1-\sqrt{3}\)
Nhỉ???

\(=>B=\dfrac{\left(\sqrt{x}-1\right)\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(B=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{x-6\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(B=\dfrac{\left(\sqrt{x}-4\right)\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-4}{\sqrt{x}}\)

Ta có: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{5\sqrt{x}-8}{x-2\sqrt{x}}\)

\(=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\cdot\left(\sqrt{x}-2\right)}\)

\(=\dfrac{x-6\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)

b) ĐKXĐ : \(x\ne\pm1\)

\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)

\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)-\left(6x-4\right)}{\left(x-1\right)\left(x+1\right)}\)

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

c) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

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

a) ĐKXĐ : \(x\ge0;x\ne16\)

\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x-4}}\right):\dfrac{x+16}{\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)

\(=\dfrac{x+16}{x-16}:\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{x-16}\)

a: \(B=\dfrac{x-4\sqrt{x}+4\sqrt{x}+16}{x-4}\cdot\dfrac{\sqrt{x}+2}{x+16}=\dfrac{1}{\sqrt{x}-2}\)

b: Khi x=9 thì B=1/(3-2)=1