a: ĐKXĐ: x>=0; x<>4

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

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

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

b: \(M=P\cdot Q=\dfrac{\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{1-5\sqrt{x}}{\sqrt{x}+1}=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

\(M\left(M-1\right)=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-5x-x-3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(1-5\sqrt{x}\right)\left(-6x-2\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\cdot\left(\sqrt{x}+1\right)^2}\)

\(=\dfrac{\sqrt{x}\left(5\sqrt{x}-1\right)\left(6x+2\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}+1\right)^2}\)

TH1: M>=căn M

=>M^2>=M

=>M^2-M>=0

=>5*căn x-1>=0

=>x>=1/25 và x<>4

TH2: M<căn M

=>5căn x-1<0

=>x<1/25

Kết hợp ĐKXĐ, ta được: 0<=x<1/25

Q = (1 + √x)/√x

= 1/√x + √x/√x

= 1/√x + 1 > 1

Vậy Q > 1

a: Ta có: \(P=\left(\dfrac{x-2\sqrt{x}+4}{\sqrt{x}-2}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x+4}{x-4}\right)\)

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

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

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

b: \(P-2=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}}>0\forall x\) thỏa mãn ĐKXĐ

nên P>2

\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\left(ĐKXĐ:x\ge0;x\ne9\right)\)

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

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{-3\sqrt{x}-3}{x-9}\)

\(b,M=P:Q\)

\(=\dfrac{-3\sqrt{x}-3}{x-9}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

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

\(=\dfrac{-3}{\sqrt{x}+3}\)

Ta thấy: \(\sqrt{x}\ge0\forall x\)

\(\Rightarrow\sqrt{x}+3\ge3\forall x\)

\(\Rightarrow\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\forall x\)

\(\Rightarrow\dfrac{-3}{\sqrt{x}+3}\ge\dfrac{-3}{3}=-1\)

hay \(M\ge-1\)

#Toru

\(Q=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)+9\sqrt{x}-4+\left(4x-4\sqrt{x}\right)\left(\sqrt{x}+4\right)}{x-16}\)

\(=\dfrac{x+4\sqrt{x}+4x\sqrt{x}+16x-4x-16\sqrt{x}}{x-16}\)

\(=\dfrac{13x+4x\sqrt{x}-12\sqrt{x}}{x-16}\)

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

\(=\dfrac{x-\sqrt{x}-5\sqrt{x}+8}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

2: \(P=A\cdot B=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

\(\Leftrightarrow P-2=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}>0\)

=>P>2

Sửa đề: 

Ta có: \(P=\frac{\sqrt{x}-4}{\sqrt{x}}\times\frac{x+\sqrt{x}+1}{\sqrt{x}-4}\)

 \(P=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)\(\left(ĐK:x>0\right)\)

Ta lấy \(P-2=\frac{x+\sqrt{x}+1}{\sqrt{x}}-2\)

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

                       \(=\frac{x-\sqrt{x}+1}{\sqrt{x}}\)

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

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

Vì \(x>0\Rightarrow\sqrt{x}>0\)

 \(\left(\sqrt{x}-\frac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}>0\)

\(\Rightarrow\frac{\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{3}{4}}{\sqrt{x}}>0\)

\(\Rightarrow P-2>0\)

\(\Rightarrow P>2\)

Học tốt 

a: \(Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

\(=\left(\dfrac{4}{\sqrt{x}+2}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)

\(=\dfrac{4\left(\sqrt{x}-2\right)-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\)

\(=\dfrac{-4\sqrt{x}-8}{\sqrt{x}+2}\cdot\dfrac{\sqrt{x}}{-\left(\sqrt{x}-3\right)}=\dfrac{4\sqrt{x}}{\sqrt{x}-3}\)

b: Q<4

=>Q-4<0

=>\(\dfrac{4\sqrt{x}}{\sqrt{x}-3}-4< 0\)

=>\(\dfrac{4\sqrt{x}-4\sqrt{x}+12}{\sqrt{x}-3}< 0\)

=>\(\dfrac{12}{\sqrt{x}-3}< 0\)

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

=>0<=x<9

Kết hợp ĐKXĐ, ta được: 0<x<9 và x<>4

\(a,Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\\ =\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right)\\ =\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{4x-8\sqrt{x}-8x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)

\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\\ =\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\\ =\dfrac{-4\sqrt{x}}{3-\sqrt{x}}\)

`b,` Để `Q<4` ta có :

\(\dfrac{-4\sqrt{x}}{3-\sqrt{x}}< 4\\ \Leftrightarrow\dfrac{-4\sqrt{x}}{3-\sqrt{x}}-4< 0\\ \Leftrightarrow\dfrac{-4\sqrt{x}-4\left(3-\sqrt{x}\right)}{3-\sqrt{x}}< 0\\ \Leftrightarrow-4\sqrt{x}-12+4\sqrt{x}< 0\\ \Leftrightarrow-12< 0\left(luon.dung\right)\)