Từ zyz = 4 => \(\sqrt{xyz}=\sqrt{4}=2\)

Ta có:A = \(\frac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+2}+\frac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\frac{2\sqrt{z}}{\sqrt{zx}+2\sqrt{z}+2}\)

A = \(\frac{\sqrt{xz}}{\sqrt{xyz}+\sqrt{xz}+2\sqrt{z}}+\frac{\sqrt{xyz}}{\sqrt{xy^2z}+\sqrt{xyz}+\sqrt{xz}}+\frac{2\sqrt{z}}{\sqrt{xz}+2\sqrt{z}+2}\)

A = \(\frac{\sqrt{xz}}{\sqrt{xz}+2\sqrt{z}+2}+\frac{2}{2\sqrt{z}+\sqrt{xz}+2}+\frac{2\sqrt{z}}{\sqrt{xz}+2\sqrt{z}+2}\)

A = \(\frac{\sqrt{xz}+2\sqrt{z}+2}{\sqrt{xz}+2\sqrt{z}+2}=1\)

a) \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)

b) \(\sqrt{18-2\sqrt{65}}+\sqrt{18+2\sqrt{65}}=\sqrt{\left(\sqrt{13}-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{13}+\sqrt{5}\right)^2}\)

\(=\sqrt{13}-\sqrt{5}+\sqrt{13}+\sqrt{5}=2\sqrt{13}\)

c) \(\sqrt{12+2\sqrt{35}}-\sqrt{12-2\sqrt{35}}=\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}\)

\(=\sqrt{7}+\sqrt{5}-\sqrt{7}+\sqrt{5}=2\sqrt{5}\)

d) \(\sqrt{9-4\sqrt{5}}+\sqrt{6+2\sqrt{5}}=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\sqrt{5}-2+\sqrt{5}+1=2\sqrt{5}-1\)

e) \(\sqrt{11+6\sqrt{2}}+\sqrt{6+4\sqrt{2}}-2\sqrt{2}=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(2+\sqrt{2}\right)^2}-2\sqrt{2}\)

\(=3+\sqrt{2}+2+\sqrt{2}-2\sqrt{2}=5\)

f) \(\sqrt{17-4\sqrt{9+4\sqrt{5}}}=\sqrt{17-4\sqrt{\left(\sqrt{5}+2\right)^2}}=\sqrt{17-4\sqrt{5}-8}\)

\(=\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)

ĐK : x > 0

Với x > 0 thì √x > 0

nên để \(\frac{\sqrt{x}-2}{\sqrt{x}}< 0\) thì √x - 2 < 0 <=> √x < 2 <=> x < 4

Kết hợp với ĐK => Với 0 < x < 4 thì \(\frac{\sqrt{x}-2}{\sqrt{x}}< 0\)

\(D=ℝ\)

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

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

\(\Leftrightarrow\left|x-2\right|-\left|x+1\right|=-3\)

Ta có: \(\left|x-2\right|-\left|x+1\right|=\left|x-2\right|-\left|x-2+3\right|\ge\left|x-2\right|-3-\left|x-2\right|=-3\)

Dấu \(=\)khi \(3\left(x-2\right)\ge0\Leftrightarrow x\ge2\).

Vậy nghiệm của phương trình đã cho là \(x\ge2\).

\(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\frac{5}{\sqrt{6}+1}=1\)

\(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\frac{5}{\sqrt{6}+1}=\frac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\frac{5\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}\)

\(=\sqrt{6}-\frac{5\left(\sqrt{6}-1\right)}{6-1}=\sqrt{6}-\sqrt{6}+1=1\)

\(\frac{4}{3+\sqrt{5}}-\frac{8}{1+\sqrt{5}}+\frac{15}{\sqrt{5}}=\frac{4\left(3-\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{8\left(\sqrt{5}-1\right)}{\left(1+\sqrt{5}\right)\left(\sqrt{5}-1\right)}+3\sqrt{5}\)

\(=\frac{4\left(3-\sqrt{5}\right)}{9-5}-\frac{8\left(\sqrt{5}-1\right)}{5-1}+3\sqrt{5}=3-\sqrt{5}-2\sqrt{5}+2+3\sqrt{5}=5\)

Trả lời :

= 5

Nhs

Hok T~

Ta có: \(\left(x-y\right)^2\ge0\Leftrightarrow x^2+y^2\ge2xy\)

\(\left(y-z\right)\ge0\Leftrightarrow y^2+z^2\ge2yz\)

\(\left(z-x\right)^2\ge0\Leftrightarrow z^2+x^2\ge2zx\)

\(\left(x-1\right)^2\ge0\Leftrightarrow x^2+1\ge2x\)

\(\left(y-1\right)^2\ge0\Leftrightarrow y^2+1\ge2y\)

\(\left(z-1\right)^2\ge0\Leftrightarrow z^2+1\ge2z\)

Cộng lại vế với vế ta được: 

\(3\left(x^2+y^2+z^2\right)+3\ge2xy+2yz+2zx+2x+2y+2z\)

\(\Leftrightarrow Q\ge\frac{2\left(x+y+yz+xy+yz+zx\right)-3}{3}=3\)

Dấu \(=\)khi \(x=y=z=1\).

Ta có: \(x+y+z+xy+yz+xz\le x+y+z+\frac{\left(x+y+z\right)^2}{3}\)

=> \(\left(x+y+z\right)^2+3\left(x+y+z\right)\ge3.6=18\)

<=> \(\left(x+y+z\right)^2+3\left(x+y+z\right)-18\ge0\)

<=> \(\left(x+y+z-3\right)\left(x+y+z+6\right)\ge0\)

<=> \(x+y+z\ge3\)(vì x + y + z + 6 > 0 vì x,y,z > 0)

Do đó: \(Q=x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=\frac{3^2}{3}=3\)

Dấu "=" xảy ra<=> x  = y= z và x + y + z = 3 <=> x = y = z = 1

Vậy MinQ = 3 <=> x = y= z = 1