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a) \(\sqrt{\frac{3a}{4}}.\sqrt{\frac{4a}{27}}=\frac{\sqrt{3a}}{2}.\frac{\sqrt{4a}}{3\sqrt{3}}=\frac{\sqrt{3}.\sqrt{a}.2.\sqrt{a}}{6\sqrt{3}}=\frac{a.2\sqrt{3}}{6\sqrt{3}}=\frac{a}{3}\)
b) \(\sqrt{15x}.\sqrt{\frac{60}{x}}=\sqrt{15x}.\frac{2\sqrt{15}}{\sqrt{x}}=\frac{30\sqrt{x}}{\sqrt{x}}=30\)
a) \(\sqrt{\frac{3a}{4}}.\sqrt{\frac{4a}{27}}=\sqrt{\frac{3a}{4}.\frac{4a}{27}}=\sqrt{\frac{1}{9}.a^2}=\sqrt{\frac{1}{9}}.\sqrt{a^2}=\frac{1}{3}.a\)( Vì \(a\ge0\)nên \(\sqrt{a^2}=\left|a\right|=a\))
b) \(\sqrt{15x}.\sqrt{\frac{60}{x}}=\sqrt{15x.\frac{60}{x}}=\sqrt{900}=30\)

a) \(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=2\sqrt{\left(a-3\right)^2}=2.\left|a-3\right|=2\left(a-3\right)=2a-6\) (Vì \(a\ge3\) )
b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=3\sqrt{\left(b-2\right)^2}=3\left|b-2\right|=3\left(2-b\right)\)
\(=6-3b\) (vì b < 2 )
b) \(\sqrt{27.48\left(1-a\right)^2}=\sqrt{27.3.16.\left(1-a\right)^2}=\sqrt{81.16.\left(1-a\right)^2}\)
\(=\sqrt{9^2.4^2.\left(1-a\right)^2}=9.4\sqrt{\left(1-a\right)^2}=36.\left|1-a\right|=36\left(1-a\right)=36-36a\) (vì a > 1)

a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{3-2\sqrt{3}+1}-\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\)
b) \(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{9+6\sqrt{2}+2}-3+\sqrt{2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
c) \(\sqrt{25x^2}-2x=-5x-2x=-7x\)(vì x < 0)
d) \(x-5+\sqrt{25-10x+x^2}=x-5+\sqrt{\left(5-x\right)^2}=x-5+x-5=2x-10\) (vì x > 5)

Bài làm:
Ta có: \(\sqrt{7,2\cdot0,8\cdot\left(x^2-4x+4\right)}\)
\(=\sqrt{\frac{36}{5}\cdot\frac{4}{5}\cdot\left(x-2\right)^2}\)
\(=\sqrt{\frac{144}{25}\cdot\left(x-2\right)^2}\)
Vì \(x\ge2\) => \(Bt=\frac{12}{5}\cdot\left(x-2\right)=\frac{12x-24}{5}\)
\(\sqrt{7,2\cdot0,8\left(x^2-4x+4\right)}\)
\(=\sqrt{\frac{36}{5}\cdot\frac{4}{5}\cdot\left(x-2\right)^2}\)
\(=\sqrt{\frac{144}{25}\cdot\left(x-2\right)^2}\)
\(=\sqrt{\left(\frac{12}{5}\right)^2\cdot\left(x-2\right)^2}\)
\(=\left|\frac{12}{5}\right|\cdot\left|x-2\right|\)
Với x >= 2
Bt <=> \(\frac{12}{5}\cdot\left(x-2\right)=\frac{12x-24}{5}\)

a) \(A=\sqrt{11+6\sqrt{2}}-3+\sqrt{2}=\sqrt{9+2.3\sqrt{2}+2}-3+\sqrt{2}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-3+\sqrt{2}=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}\)
b) x<0
\(B=\sqrt{9x^2}-2x=\left|3x\right|-2x=-3x-2x=-5x\)
c) x>4
\(C=x-4+\sqrt{16-8x+x^2}=x-4+\sqrt{\left(4-x\right)^2}\)
\(=x-4+\left|4-x\right|=x-4+x-4=2x-8\)

#)Giải :
a) Câu trc của bn mk có giải rùi, thắc mắc vô Thống kê hđ của mk xem lại nhé !
b) Để \(P>0\Rightarrow\frac{x-1}{\sqrt{x}}>0\Rightarrow x-1>0\left(\sqrt{x}>0\right)\Rightarrow x>1\)
c) Bó tay @@
\(a,P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{1}{x-\sqrt{x}}\right):\left(\frac{1}{\sqrt{x}+1}+\frac{2}{x-1}\right)\)
\(=\left(\frac{x}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\left(\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\frac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{\sqrt{x}-1+2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(x-1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x-1}{\sqrt{x}}\)
Vậy với \(x>0;x\ne1\)thì \(P=\frac{x-1}{\sqrt{x}}\)
\(b,\)Để \(P>0\Leftrightarrow\frac{x-1}{\sqrt{x}}>0\Leftrightarrow x-1>0\Leftrightarrow x>1\left(\sqrt{x}>0\right)\)
\(\sqrt{2x}.\sqrt{72x}-4x=\sqrt{2x}.\sqrt{2.36.x}-4x\)
\(=\sqrt{2x}.\sqrt{36}.\sqrt{2x}-4x=\sqrt{2x}^2.6-4x\)
\(=2x.6-4x=12x-4x=8x\)
\(\sqrt{2x}.\sqrt{72x}-4x\)
\(=\sqrt{2x.72x}-4x\)
\(=\sqrt{144x^2}-4x\)
\(=12\left|x\right|-4x\)
\(=12x-4x\left(x\ge0\right)=8x\)