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1) \(\sqrt{\frac{24}{3}}\cdot\sqrt{\frac{3a}{8}}=\sqrt{\frac{72a}{24}}=\sqrt{3a}\)
2) \(\sqrt{13a}\cdot\sqrt{\frac{52}{a}}=\sqrt{\frac{13a\cdot52}{a}}=\sqrt{676}=26\)
3) \(\sqrt{5a}\cdot\sqrt{45a}-3a=\sqrt{225a^2}-3a=15a-3a=12a\)
4) \(\left(3-a\right)^2-\sqrt{0,2}\cdot\sqrt{180a^2}=a^2-6a+9-\sqrt{36a^2}=a^2-6a+9-6a=a^2-12a+9\)
\(A=\sqrt{9.3.3.16\left(1-a^2\right)}=3.3.4.\left|1-a\right|=36\left(a-1\right)\)
\(B=\frac{1}{a-b}a^2.\left|a-b\right|=\frac{a^2\left(a-b\right)}{a-b}=a^2\)
\(C=\sqrt{5.45.a^2}-3a=\sqrt{5^2.3^2.a^2}-3a=15\left|a\right|-3a=15a-3a=12a\)
\(D=\left(3-a\right)^2-\sqrt{\frac{2.180}{10}a^2}=\left(3-a\right)^2-6\left|a\right|\)
a/ \(\sqrt{\frac{2a}{3}}\cdot\sqrt{\frac{3a}{8}}\)
\(=\sqrt{\frac{2a}{3}\cdot\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\sqrt{\frac{a^2}{2^2}}=\sqrt{\left(\frac{a}{2}\right)^2}=\left|\frac{a}{2}\right|\)
mak ta có \(a\ge0\)
\(\Rightarrow\left|\frac{a}{2}\right|=\frac{a}{2}\)\(\Rightarrow\sqrt{\frac{2a}{3}}\cdot\sqrt{\frac{3a}{8}}=\frac{a}{2}\)
b/ \(\sqrt{13a}\cdot\sqrt{\frac{52}{a}}\)
\(=\sqrt{13a\cdot\frac{52}{a}}=\sqrt{\frac{13a\cdot52}{a}}=\sqrt{13\cdot52}=\sqrt{13\cdot13\cdot4}=\sqrt{13^2\cdot2^2}=\sqrt{\left(13\cdot2\right)^2}=13\cdot2=26\)
c/ \(\sqrt{5a}\cdot\sqrt{45}-3a\)
\(=\sqrt{5a\cdot45a}-3a=\sqrt{5a\cdot5a\cdot9}-3a\)
\(=\sqrt{5^2\cdot a^2\cdot3^2}-3a=\left|5\cdot a\cdot3\right|-3a\)
\(=15\left|a\right|-3a\)
Có \(a\ge0\Rightarrow\left|a\right|=a\)
\(\Rightarrow15\left|a\right|-3a=15a-3a=12a\)
\(\Rightarrow\sqrt{5a}\cdot\sqrt{45}-3a=12a\)
d/ \(\left(3-a\right)^2-\sqrt{0,2}\cdot\sqrt{180a^2}\)
\(=\left(3-a\right)^2-\sqrt{0,2\cdot180a^2}\)
\(=\left(3-a\right)^2-\sqrt{0,2\cdot9\cdot2\cdot10\cdot a^2}\)
\(=\left(3-a\right)^2-\sqrt{4\cdot9\cdot a^2}\)
\(=\left(3-a\right)^2-\sqrt{2^2\cdot3^2\cdot a^2}\)
\(=\left(3-a\right)^2-\left|2\cdot3\cdot a\right|\)
\(=\left(3-a\right)^2-6\left|a\right|=9-6a+a^2-6\left|a\right|\)
Chia làm 2 Trường Hợp:
+ TH1 : \(9-6a+a^2-6a=9-12a+a^2\left(a\ge0\right)\)
+ TH2 : \(9-6a+a^2-\left(-6a\right)=9+a^2\left(a< 0\right)\)
Bài 1:
a: \(=\sqrt{\dfrac{7-4\sqrt{3}}{2-\sqrt{3}}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
Bài 2:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
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