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\(=\left(3-a\right)^2-\sqrt{0,2.180a^2}=9-6a+a^2-\sqrt{36a^2}=9-6a+a^2-6.lal\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\sqrt{27\cdot48\left(1-a^2\right)}\)
\(=\sqrt{3^4\cdot4^2\cdot\left(1-a^2\right)}\)
\(=36\sqrt{1-a^2}\)
c) Ta có: \(\sqrt{5a}\cdot\sqrt{45a}-3a\)
\(=15a-3a=12a\)
b) Ta có: \(B=\dfrac{1}{a-b}\cdot\sqrt{a^4\cdot\left(a-b\right)^2}\)
\(=\dfrac{1}{a-b}\cdot a^2\cdot\left(a-b\right)\)
\(=a^2\)
d) Ta có: \(D=\left(3-a\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)
\(=a^2-6a+9-\sqrt{36a^2}\)
\(=a^2-6a+9-\left|6a\right|\)
\(=\left[{}\begin{matrix}a^2-6a+9-6a\left(a\ge0\right)\\a^2-6a+9+6a\left(a< 0\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}a^2-12a+9\\a^2+9\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\left|\frac{a}{2}\right|=\frac{a}{2}\)
do \(a\ge0\)
b, \(\sqrt{13a}.\sqrt{\frac{52}{a}}=\sqrt{\frac{676a}{a}}=\sqrt{676}=26\)
c, \(\sqrt{5a}.\sqrt{45a}-3a=\sqrt{225a^2}-3a=\left|15a\right|-3a\)
\(=15a-3a=12a\)do a > 0
d, \(=\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)
\(=\left(3-a\right)^2-\sqrt{36a^2}=\left(3-a\right)^2-\left|6a\right|\)
Với \(a\ge0\Rightarrow\left(3-a\right)^2-6a=a^2-6a+9-6a=a^2-12a+9\)
Với \(a< 0\Rightarrow\left(3-a\right)^2+6a=a^2-6a+9+6a=a^2+9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/ \(\left(3-a\right)^2-\sqrt{\frac{180a^2}{5}}=a^2-6a+9-6\left|a\right|\)
Nếu \(a\ge0\) thì \(a^2-6a+9-6\left|a\right|=a^2-12a+9\)
Nếu \(a< 0\) thì \(a^2-6a+9-6\left|a\right|=a^2+9\)
b/ \(\sqrt{150}-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)
\(=5\sqrt{6}-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{2}\)
\(5\sqrt{6}-20\sqrt{2}=5\sqrt{2}\left(\sqrt{3}-4\right)\)
c/ Bạn viết lại đề nhé :)
![](https://rs.olm.vn/images/avt/0.png?1311)
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) \(\left(3-a\right)^2-\sqrt{0,2}\cdot\sqrt{180a^2}=a^2-6a+9-\sqrt{36a^2}=a^2-6a+9-6a=a^2-12a+9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\sqrt{9.3.3.16\left(1-a^2\right)}=3.3.4.\left|1-a\right|=36\left(a-1\right)\)
\(B=\frac{1}{a-b}a^2.\left|a-b\right|=\frac{a^2\left(a-b\right)}{a-b}=a^2\)
\(C=\sqrt{5.45.a^2}-3a=\sqrt{5^2.3^2.a^2}-3a=15\left|a\right|-3a=15a-3a=12a\)
\(D=\left(3-a\right)^2-\sqrt{\frac{2.180}{10}a^2}=\left(3-a\right)^2-6\left|a\right|\)
rút gọn biểu thức
Ta có: \(\left(a-3\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)
\(=\left(a-3\right)^2-\sqrt{36a^2}\)
\(=a^2-6a+9-6a\)
\(=a^2-12a+9\)