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Lời giải:
Để ý rằng \(a^2+1=a^2+ab+bc+ac=(a+b)(a+c)\)
Tương tự thì \(b^2+1=(b+c)(b+a)\)
\(c^2+1=(c+a)(c+b)\)
\(\Rightarrow A=\sqrt{(a+b)^2(b+c)^2(c+a)^2}=|(a+b)(b+c)(c+a)|\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Viết lại đề cho bạn nè
\(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
= \(\sqrt{9-6\sqrt{3}+3}-\sqrt{12-12\sqrt{3}+9}\)
= \(\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}\right)^2-2.3.\sqrt{12}+3^2}\)
= \(\sqrt{\left(3-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{12}-3\right)^2}\)
= |\(3-\sqrt{3}\)| - |\(\sqrt{12}-3\)|
= \(3-\sqrt{3}-\sqrt{12}+3\)
= \(6-\sqrt{3}-2\sqrt{3}\)
= \(6-3\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
=15√20 -3√45+2√5
=15\(\sqrt{4x5}\)-3\(\sqrt{9x5}\)+2√5
=30√5 -9√5+2√5
=23√5
\(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\) =\(\left(150\sqrt{2}-45\sqrt{2}+10\sqrt{2}\right):\sqrt{10}\)
=\(115\sqrt{2}:\sqrt{10}\)
chắc vậy
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\frac{16a^4b^6}{128a^6b^6}}=\sqrt{\frac{1}{8a^2}}=\frac{\sqrt{1}}{\sqrt{8a^2}}=\frac{1}{\sqrt{2}\sqrt{4}\sqrt{a}}\)
=\(\frac{1}{2\sqrt{2}a}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{\left(x+1\right)^2}{\left(x-1\right)^2}:\frac{2\left(x+1\right)^2}{4\left(x-1\right)^2}=\frac{\left(x+1\right)^2}{\left(x-1\right)^2}.\frac{4\left(x-1\right)^2}{2\left(x+1\right)^2}=2\)