\(\sqrt{25}=5;\sqrt{169}=13;\sqrt{3600}=60;\sqrt{4}=2;\sqrt{9}=3;\sqrt{0}=0;\sqrt{81}=9\)

0 , 81 = 0 , 9   v ì   0 , 9 ≥ 0   v à   0 , 9 2 = 0 , 81

Gọi hai số là \(a\) và \(b\).

Ta có:

\({a+b=\sqrt{15}\\a-b=\sqrt{11}}\)

Nhân hai vế:

\(\left(\right. a + b \left.\right) \left(\right. a - b \left.\right) = a^{2} - b^{2}\)

\(a^{2} - b^{2} = \sqrt{15} \cdot \sqrt{11} = \sqrt{165}\)

Mà:

\(a^{2} - b^{2} = \left(\right. a - b \left.\right) \left(\right. a + b \left.\right)\)

Ta cần tích:

\(a b = \frac{\left(\right. a + b \left.\right)^{2} - \left(\right. a - b \left.\right)^{2}}{4}\) \(a b = \frac{15 - 11}{4} = \frac{4}{4} = 1\)

* Tích hai số là 1.

\(\sqrt{\sqrt{5}-\sqrt{3x}}=\sqrt{8+\sqrt{60}}\)

\(\sqrt{\sqrt{5-\sqrt{3x}=}\sqrt{8+\sqrt{60}}}\) k mk nha

Trả lời:

\(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)

\(=\sqrt{45-12\sqrt{5}+4}-\sqrt{45+12\sqrt{5}+4}\)

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

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

\(=-4\)

Học tốt 

\(\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)

\(=\sqrt{45-12\sqrt{5}+4}-\sqrt{45+12\sqrt{5}+4}\)

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

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

\(=3\sqrt{5}-3\sqrt{5}-4=-4\)