Question

[Ngô Trần Bảo Quang]

1/Tính 

a/ \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}\)

b/\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

2/ So sánh \(\sqrt{a+b}\) và \(\sqrt{a}+\sqrt{b}\)

3/CÁch phân biệt CaO với P2O5

                             SO2 và O2

Viết PTHH

Giúp anh hai mấy bài làm đc kb nhé 

Answer 1

[Ahwi]

1)

a/ \(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}=\frac{\sqrt{3\cdot2}+\sqrt{2\cdot7}}{2\sqrt{3}+2\sqrt{7}}\)

                                  \(=\frac{\sqrt{2}\left(\sqrt{3}+\sqrt{7}\right)}{2\left(\sqrt{3}+\sqrt{7}\right)}=\frac{\sqrt{2}}{2}\)

b/ \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\text{​​}\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{6}+\sqrt{8}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

​​\(=\frac{\left(\sqrt{2}+\text{​​}\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{3\cdot2}+\sqrt{4\cdot2}+\sqrt{2\cdot2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\text{​​}\sqrt{3}+\sqrt{4}\right)+\left(\sqrt{2}\cdot\sqrt{3}+\sqrt{4}\cdot\sqrt{2}+\sqrt{2}\cdot\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\text{​​}\sqrt{3}+\sqrt{4}\right)+\sqrt{2}\left(\sqrt{2}+\text{​​}\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\frac{\left(\sqrt{2}+\text{​​}\sqrt{3}+\sqrt{4}\right)\left(\sqrt{2}+1\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\sqrt{2}+1\)

Answer 2

[Ahwi]

2)

+ Ta Có :

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

\(\sqrt{a}+\sqrt{b}\Rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2=\left(\sqrt{a}\right)^2+2\sqrt{a}\cdot\sqrt{b}+\left(\sqrt{b}\right)^2\)

                                                                  \(=a+2\sqrt{a}\cdot\sqrt{b}+b\)

+ Ta Lại có \(2\sqrt{a}\cdot\sqrt{b}>0\)

Tiếp tục có    \(a+b\)  và   \(a+2\sqrt{a}\cdot\sqrt{b}+b\)

                  \(\Rightarrow a+b< a+b+2\sqrt{a}\cdot\sqrt{b}\)

\(\Rightarrow\sqrt{a+b}< \sqrt{a}+\sqrt{b}\)