a, Ta có : \(a^2+b^2\ge2ab\) ( cauchuy )

\(\Rightarrow a^2+2ab+b^2=\left(a+b\right)^2\ge4ab\)

\(\Rightarrow\dfrac{a+b}{ab}=\dfrac{a}{ab}+\dfrac{b}{ab}=\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

b, Ta có : \(a^2+b^2\ge2ab\) ( cauchuy )

\(\Rightarrow ab\le\dfrac{a^2+b^2}{2}\)

ab≤a2+b2/2

Tính nhanh 5/8+5/24+5/48+......+5/9800

Bài làm:

a) \(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)

\(A=3+2\sqrt{3}+1+\sqrt{\frac{25.48}{16}}-\frac{2}{\sqrt{4}}\)

\(A=4+2\sqrt{3}+\sqrt{25.3}-\frac{2}{2}\)

\(A=4+2\sqrt{3}+5\sqrt{3}-1\)

\(A=3+7\sqrt{3}\)

b) \(\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)

\(=\frac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)}-\frac{\sqrt{2}+1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}\)

\(A=\frac{4\left(3+\sqrt{5}\right)}{9-5}-\frac{3\left(\sqrt{5}-\sqrt{2}\right)}{5-2}-\frac{\sqrt{2}+1}{2-1}\)

\(A=3+\sqrt{5}-\sqrt{5}+\sqrt{2}-\sqrt{2}-1\)

\(A=2\)

Phần b mình viết nhầm tên thành A, bn sửa thành B nhé

c) \(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

\(C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)

\(C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

\(C=\sqrt{3}-1-2-\sqrt{3}\)

\(C=-3\)

tối nay mk sẽ trả lời , đợi nha, mk đi hk đã

ta có:

\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\),

 \(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}...\)

\(\frac{1}{10^2}=\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)

Từ trên => A < \(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)

=> \(A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...+\frac{1}{9}-\frac{1}{10}\)

=> \(A< \frac{1}{2}-\frac{1}{10}=\frac{5}{10}-\frac{1}{10}=\frac{4}{10}=\frac{2}{5}\)

=> \(A< \frac{2}{5}\)mà \(\frac{2}{5}< \frac{1}{2}\)

=> \(A< \frac{1}{2}\)=> \(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}< \frac{1}{2}\)

Chúc bn học tốt !

thanks, mình quên mk