Question

a)Tính giá trị biểu thức:p= \(\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{3}+\sqrt{2}}\)

b)Chứng minh rằng nếu a,b,c là các số dương thỏa mãn a+c =2b thì ta luôn có

\(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}=\dfrac{2}{\sqrt{a}+\sqrt{c}}\)

Answer 1

\(P=\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{3}+\sqrt{2}}\)

=\(\dfrac{\left(3+2\sqrt{2.3}+2\right)\sqrt{3-2\sqrt{3.2}+2}}{\sqrt{3}+\sqrt{2}}\)

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

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

=\(3-2=1\)

Answer 2

ta có : \(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\ge2\sqrt{\dfrac{1}{\sqrt{ab}+\sqrt{ac}+\sqrt{bc}+b}}\)

\(\ge\dfrac{2}{\sqrt{a+b+c+b}}=\dfrac{2}{\sqrt{4b}}=\dfrac{2}{2\sqrt{b}}=\dfrac{1}{\sqrt{b}}=\dfrac{2}{\sqrt{a+c}}\ge\dfrac{2}{\sqrt{a}+\sqrt{b}}\)

dấu "=" xảy ra khi \(a=b=c\Leftrightarrow a+c=2b\Rightarrow\left(đpcm\right)\)