a, Với a ; b >= 0 ; a khác b
\(=\dfrac{a+2\sqrt{ab}+b-4\sqrt{ab}+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a-b}\)
\(=\dfrac{a-2\sqrt{ab}+b+\sqrt{ab}-b}{a-b}=\dfrac{a-\sqrt{ab}}{a-b}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)
b, Ta có : \(b=4-2\sqrt{3}\Rightarrow\sqrt{b}=\sqrt{3}-1\)
Thay a = 3 ; b = căn 3 - 1 ta được :
\(\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{3}-1}=\dfrac{\sqrt{3}}{2\sqrt{3}-1}=\dfrac{\sqrt{3}.\left(2\sqrt{3}+1\right)}{11}=\dfrac{6+\sqrt{3}}{11}\)
a) \(M=\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{a-b}+\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a-\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)
b) Ta có: \(\sqrt{b}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
\(\Rightarrow M=\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{3}-1}=\dfrac{\sqrt{3}}{2\sqrt{3}-1}=\dfrac{\sqrt{3}\left(2\sqrt{3}+1\right)}{\left(2\sqrt{3}-1\right)\left(2\sqrt{3}+1\right)}=\dfrac{6+\sqrt{3}}{11}\)
a) Ta có: \(M=\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{4\sqrt{ab}}{a-b}+\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}\)
\(=\dfrac{a+2\sqrt{ab}+b-4\sqrt{ab}+\sqrt{ab}-b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{a-\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}+\sqrt{b}}\)
b) Thay a=3 và \(b=4-2\sqrt{3}\) vào M, ta được:
\(M=\dfrac{\sqrt{3}}{\sqrt{3}+\sqrt{3}-1}=\dfrac{\sqrt{3}}{2\sqrt{3}-1}=\dfrac{6+\sqrt{3}}{11}\)
