a: \(=\sqrt{3a}:\sqrt{b}\)

b: \(=\sqrt{a}:\sqrt{xy}\)

\(\sqrt{\dfrac{3+\sqrt{5}}{2}}=\sqrt{\dfrac{6+2\sqrt{5}}{2}}=\sqrt{\dfrac{\left(1+\sqrt{5}\right)^2}{2}=\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}}\)

\(\sqrt{\dfrac{3+\sqrt{5}}{2}}=\sqrt{\dfrac{6+2\sqrt{5}}{4}}=\dfrac{\sqrt{5}+1}{2}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}\sqrt{5}\)

Muốn chia các căn bậc hai của số a không âm cho căn bậc hai của số b dương ta có thể chia a cho cho b rồi khai phương kết quả đó

Nếu a lớn hoặc bằng không và b lớn hơn không thì ta có căn của a phần b bằng căn a phần căn b.

a) 

\(P=\left(\dfrac{b-a}{\sqrt{b}-\sqrt{a}}-\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}\right):\dfrac{\left(\sqrt{b}-\sqrt{a}\right)^2+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)

\(=\left[\sqrt{b}+\sqrt{a}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]:\dfrac{b-\sqrt{ab}+a}{\sqrt{a}+\sqrt{b}}\)

\(=\left(\sqrt{b}+\sqrt{a}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\right).\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

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

\(=\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)\(=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}\)

b) \(P=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\)

Vì \(\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b>0;\forall a\ge0;b\ge0;a\ne b\)

\(\sqrt{ab}\ge0\)\(\forall a\ge0;b\ge0\)

\(\Rightarrow P=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\ge0\)

Vậy...

cảm ơn tất cả mọi người

Ta có: \(M=\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}-\dfrac{a}{\sqrt{a}+\sqrt{b}}+\dfrac{b}{\sqrt{a}-\sqrt{b}}\)

\(=\dfrac{a\sqrt{a}-b\sqrt{b}-a\sqrt{a}+a\sqrt{b}+b\sqrt{a}+b\sqrt{b}}{\left(\sqrt{a}+b\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\dfrac{\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)

a: \(=\dfrac{\sqrt{a}-1}{\sqrt{a}\left(a-\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{1}\)

\(=a-1\)

b: \(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}\right)\)

\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{\sqrt{ab}+b+\sqrt{ab}-b}{\sqrt{a}\left(a-b\right)}\)

\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{1}{\sqrt{a}}\)

c: \(=\dfrac{a\sqrt{b}+b}{a-b}\cdot\sqrt{\dfrac{ab+b^2-2b\sqrt{ab}}{a^2+2a\sqrt{b}+b}}\cdot\left(\sqrt{a}+\sqrt{b}\right)\)

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

\(=\dfrac{\sqrt{b}\left(a+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\cdot\dfrac{\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)}{a+\sqrt{b}}=b\)

\(B=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}=\dfrac{11}{a-9}\)

a: Ta có: \(P=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right)\cdot\dfrac{\sqrt{a}-1}{a^2}\)

\(=\dfrac{4a-1}{\sqrt{a}-1}\cdot\dfrac{\sqrt{a}-1}{a^2}\)

\(=\dfrac{4a-1}{a^2}\)

b: Để P=3 thì \(4a-1=3a^2\)

\(\Leftrightarrow3a^2-4a+1=0\)

\(\Leftrightarrow\left(3a-1\right)\left(a-1\right)=0\)

hay \(a=\dfrac{1}{9}\)

a) ĐK: a>0; a≠1 

Ta có: \(P=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}}\right).\dfrac{\sqrt{a}-1}{a^2}\)

                  \(=\left(\dfrac{4a}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}-1}\right).\dfrac{\sqrt{a}-1}{a^2}\)

                  \(=\dfrac{4a-1}{\sqrt{a}-1}.\dfrac{\sqrt{a}-1}{a^2}=\dfrac{4a-1}{a^2}\)

b) Ta có: \(P=3\Leftrightarrow\dfrac{4a-1}{a^2}=3\Leftrightarrow3a^2=4a-1\Leftrightarrow3a^2-4a+1=0\)

               \(\Leftrightarrow\left(a-1\right)\left(3a-1\right)=0\)

               \(\Leftrightarrow\left[{}\begin{matrix}a=1\left(loại\right)\\a=\dfrac{1}{3}\left(tm\right)\end{matrix}\right.\)