\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)

\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)

\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)

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

Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)

\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)

1:

\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

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

\(=\dfrac{20-6}{2}=7\)

Ta có: M=A+B

\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{11\sqrt{x}-3}{x-9}\)

\(=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)

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

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

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

d. \(=\dfrac{\sqrt{a}\left(\sqrt{a}-3\right)}{\sqrt{a}-3}=\sqrt{a}\)

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

\(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)

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

\(=\sqrt{a}-\sqrt{b}-\sqrt{a}+\sqrt{b}=0\)

   Đk: \(a,b\ge0\) và \(a\ne b\)

 \(\dfrac{a+b-\sqrt{2ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)

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

\(=\sqrt{a}-\sqrt{b}-\left(\sqrt{a}-\sqrt{b}\right)=0\)

Câu a, bạn coi lại đề xem $a^2=6-3\sqrt{3}$ hay $a=6-3\sqrt{3}$???

b.

\(B=\frac{\sqrt{(x-2)+(x+2)+2\sqrt{(x-2)(x+2)}}}{\sqrt{x^2-4}+x+2}\)

\(=\frac{\sqrt{(\sqrt{x-2}+\sqrt{x+2})^2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}=\frac{1}{\sqrt{x+2}}\)

\(=\frac{1}{\sqrt{3+\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{(\sqrt{5}+1)^2}}=\frac{\sqrt{2}}{\sqrt{5}+1}\)

Nguyễn Hoàng trung: Chả qua nếu $a=6-3\sqrt{3}; b=2+\sqrt{3}$ thì kết quả sẽ đẹp hơn. Còn như đề thì vẫn rút gọn được.

\(A=\frac{a-\sqrt{ab}+b}{(\sqrt{a}-\sqrt{b})^2}=\frac{a-\sqrt{ab}+b}{a-2\sqrt{ab}+b}\)

\(2a^2=12-6\sqrt{3}=(3-\sqrt{3})^2\Rightarrow a=\frac{3-\sqrt{3}}{\sqrt{2}}\) (do $a\geq 0$)

\(2b^2=4+2\sqrt{3}=(\sqrt{3}+1)^2\Rightarrow b=\frac{\sqrt{3}+1}{\sqrt{2}}\) (do $b\geq 0$)

\(\Rightarrow a+b=2\sqrt{2}; ab=\frac{\sqrt{3}(\sqrt{3}-1)(\sqrt{3}+1)}{2}=\sqrt{3}\)

Do đó: $A=\frac{2\sqrt{2}-\sqrt[4]{3}}{2\sqrt{2}-2\sqrt[4]{3}}$

a: \(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\)

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

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

b: \(\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}\)

\(=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{15}\left(\sqrt{2}-1\right)}\)

\(=\dfrac{\sqrt{6}}{\sqrt{15}}=\sqrt{\dfrac{6}{15}}=\sqrt{\dfrac{2}{5}}=\dfrac{\sqrt{10}}{5}\)

c: \(\sqrt{9a}+\sqrt{81a}+3\sqrt{25a}-16\sqrt{49a}\)

\(=3\sqrt{a}+9\sqrt{a}+3\cdot5\sqrt{a}-16\cdot7\sqrt{a}\)

\(=27\sqrt{a}-112\sqrt{a}=-85\sqrt{a}\)

d: \(\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}=\dfrac{\left(\sqrt{ab}-\sqrt{bc}\right)\left(\sqrt{ab}+\sqrt{bc}\right)}{\sqrt{ab}-\sqrt{bc}}\)

\(=\sqrt{ab}+\sqrt{bc}\)

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

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

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

\(=a\cdot\sqrt{a^2+3\cdot a\cdot\sqrt{a}\cdot b\cdot\sqrt{b}}\)

e: ĐKXĐ: a>=0 và a<>1

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

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

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

\(=\left(\sqrt{a}+1\right)^2\cdot\left(a-\sqrt{a}+1\right)\)