\(a,\frac{6}{4+\sqrt{4-2\sqrt{3}}}=\frac{6}{4+\sqrt{\sqrt{3}^2-2\sqrt{3}+\sqrt{1}^2}}\)

\(=\frac{6}{4+\sqrt{\left(\sqrt{3}-\sqrt{1}\right)^2}}=\frac{6}{4+|\sqrt{3}-1|}=\frac{6}{3+\sqrt{3}}\)

\(=\frac{6}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{36}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{3}.\sqrt{12}}{\sqrt{3}\left(\sqrt{3}+1\right)}=\frac{\sqrt{12}}{\sqrt{3}+1}\)

\(d,\frac{1}{\sqrt{7-2\sqrt{10}}}+\frac{1}{\sqrt{7+2\sqrt{10}}}\)

\(=\frac{1}{\sqrt{\sqrt{5}^2-2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}+\frac{1}{\sqrt{\sqrt{5}^2+2.\sqrt{2}.\sqrt{5}+\sqrt{2}^2}}\)

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

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

\(=\frac{2\sqrt{5}}{\sqrt{5}^2-\sqrt{2}^2}=\frac{\sqrt{5.4}}{5-2}=\frac{\sqrt{20}}{3}\)

a)\(\frac{1}{\sqrt{2}+\sqrt{1}}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}+...+\frac{1}{\sqrt{1936}+\sqrt{1935}}=\)

\(\frac{\sqrt{2}-1}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}\)\(+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}-\sqrt{3}\right)\left(\sqrt{4}+\sqrt{3}\right)}+...\)\(+\frac{\sqrt{1936}-\sqrt{1935}}{\left(\sqrt{1936}-\sqrt{1935}\right)\left(\sqrt{1936}+\sqrt{1935}\right)}\)= \(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{1936}-\sqrt{1935}\)= \(-1-\sqrt{1935}\)

b)đề hơi sai bạn ạ mẫu thức số một bằng 0 còn đâu sửa lại đề đi nhé sau đó trục căn thức tương tự như mk làm nha

cảm ơn bạn nha mik ghi dề sai đề đúng là như thế này nè\(\frac{1}{\sqrt{1}-\sqrt{2}}\)  bạn giải giúp mik lun đi mik cảm ơn b nhìu lắm

a) \(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}}}{\sqrt{3}}\)

\(=\dfrac{\sqrt{3}}{3}+\dfrac{\sqrt{2}}{6}+\dfrac{\sqrt{\left(\dfrac{5}{12}-\dfrac{\sqrt{6}}{6}\right)\cdot3}}{3}\)

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

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

b) \(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}=...\)

c) \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=...\)

d) \(\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{2\sqrt{3-\sqrt{3+\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}-\sqrt{2}}\)

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

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

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

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

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

\(=\dfrac{2\sqrt{3-\left(\sqrt{3}+1\right)}\cdot\left(\sqrt{6}+\sqrt{2}\right)}{4}\)

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

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

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

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(6+2\sqrt{12}+2\right)}}{2}\)

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

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot\left(8+4\sqrt{3}\right)}}{2}\)

\(=\dfrac{\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}}{2}\)

\(=\dfrac{\sqrt{\left(4-3\right)\cdot4}}{2}\)

\(=\dfrac{\sqrt{1\cdot4}}{2}\)

\(=\dfrac{2}{2}\)

\(=1\)

bài 2 là bài 21 trong nâng cao phát triển toán 9, chắc bạn có chứ

Bài 1: Ta có:

\(x^2=5+\sqrt{13+\sqrt{5+\sqrt{13+\sqrt{5+\sqrt{13+...}}}}}=5+\sqrt{13+x}\)

\(\Rightarrow x^2-5=\sqrt{13+x}\Rightarrow x^4-10x^2+25=13+x\Leftrightarrow x^4-10x^2-x+12=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2-x-4\right)=0\)

Pt này có 1 nghiệm x = 3 và 3 nghiệm nhỏ hơn 2.

Vì \(x>\sqrt{4}=2\)

Vậy x = 3.