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

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

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

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

Lời giải:

$\sqrt{7+2\sqrt{10}}=\sqrt{2+5+2\sqrt{2.5}}=\sqrt{(\sqrt{2}+\sqrt{5})^2}=\sqrt{2}+\sqrt{5}$

\(\sqrt[3]{3\sqrt[3]{3}-3\sqrt[3]{2}-1}=\sqrt[3]{(1-\sqrt[3]{2})^3}=1-\sqrt[3]{2}\)

Do đó:

\(\text{TS}=\sqrt[3]{2}+\sqrt{2}+\sqrt{5}+1-\sqrt[3]{2}=\sqrt{2}+\sqrt{5}+1=\text{MS}\)

\(A=\frac{\text{TS}}{\text{MS}}=1\)

\(A=\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

\(=\dfrac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-\sqrt{1}\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{\sqrt{100}-\sqrt{99}}{\left(\sqrt{100}-\sqrt{99}\right)\left(\sqrt{100}+\sqrt{99}\right)}\)

\(=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}=\sqrt{100}-\sqrt{1}=10-1=9\)

cả 2 ý bạn trục căn thức ở mấu là xong nhé:

vd: \(\dfrac{1}{\sqrt{1}+\sqrt{2}}=\dfrac{\sqrt{1}-\sqrt{2}}{-1}\). Rồi tương tự như vậy

\(\dfrac{\sqrt[3]{4}+\sqrt[3]{2}+2}{\sqrt[3]{4}+\sqrt[3]{2}+1}\)​

\(=\dfrac{\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{8}}{\sqrt[3]{4}+\sqrt[3]{2}+1}=\dfrac{\sqrt[3]{2}\left(\sqrt[3]{2}+1+\sqrt[3]{4}\right)}{\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1}}=\sqrt[3]{2}\)

Bài 1: 

a) \(\dfrac{a+\sqrt{a}}{\sqrt{a}}=\sqrt{a}+1\)

b) \(\dfrac{\sqrt{\left(x-3\right)^2}}{3-x}=\dfrac{\left|x-3\right|}{3-x}=\pm1\)

Bài 2: 

a) \(\dfrac{\sqrt{9x^2-6x+1}}{9x^2-1}=\dfrac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}=\pm\dfrac{1}{3x+1}\)

b) \(4-x-\sqrt{x^2-4x+4}=4-x-\left|x-2\right|=\left[{}\begin{matrix}6-2x\left(x\ge2\right)\\2\left(x< 2\right)\end{matrix}\right.\)

Với n\(\in N\)^* có: \(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)\(=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left(n+1-n\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}\)\(=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

\(\Rightarrow\)\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\) (*)

a) Áp dụng (*) vào T

\(\Rightarrow T=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}\)\(=1-\dfrac{1}{10}=\dfrac{9}{10}\)

b) Có \(VT=1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)\(=1-\dfrac{1}{\sqrt{n+1}}=\dfrac{4}{5}\)

\(\Leftrightarrow\sqrt{n+1}=5\Leftrightarrow n=24\) (tm)

Vậy n=24.

a: \(=\dfrac{6+4\sqrt{2}}{\sqrt{2}+2+\sqrt{2}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}-2+\sqrt{2}}\)

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

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

=căn 2+1+căn 2-1=2căn 2

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

\(a,=\dfrac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\\ =\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{2}=\dfrac{-2-6\sqrt{15}}{2}=-1-3\sqrt{15}\)

\(b,=x\sqrt{2\left(x+1\right)}+\sqrt{\dfrac{2\left(x+1\right)^2}{x+1}}-\sqrt{\dfrac{16\left(x+1\right)}{2}}\\ =x\sqrt{2\left(x+1\right)}+\sqrt{2\left(x+1\right)}-2\sqrt{2\left(x+1\right)}\\ =\sqrt{2\left(x+1\right)}\left(x+1-2\right)=\left(x-1\right)\sqrt{2\left(x+1\right)}\)

a.\(=\dfrac{\left(\sqrt{5}-2\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}-\dfrac{\left(2\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}\)

\(=\dfrac{5-\sqrt{15}-2\sqrt{15}+6}{5-3}-\dfrac{10+2\sqrt{15}+\sqrt{15}+3}{5-3}\)

=\(\dfrac{11-3\sqrt{15}-13-3\sqrt{15}}{2}=\dfrac{-2-6\sqrt{15}}{2}\)

=\(-1-3\sqrt{15}\)

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

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

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

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