a)

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

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

b)

\(\left(2+\dfrac{11-\sqrt{11}}{1-\sqrt{11}}\right)\left(2+\dfrac{\sqrt{11}+11}{\sqrt{11}+1}\right)\\ =\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{-\left(\sqrt{11}-1\right)}\right)\left(2+\dfrac{\sqrt{11}\left(1+\sqrt{11}\right)}{\sqrt{11}+1}\right)\\ =\left(2-\sqrt{11}\right)\left(2+\sqrt{11}\right)\\ =4-11\\ =-7\)

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

=(căn 3-căn 2)(căn 3+căn 2)

=3-2=1

b: \(=\left(2-\dfrac{\sqrt{11}\left(\sqrt{11}-1\right)}{\sqrt{11}-1}\right)\left(2+\dfrac{\sqrt{11}\left(\sqrt{11}+1\right)}{\sqrt{11}+1}\right)\)

=(2-căn 11)(2+căn 11)

=4-11

=-7

Bài 1:

ĐKXĐ: \(\dfrac{5}{x^2+6}>=0\)

=>\(x^2+6>0\)

mà \(x^2+6>=6>0\forall x\)

nên \(x\in R\)

Bài 2:

a: Sửa đề: \(\dfrac{3}{\sqrt{2}}+\sqrt{\dfrac{1}{2}}-2\cdot\sqrt{18}+\sqrt{\left(1-\sqrt{2}\right)^2}\)

\(=\dfrac{3}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-2\cdot3\sqrt{2}+\left|1-\sqrt{2}\right|\)

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

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

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

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

c: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\sqrt[3]{\dfrac{3}{4}\cdot\dfrac{9}{16}}=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)

d: \(\sqrt[3]{54}=\sqrt[3]{27\cdot2}=3\sqrt[3]{2}\)

e: \(\dfrac{\sqrt[3]{54}}{\sqrt[3]{-2}}=\sqrt[3]{\dfrac{54}{-2}}=\sqrt[3]{-27}=-3\)

f: \(\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

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

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

1) 

\(\left(\dfrac{6-2\sqrt{2}}{3-\sqrt{2}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{2+\sqrt{5}}\)

\(=\left[\dfrac{2\left(3-\sqrt{2}\right)}{3-\sqrt{2}}-\sqrt{5}\right]\left(2+\sqrt{5}\right)\)

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

\(=4-5\)

\(=-1\)

\(---\)

2) \(\sqrt{\left(2x+3\right)^2}=9\)

\(\Rightarrow\left|2x+3\right|=9\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=9\left(đk:x\ge-\dfrac{3}{2}\right)\\2x+3=-9\left(đk:x< -\dfrac{3}{2}\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{-6;3\right\}\)

\(Toru\)

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

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

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

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

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

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

Lời giải:
a.

\(=\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}+\frac{4(\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)}=\frac{\sqrt{5}+2}{5-2^2}+\frac{4(\sqrt{5}-1)}{5-1}\)

$=\sqrt{5}+2+(\sqrt{5}-1)=2\sqrt{5}+1$
b.

$=\frac{4(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}+\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}-2\sqrt{3}$

$=\frac{4(\sqrt{3}+1)}{2}+\frac{7(3+\sqrt{2})}{1}-2\sqrt{3}$
$=2(\sqrt{3}+1)+7(3+\sqrt{2})-2\sqrt{3}$
$=23+7\sqrt{2}$
c.

$=(\frac{4(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}-\frac{\sqrt{5}+2}{(\sqrt{5}-2)(\sqrt{5}+2)}).\frac{7(3+\sqrt{2})}{(3-\sqrt{2})(3+\sqrt{2})}$

$=[(3+\sqrt{5})-(\sqrt{5}+2)].(3+\sqrt{2})$

$=1(3+\sqrt{2})=3+\sqrt{2}$

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

\(=\sqrt{3}-3\sqrt{3}+\sqrt{3}=-\sqrt{3}\)

b: \(=\left(\left(2-2\sqrt{5}\right)\left(\sqrt{5}+2\right)+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(2\sqrt{5}+4-10-4\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

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

\(=-20+2\sqrt{15}+\sqrt{15}-3-6\sqrt{5}+6\sqrt{3}\)

\(=-23+3\sqrt{15}-6\sqrt{5}+6\sqrt{3}\)

a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)

\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)

\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)

b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)

\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)

c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

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

a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)

= \(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)

= \(-33\sqrt{2}\)

b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)

= \(10-2\sqrt{21}+14\sqrt{21}\)

= \(10+12\sqrt{21}\)