Đặt \(A=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)\left(11-\sqrt{113}\right)....\left(11-\sqrt{104}\right)\)

\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-\sqrt{121}\right)....\left(11-\sqrt{104}\right)\)

\(=\left(11-\sqrt{103}\right)\left(11-\sqrt{109}\right)....\left(11-11\right)....\left(11-\sqrt{104}\right)\)

\(=0\)

Do đó biểu thức trên đầu bài bằng 0

bạn ơi, trong dãy này không có số \(\sqrt{121}\)đâu

\(=\left(3\sqrt{11}-\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}-\sqrt{66}\)

\(=22-3\sqrt{22}-\sqrt{66}\)

\(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\cdot\sqrt{11}-\sqrt{3\cdot22}\)

\(=\left(\sqrt{3^2\cdot11}-\sqrt{3^2\cdot2}-\sqrt{11}\right)\cdot\sqrt{11}-\sqrt{66}\)

\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\cdot\sqrt{11}-\sqrt{66}\)

\(=3\cdot11-3\sqrt{22}-11-\sqrt{66}\)

\(=33-3\sqrt{22}-11-\sqrt{66}\)

\(=22-3\sqrt{22}-\sqrt{66}\)

a/ Đề sai

b/ \(\sqrt{125}-4\sqrt{45}+3\sqrt{2}-\sqrt{80}=5\sqrt{5}-12\sqrt{5}+3\sqrt{2}-4\sqrt{5}\)

\(=-11\sqrt{5}+3\sqrt{2}\)

c/ \(2\sqrt{\frac{27}{4}}-\sqrt{\frac{48}{9}}-\frac{2}{5}\sqrt{\frac{75}{16}}=2.\frac{3\sqrt{3}}{2}-\frac{4\sqrt{3}}{3}-\frac{2}{5}.\frac{5\sqrt{3}}{4}\)

\(=3\sqrt{3}-\frac{4\sqrt{3}}{3}-\frac{\sqrt{3}}{2}=\sqrt{3}\left(3-\frac{4}{3}-\frac{1}{2}\right)=\frac{7\sqrt{3}}{6}\)

d/ \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\cdot\sqrt{11}+3\sqrt{22}=33-3\sqrt{22}-11+3\sqrt{22}=22\)

Hoàng Anh Tuấn : mik vẫn chưa hiểu câu 2 , 3 b ra thế nào ? xin b hãy giải theo 1 cách dễ hiểu hay giảng cho mik đc ko ạ !!!

câu c bạn làm nhầm đề bài r kìa Hoàng Anh Tuấn 

\(\sqrt{18}=3\sqrt{2}\) chứ sao lại bằng \(3\sqrt{3}\)đc

câu 2 sao có thể rút gọn đc hả b ? trên tử là phép cộng trừ mà , vậy ls có thể làm vậy ?

`a, (sqrt 28 - sqrt 12 - sqrt 7) sqrt 7 + 2 sqrt 21`.

`= sqrt(28.7) - sqrt(12.7) - sqrt(7.7) + 2 sqrt 21`.

`= sqrt(4. 7.7) - sqrt (12.7) - 7 + 2 sqrt 21`.

`= 14 - sqrt(4.3.7) - 7 + 2 sqrt 21`.

`= 7`.

`b, (sqrt99-sqrt18-sqrt11)sqrt11+3sqrt22`

`= sqrt(99.11)- sqrt(18.11)-sqrt(11.11) +3sqrt22`

`= sqrt(9.11.11)-sqrt(2.9.11)-11+3sqrt22`

`= 33 - 11 = 22`.

a: \(\left(\sqrt{7}+\sqrt{15}\right)^2=22+2\sqrt{105}=7+15+2\sqrt{105}\)

\(7^2=49=7+42\)

mà \(15+2\sqrt{105}< 42\)

nên \(\sqrt{7}+\sqrt{15}< 7\)

b: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)

\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)

mà \(2\sqrt{22}< 15+10\sqrt{3}\)

nên \(\sqrt{2}+\sqrt{11}< 5+\sqrt{3}\)

Giải:

\(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=33-3\sqrt{22}-11+3\sqrt{22}\)

\(=22\)

Vậy ...

a) \(\sqrt{33-12\sqrt{6}}+\sqrt{15+6\sqrt{6}}=\sqrt{24-2.2\sqrt{6}.3+9}+\sqrt{6+2.\sqrt{6}.3+9}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}+3\right)^2}=\left|2\sqrt{6}-3\right|+\left|\sqrt{6}+3\right|=2\sqrt{6}-3+\sqrt{6}+3=3\sqrt{6}\)

b) \(\dfrac{\sqrt{99}}{\sqrt{11}}+\dfrac{\sqrt{28}}{\sqrt{7}}-\sqrt{\sqrt{81}}=\sqrt{\dfrac{99}{11}}+\sqrt{\dfrac{28}{7}}-\sqrt{9}=\sqrt{9}+\sqrt{4}-\sqrt{9}=\sqrt{4}=2\)

a) \(\sqrt{33-12\sqrt{6}}\) + \(\sqrt{15+6\sqrt{6}}\)

= \(\sqrt{9-2.3.2\sqrt{6}+24}\)+\(\sqrt{9+2.3\sqrt{6}+6}\)

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

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

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

=\(\sqrt{6}\)

b)\(\dfrac{\sqrt{99}}{\sqrt{11}}\)+\(\dfrac{\sqrt{28}}{\sqrt{7}}\)\(-\sqrt{\sqrt{81}}\)

= \(\sqrt{\dfrac{99}{11}}+\sqrt{\dfrac{28}{7}}-3\)

=\(\sqrt{9}+\sqrt{4}-3\)

= 3+2-3

= 2

a) \(\sqrt{33-12\sqrt{6}}+\sqrt{15+6\sqrt{6}}\)

=\(\sqrt{24-2.2\sqrt{6}.3+9}\)+\(\sqrt{9+2.3\sqrt{6}+6}\)

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

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

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

\(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

= \(14-2\sqrt{21}-7+2\sqrt{21}\) = \(7\)

\(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{21}\)

= \(33-3\sqrt{22}-11+3\sqrt{21}\) = \(22-3\sqrt{22}+3\sqrt{21}\)

Mình sẽ làm cụ thể một tí nhé:

a) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

\(=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

\(=\left(\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}\)

\(=7-2\sqrt{21}+2\sqrt{21}\)

\(=7\)

b) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{21}\)

\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{21}\)

\(=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{21}\)

\(=22-3\sqrt{22}+3\sqrt{21}\)