\(C=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)

`C=(4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}`

`C=(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10})\sqrt{4-\sqrt{15}}`

`C=(\sqrt{10}+\sqrt{6})\sqrt{4-\sqrt{15}}`

`C=\sqrt{(\sqrt{10}+\sqrt{6})^2 .(4-\sqrt{15})}`

`C=\sqrt{(10+6+2\sqrt{60})(4-\sqrt{15})}`

`C=\sqrt{(16+4\sqrt{15})(4-\sqrt{15})}`

`C=\sqrt{64-16\sqrt{15}+16\sqrt{15}-60}`

`C=\sqrt{4}=2`

a) \(\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}\)

\(=\sqrt{14}\cdot\sqrt{5-\sqrt{21}}+\sqrt{6}\cdot\sqrt{5-\sqrt{21}}\)

\(=\sqrt{14\cdot\left(5-\sqrt{21}\right)}+\sqrt{6\cdot\left(5-\sqrt{21}\right)}\)

\(=\sqrt{70-14\sqrt{21}}+\sqrt{30-6\sqrt{21}}\)

\(=\sqrt{7^2-2\cdot7\cdot\sqrt{21}+\left(\sqrt{21}\right)^2}+\sqrt{\left(\sqrt{21}\right)^2-2\cdot3\cdot\sqrt{21}+3^2}\)

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

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

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

\(=4\)

b) \(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left[4\cdot\left(\sqrt{10}-\sqrt{6}\right)+\sqrt{15}\cdot\left(\sqrt{10}-\sqrt{6}\right)\right]\cdot\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+\sqrt{150}-\sqrt{90}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4\sqrt{10}-4\sqrt{6}+5\sqrt{6}-3\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{10}+\sqrt{6}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

\(=\sqrt{10\cdot\left(4-\sqrt{15}\right)}+\sqrt{6\cdot\left(4-\sqrt{15}\right)}\)

\(=\sqrt{40-10\sqrt{15}}+\sqrt{24-6\sqrt{15}}\)

\(=\sqrt{5^2-2\cdot5\cdot\sqrt{15}+\left(\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}\right)^2-2\cdot3\cdot\sqrt{15}+3^2}\)

\(=\sqrt{\left(5-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)

\(=\left|5-\sqrt{15}\right|+\left|\sqrt{15}-3\right|\)

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

\(=2\)

(4+căn 15) nhân (căn 10 -căn 6) nhân (căn tất cả của 4-căn 15) 
= (4+ căn 15) (căn 5 - căn 3) (căn 2) (căn (4- căn 15)) 
= (4+ căn 15) (căn 5 - căn 3) (căn (8- 2 căn 15)) 
= (4+ căn 15) (căn 5 - căn 3) (căn 5 - căn 3) 
= (4+ căn 15) (5 + 3 - 2 căn 15) 
= (4+ căn 15) (4.2 - 2 . căn 15) 
= 2.(4+ căn 15) (4- căn 15) 
= 2. (16-15) 
=2 

(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

ai nhanh nhat mk se k (neu dung). mk cần gấp

tích mình đi

ai tích mình 

mình tích lại 

thanks

bn trả lời đúng đi rồi mk k. thank

`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`

`=`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`

`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`

`=2(\sqrt5-1)sqrt{6+2\sqrt5}`

`=2(\sqrt5-1)(\sqrt5+1)`

`=2(5-1)`

`=8`

`2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{6-2\sqrt5}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt{(\sqrt5-1)^2}}`

`=2(\sqrt{10}-\sqrt2)\sqrt{4+\sqrt5-1}`

`=2(\sqrt{10}-\sqrt2)\sqrt{3+\sqrt5)`

`=2\sqrt2(\sqrt5-1)\sqrt{3+\sqrt5}`

`=2(\sqrt5-1)sqrt{6+2\sqrt5}`

`=2(\sqrt5-1)(\sqrt5+1)`

`=2(5-1)`

`=8`

`(4\sqrt2+\sqrt{30})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`

`=\sqrt2(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{4-\sqrt{15}}`

`=(4+\sqrt{15})(\sqrt5-\sqrt3)\sqrt{8-2\sqrt{15}}`

`=(4+\sqrt{15})(\sqrt5-\sqrt3)(\sqrt5-\sqrt3)`

`=(4+\sqrt{15})(8-2\sqrt{15})`

`=2(4+\sqrt{15})(4-\sqrt{15})`

`=2(16-15)`

`=2`

a) \(2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{6-2\sqrt{5}}}\)

\(=2\left(\sqrt{10}-\sqrt{2}\right)\sqrt{4+\sqrt{5}-1}\)

\(=\dfrac{2\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{5}+1\right)}{\sqrt{2}}\)

\(=2\cdot4=8\)

=2 nha bạn

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

 ~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

~~~~~~~~~~~ Và chúc các bạn trả lời câu hỏi này kiếm được nhiều k hơn ~~~~~~~~~~~~

chac chan de bai dung ko

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}.\sqrt{2}-\sqrt{3}.\sqrt{2}\right)\left(\sqrt{4-\sqrt{15}}\right)\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{2}\sqrt{4-\sqrt{15}}\)

=\(\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{5-2\sqrt{15}+3}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)|^{ }_{ }\sqrt{5}-\sqrt{3|^{ }_{ }}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)(vi \(\sqrt{5}-\sqrt{3}\)>0)

\(=\left(4+\sqrt{15}\right)\left(5+2\sqrt{15}+3\right)\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=32-8\sqrt{15}+8\sqrt{15}-30\)

\(=2\)

\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right).\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)|\sqrt{5}-\sqrt{3}|\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)^2\)

\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)

\(=2.[\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)]\)

\(=2.\left(4^2-\sqrt{15}^2\right)\)

\(=2.1=2\)

\(A=\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}-\frac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}+\frac{1}{\sqrt{a}}\right)\)

\(A=\)\(\left[\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right]\left[\frac{a+1}{\sqrt{a}}\right]\)

\(A=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{a-1}.\)  \(\frac{a+1}{\sqrt{a}}\)

\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}\)

\(A=\frac{4a\left(a+1\right)}{a-1}\)

ta có \(a=\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)

\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8-2\sqrt{15}}\)

\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

\(a=\left(4+\sqrt{15}\right)\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)\)

\(a=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)

\(a=\left(4+\sqrt{15}\right).2\left(4-\sqrt{15}\right)\)

\(a=2\left(16-15\right)\)

\(a=2\)

khi đó \(A=\frac{4.2.\left(2+1\right)}{2-1}=8.3=24\)

vậy.....

bạn có thể phân tích 4 + căn 5 = căn ( 4 + căn 5) . căn ( 4 + căn 5) 
và căn 10 - căn 6 = căn 2( căn 5 - căn 3) 
Khi đó Biểu thức rút gọn trở thành 
căn(4 - căn 15).căn(4+15) . căn (4 + căn 15) . căn 2(căn 5 - căn 3) 
= căn (16 - 15) . căn (8 + 2.căn 15).(căn 5 - căn 3) = căn (3 + 2.căn 3. căn 5 + 5). (căn 5 - căn 3) 
= căn [ (căn 3 + căn 5)^2 ] . (căn 5 - căn 3) = (căn 5 + căn 3)(căn 5 - căn 3) = 5 - 3 = 2

5 - 3 = 2

5 - 3 = 2

áp số : 2