\(=\left(3\sqrt{2}-2\sqrt{2}+\sqrt{14}\right).\sqrt{2}-\sqrt{7}\\ =\left(\sqrt{2}+\sqrt{14}\right).\sqrt{2}-\sqrt{7}\\ =2+2\sqrt{7}-\sqrt{7}\\ =2+\sqrt{7}\)

Bạn ghi lại đề dùm

\(\sqrt{8-2\sqrt{7}-\left[\left(\sqrt{7}+1\right)^2\right]}\)

\(\sqrt{8-2\sqrt{7}-\sqrt{7}-1}\)

\(\Leftrightarrow\sqrt{7-\sqrt{7}}\)

\(\sqrt{8-2\sqrt{7}}-\sqrt{23-8\sqrt{7}}=\) \(\sqrt{1-2\sqrt{7}+7}-\sqrt{7-2.4.\sqrt{7}+16}\)

\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-4\right)^2}\)

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

\(=\sqrt{7}-1+\sqrt{7}-4\)\(=2\sqrt{7}-5\)

chúc bn học tốt

=\(\sqrt{\left(\sqrt{7}-1\right)^2}\)- \(\sqrt{\left(4-\sqrt{7}\right)^2}\)

= \(\sqrt{7}\)- 1 - 4 + \(\sqrt{7}\)

= \(2\sqrt{7}\)-5

đ/á ra hơi kì

#mã mã#

mình nghĩ là hai bn làm đúng đó

`\sqrt(((2-\sqrt5)^2)/8)`

`= (\sqrt((2-\sqrt5)^2))/(\sqrt8)`

`= (|2-\sqrt5|)/(2\sqrt2)`

`=(\sqrt5-2)/(2\sqrt2)`

`=(\sqrt10-2\sqrt2)/4`

.

`7/(3\sqrt14) = (\sqrt7 .\sqrt7)/(3.\sqrt7 .\sqrt2)`

`=(\sqrt7)/(3\sqrt2)`

`=(\sqrt14)/(3.2)`

`=(\sqrt14)/6`

\(\sqrt{\dfrac{\left(2−\sqrt{5}\right)^2}{8}}\)= \(\dfrac{\sqrt{5}-2}{2\sqrt{2}}\)

\(\dfrac{7}{3\sqrt{14}}\) = \(\dfrac{\sqrt{7}}{3\sqrt{2}}\)

\(\sqrt{\dfrac{\left(2-\sqrt{5}\right)^2}{8}}=\dfrac{\sqrt{5}-2}{2\sqrt{2}}=\dfrac{\sqrt{10}-2\sqrt{2}}{4}\)

\(\dfrac{7}{3\sqrt{14}}=\dfrac{7\sqrt{14}}{42}=\dfrac{\sqrt{14}}{6}\)

\(\left(\sqrt{8+3\sqrt{7}}+\sqrt{8-3\sqrt{7}}\right)^2\)

\(=8+3\sqrt{7}+8-3\sqrt{7}+2\sqrt{64-63}\)

\(=16+2=18\)

\(\left(\sqrt{8+3\sqrt{7}+\sqrt{8-3\sqrt{7}}}\right)^2\)

=\(8+3\sqrt{7}+8-3\sqrt{7}+2\sqrt{64-63}\)

=16+2=18

\(\left(\sqrt{8+3\sqrt{7}}+\sqrt{8-3\sqrt{7}}\right)^2=16+2\sqrt{8^2-\left(3\sqrt{7}\right)^2}=16+2=18\)

\(\left(\sqrt{8+3\sqrt{7}}+\sqrt{8-3\sqrt{7}}\right)^2\)

\(=8+3\sqrt{7}+2\sqrt{\left(8+3\sqrt{7}\right)\left(8-3\sqrt{7}\right)}+8-3\sqrt{7}\)

\(=16\cdot\sqrt{8^2-\left(3\sqrt{7}\right)^2}=16\cdot1=16\)

\(\sqrt{\frac{2}{8-3\sqrt{7}}}=\frac{2\left(8+3\sqrt{7}\right)}{\left(8-3\sqrt{7}\right)\left(8+3\sqrt{7}\right)}=2\left(8+3\sqrt{7}\right)\)

a, \(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)

\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{7}-1-\sqrt{7}-1=-2\)

b, \(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)

\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{4-2.2\sqrt{2}+2}\)

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

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

câu 1 đã làm 

câu 2

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

\(\Leftrightarrow\sqrt{2}+1+\sqrt{2}-2\Leftrightarrow2\sqrt{2}-1\)

\(2-\sqrt{2}< 0\)  thì căn bình phương của nó sẽ ra \(\sqrt{2}-2\)

bạn làm bên dưới bỏ dấu căn mà k biết bên trong biểu thức âm hay dương , giữu nguyên \(2-\sqrt{2}\) thic cx cạn cmn lời 

a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)

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

\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)

\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)

b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)

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

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

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

c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)