\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{7}\)

\(=\sqrt{196}-2\sqrt{98}+7+7\sqrt{7}\)

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

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

\(\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

\(=\sqrt{4.7}-\sqrt{9.7}+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)

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

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

Bài 1:

a. Ta có \(\sqrt{\dfrac{2}{x^2}}=\dfrac{\sqrt{2}}{\left|x\right|}=\dfrac{\sqrt{2}}{x}\) ,để biểu thức có nghĩa thì \(x>0\)

b. Để biểu thức \(\sqrt{\dfrac{-3}{3x+5}}\) có nghĩa thì \(\dfrac{-3}{3x+5}\ge0\) 

mà \(-3< 0\Rightarrow3x+5< 0\) \(\Rightarrow x< \dfrac{-5}{3}\)

Bài 2:

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

b. \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

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

\(=21\)

c. \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)

\(=14-6\sqrt{28}+18+6\sqrt{28}\)

\(=32\)

\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

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

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

=21

a: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)

\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)

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

=0

b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)

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

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

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

=2

c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)

\(=\left|\sqrt{7}-4\right|-2\sqrt{7}+3\sqrt{7}\)

\(=4-\sqrt{7}+\sqrt{7}\)

=4

d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)

\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)

\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)

\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)

\(=16\sqrt{5}\)

e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)

\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)

\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)

\(=-16\sqrt{3}\)

a) \(\left(\sqrt{28}-2\sqrt{3}+\sqrt{7}\right)\sqrt{7}+\sqrt{84}\)

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

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

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

\(=21\)

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

\(=\dfrac{\sqrt{\left(\sqrt{3}\right)^2-2\sqrt{3}\cdot1+1^2}}{\sqrt{6}-\sqrt{2}}\)

\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{6}-\sqrt{2}}\)

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

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

\(=\dfrac{1}{\sqrt{2}}\)

Bạn ơi ! Cho mình hỏi,đề bài yêu cầu gì vậy bạn ?

\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=\left(2\sqrt{7}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

\(=14-14\sqrt{2}+7+7\sqrt{8}\)

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