a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)

=3căn 6-6-3căn 6=-6

b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)

\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)

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

\(=-\sqrt{6}\)

Học tốt!!!!!!!!!!!!!!!!!!!!!!!

a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)

\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)

\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)

\(=5\sqrt{6}\)

b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)

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

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

\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)

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

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

9√6 + 3√6 - √6 = 11√6

c) (√28 - 2√3 + √7)√7 + √84

= (√4.7 - 2√3 + √7)√7 + √4.21

= (2√7 - 2√3 + √7)√7 + 2√21

= (3√7 - 2√3)√7 + 2√21

= 3.7 - 2√21 + 2√21 = 21

M = \(\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)

= \(\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}+\sqrt{6}\right)\left(\sqrt{7}-\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)

= \(2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)

= \(5\sqrt{6}\)

a) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)

\(=20\sqrt{3}-12\sqrt{3}-2\sqrt{57}+6\sqrt{3}\)

\(=\left(20-12+6\right)\sqrt{3}-2\sqrt{57}\)

\(=14\sqrt{3}-2\sqrt{57}\)

b) \(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)

\(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}\)

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

\(=-4\sqrt{6}\)

`a)(\sqrt{14}-3\sqrt{2})^2+6\sqrt{28}`

`=14-12\sqrt{7}+18+12\sqrt{7}=32`

`b)2\sqrt{20}-3\sqrt{20}+\sqrt{125}`

`=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}`

`=3\sqrt{5}`.

a) \(\left(\sqrt{14}-3\sqrt{2}\right)^2-6\sqrt{28}\)

\(=\left(\sqrt{14}\right)^2-2\cdot\sqrt{14}\cdot3\sqrt{2}+\left(3\sqrt{2}\right)^2+6\sqrt{28}\)

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

\(=14+18\)

\(=32\)

b) \(2\sqrt{20}-3\sqrt{20}+\sqrt{125}\)

\(=2\cdot2\sqrt{5}-3\cdot2\sqrt{5}+5\sqrt{5}\)

\(=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}\)

\(=3\sqrt{5}\)

`a)2\sqrt{28}+\sqrt{63}-\sqrt{112}`

`=4\sqrt{7}+3\sqrt{7}-4\sqrt{7}`

`=3\sqrt{7}`.

`b)3\sqrt{48}-5\sqrt{108}+6\sqrt{1/3}`

`=12\sqrt{3}-30\sqrt{3}+2\sqrt{[3^2]/3}`

`=-18\sqrt{3}+2\sqrt{3}`

`=-16\sqrt{3}`.

\(a^2-2a+b^2+4b+4c^2-4c+6=0\)'

\(\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)=0\)

\(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)

b tự làm nốt nhé~

\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54-x\right)\)

\(M=x^3+3^3-x^3-54+x\)

\(M=x+27-54\)

\(M=x+27-54\)

\(M=7-27\)

\(M=-20\)

\(M=\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54-x\right)\)

\(M=x^3+3^3-x^3-54+x\)

\(M=27-54+x\)

\(M=-27+x\)

thay x =7 vào M ta có :

\(M=-27+7=-20\)