\(2\sqrt{27}-\sqrt{\dfrac{16}{3}}-\sqrt{48}-\sqrt{8\dfrac{1}{3}}\)

\(=6\sqrt{3}-4\sqrt{\dfrac{1}{3}}-4\sqrt{3}-5\sqrt{\dfrac{1}{3}}\)

\(=2\sqrt{3}-9\sqrt{\dfrac{1}{3}}\)

\(=2\sqrt{3}-3\sqrt{9\cdot\dfrac{1}{3}}\)

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

\(=-\sqrt{3}\)

________________________

\(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\)

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

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

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

\(=15-12\)

\(=3\)

bạn viết lại đề câu a.

\(-\left(x+3\right)^3+\left(x+9\right)\left(x^2+27\right)=-\left(x^3+27+9x^2+27x\right)+x^3+27x+9x^2+243\)

\(=-x^3-27-9x^2-27x+x^3+27x+9x^2+243=243-27=216\)

nha . Cảm ơn . Chúc bạn học tốt

1.
A= \(2\sqrt{6}\) + \(6\sqrt{6}\) - \(8\sqrt{6}\)
A= 0
2.
A= \(12\sqrt{3}\) + \(5\sqrt{3}\) - \(12\sqrt{3}\)
A= 0
3.
A= \(3\sqrt{2}\) - \(10\sqrt{2}\) + \(6\sqrt{2}\)
A= -\(\sqrt{2}\)
4.
A= \(3\sqrt{2}\) + \(4\sqrt{2}\) - \(\sqrt{2}\)
A= \(6\sqrt{2}\)
5.
M= \(2\sqrt{5}\) - \(3\sqrt{5}\) + \(\sqrt{5}\)
M= 0
6.
A= 5 - \(3\sqrt{5}\) + \(3\sqrt{5}\)
A= 5

This literally took me a while, pls sub :D
https://www.youtube.com/channel/UC4U1nfBvbS9y_Uu0UjsAyqA/featured

a, `(x-3)(x^2+3x+9)-(x^2-1)(9x+27)`

`=x^3-3^3-(9x^3+27x^2-9x-27)`

`=x^3-3^3-9x^3-27x^2+9x+27`

`=-8x^3-27x^2+9x`

b, `(x-2)(x^2+2x+4)-x(x-3)(x+3)`

`=x^3-2^3-x(x^2-9)`

`=x^3-8-x^3+9x`

`=9x-8`

a) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x^2-1\right)\left(9x+27\right)\)

\(=x^3-27-\left(9x^3+27x^2-9x-27\right)\)

\(=x^3-27-9x^3-27x^2+9x+27\)

\(=-8x^3-27x^2+9x\)

b) Ta có: \(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)\)

\(=x^3-8-x\left(x^2-9\right)\)

\(=x^3-8-x^3+9x\)

\(=9x-8\)

a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)

yggucbsgfuyvfbsudy

????????

\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)

Ta có: \(C=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)

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

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

Ta có: \(B=\dfrac{\sqrt{2-\sqrt{3}}+\sqrt{4-\sqrt{15}}+\sqrt{10}}{\sqrt{23-3\sqrt{5}}}\)

\(=\dfrac{\sqrt{4-2\sqrt{3}}+\sqrt{8-2\sqrt{15}}+2\sqrt{5}}{3\sqrt{5}-1}\)

\(=\dfrac{\sqrt{3}-1+\sqrt{5}-\sqrt{3}+2\sqrt{5}}{3\sqrt{5}-1}\)

=1