\(3.\sqrt{3x}-5\sqrt{12x}+7\sqrt{27x}=28\\ \Leftrightarrow3\sqrt{3x}-10\sqrt{3x}+21.\sqrt{3x}=28\\ \Leftrightarrow\sqrt{3x}\left(3-10+21\right)=28\\ \Leftrightarrow\sqrt{3x}=2\\ \Leftrightarrow3x=4\\ \Leftrightarrow x=\dfrac{4}{3}\)

ta có \(3\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=28\)

<=>\(14\sqrt{3x}=28\)

<=>\(\sqrt{3x}=2< =>x=\dfrac{4}{3}\)

a/ 3x(12x-4)-9x(4x-3)

=36x²-12x-36x²+27x

=(36x²-36x²)-12x+27x

=15x

b/ x(5-2x)+2x(x-1)

=5x-2x²+2x²-2x

=(5x-2x)-(-2x²+2x²)

=3x

c/ 5x(12x+7)-3x(20x-5)

=60x²+35x-60x²+15x

=(60x²-60x²)+(35x+15x)\

=50x

d/ 3x(2x-7)+2x(5-3x)

=6x²-21x+10x-6x²

=(6x²-6x²)+(10x-21x)

=-11x

e/ đề sai hay sao ý ra số to lắm @@

a,

128-3x-12=23

3x=128-12-23

3x=93

x=93:3

= 31

b,

(12x+84+55):5=35

12x+84+55=35.5

12x+84+55=175

12x=175-55-84

12x=36

x=36:12

x=3

9) Sửa: \(2\sqrt{8\sqrt{3}}-2\sqrt{5\text{ }\sqrt{3}}-3\sqrt{20\sqrt{3}}\)

\(=2\sqrt{2^2\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{2^2\cdot5\sqrt{3}}\)

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

\(=4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)

\(=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)

10) \(\sqrt{12x}-\sqrt{48x}-3\sqrt{3x}+27\)

\(=\sqrt{2^2\cdot3x}-\sqrt{4^2\cdot3x}-3\sqrt{3x}+27\)

\(=2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}+27\)

\(=-5\sqrt{3x}++27\)

11) \(\sqrt{18x}-5\sqrt{8x}+7\sqrt{18x}+28\)

\(=\sqrt{3^2\cdot2x}-5\sqrt{2^2\cdot2x}+7\sqrt{3^2\cdot2x}+28\)

\(=3\sqrt{2x}-5\cdot2\sqrt{2x}+7\cdot3\sqrt{2x}+28\)

\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28\)

\(=14\sqrt{2x}+28\)

12) \(\sqrt{45a}-\sqrt{20a}+4\sqrt{45a}+\sqrt{a}\)

\(=\sqrt{3^2\cdot5a}-\sqrt{2^2\cdot5a}+4\sqrt{3^2\cdot5a}+\sqrt{a}\)

\(=3\sqrt{5a}-2\sqrt{5a}+4\cdot3\sqrt{5a}+\sqrt{a}\)

\(=3\sqrt{5a}-2\sqrt{5a}+12\sqrt{5a}+\sqrt{a}\)

\(=13\sqrt{5a}+\sqrt{a}\)

\(5x\left(12x+7\right)-3x\left(20x-5\right)=-100\\ \Leftrightarrow60x^2+35x-60x^2+15x=-100\\ \Leftrightarrow50x=-100\\ \Leftrightarrow x=-2\)

Vậy x = -2 là nghiệm của pt.