Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{3x}-5\sqrt{12x}+7\cdot\sqrt{27x}=12\)

=>\(\sqrt{3x}-5\cdot2\sqrt{3x}+7\cdot3\sqrt{3x}=12\)

=>\(12\sqrt{3x}=12\)

=>\(\sqrt{3x}=1\)

=>3x=1

=>x=1/3(nhận)

Bài 1:

a: \(A=\left(\sqrt{\frac23}+\sqrt{\frac{50}{3}}-\sqrt{24}\right)\cdot\sqrt6\)

\(=\left(\frac{2\sqrt6}{6}+\sqrt{\frac{100}{6}}-2\sqrt6\right)\cdot\sqrt6\)

\(=2+\sqrt{100}-2\cdot6=2+10-12=0\)

b: \(B=\left(\frac{\sqrt{14}-\sqrt7}{\sqrt2-1}+\frac{\sqrt{15}-\sqrt5}{\sqrt3-1}\right):\frac{1}{\sqrt7-\sqrt5}\)

\(=\left(\frac{\sqrt7\left(\sqrt2-1\right)}{\sqrt2-1}+\frac{\sqrt5\left(\sqrt3-1\right)}{\sqrt3-1}\right)\cdot\left(\sqrt7-\sqrt5\right)\)

\(=\left(\sqrt7+\sqrt5\right)\left(\sqrt7-\sqrt5\right)\)

=7-5

=2

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

\(\dfrac{1}{\sqrt{5}-2}+\dfrac{10}{\sqrt{5}}\)

\(=\dfrac{1\cdot\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}+\dfrac{2\sqrt{5}\cdot\sqrt{5}}{\sqrt{5}}\)

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

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

\(=\sqrt{5}+2+2\sqrt{5}\)

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

\(A=\sqrt{5}.\left(\sqrt{20}-3\right)+\sqrt{45}.\)

\(=\sqrt{5}.\left(\sqrt{4.5}-3\right)+\sqrt{9.5}\)

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

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

\(=2.\sqrt{5}^2=2.5=10\)

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

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

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

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

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

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

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

Cảm ơn nhiều ạ