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

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

=0

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

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

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

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

Cảm ơn nhiều ạ

Lời giải:

$\sqrt{(3-\sqrt{3})^2}=|3-\sqrt{3}|=3-\sqrt{3}$ do $3-\sqrt{3}>0$

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

Ta có: \(B=\frac{2+\sqrt3}{2-\sqrt3}\)

\(=\frac{\left(2+\sqrt3\right)\left(2+\sqrt3\right)}{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}\)

\(=\frac{\left(2+\sqrt3\right)^2}{4-3}=\left(2+\sqrt3\right)^2\)

\(=7+4\sqrt3\)

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