\(=\left(\dfrac{\sqrt{10}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{6^2}}{\sqrt{6}}\right)\sqrt{4+\sqrt{15}}\)

\(=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\sqrt{4+\sqrt{15}}\)

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

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

 \(VT\Leftrightarrow\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)=5-3=2=VP\left(dpcm\right)\)

\(\left(\dfrac{\sqrt{30}-\sqrt{20}}{\sqrt{3}-\sqrt{2}}-\dfrac{6}{\sqrt{6}}\right)\sqrt{4+\sqrt{15}}=\left(\dfrac{\sqrt{10}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\sqrt{6}\right)\left(4+\sqrt{15}\right)\)

\(=\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4+\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)\sqrt{8+2\sqrt{15}}\)

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

\(=5-3=2\)

\(\sqrt{4x+8}+3\sqrt{x+2}=3+\dfrac{4}{5}\sqrt{25x+50}\left(x\ge-2\right)\)

\(\Rightarrow2\sqrt{x+2}+3\sqrt{x+2}-4\sqrt{x+2}=3\Rightarrow\sqrt{x+2}=3\Rightarrow x=7\)

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

\(=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}+\dfrac{\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

a: Ta có: \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)

\(=\left(\dfrac{\sqrt{6}}{2}-\dfrac{4\sqrt{6}}{2}\right)\cdot\dfrac{1}{\sqrt{6}}\)

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

`e)(3/2sqrt6+2sqrt{2/3}-4sqrt{3/2})(3/2sqrt6+2sqrt{2/3}+4sqrt{3/2})`

`=(3/2sqrt6+2sqrt{2/3})^2-(4\sqrt{3/2})^2`

`=((3sqrt6)/2+(2sqrt2)/3)^2-16*3/2`

`=((9sqrt6)/6+(4sqrt6)/6)^2-24`

`=((13sqrt6)/6)^2-24`

`=13^2/6-24`

`=25/6`

giả sử 2 vế bằng nhau, nhân tích chéo, rồi được 2 vế = nhau là kết luận thỏa mãn

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=\frac{1}{1+\sqrt{2}}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1=vp\)

Ban co the giai chi tiet hon ko

\(\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}=\frac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}\)

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

ta tính VT ra xong rồi nói VT = VP

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