a: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=-2\cdot3=-6\)

\(\sqrt[3]{\left(-8\right)\cdot27}=\sqrt[3]{-216}=-6\)

Do đó: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=\sqrt[3]{\left(-8\right)\cdot27}\)

b: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=-\dfrac{2}{3}\)

\(\sqrt[3]{-\dfrac{8}{27}}=-\dfrac{2}{3}\)

Do đó: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)

Sqrt ở đây làm căn bậc hai nhé

\(D=\left|\sqrt{8}-3\right|+\left|\sqrt{19}-4\right|-\left(\sqrt{19}-\sqrt{8}\right)\)

\(D=\left(3-2\sqrt{2}\right)+\sqrt{19}-4-\left(\sqrt{19}-\sqrt{8}\right)\)

\(D=\left(3-2\sqrt{2}\right)+\sqrt{19}-4-\left(\sqrt{19}-2\sqrt{2}\right)\)

\(D=-2\sqrt{2}+3+\sqrt{19}-4-\left(\sqrt{19}-2\sqrt{2}\right)\)

\(D=-2\sqrt{2}+3+\sqrt{19}-4-\sqrt{19}+2\sqrt{2}\)

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

\(D=3-4\)

\(D=-1\)

Mình dùng máy casio nhé bạn.

KQ; 0,6151214812.

Bạn có cần cách làm không?

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

\(=\sqrt{\sqrt{6}-\sqrt{2}}.\sqrt{\sqrt{6}+\sqrt{2}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\sqrt{2}\right)}\)

\(=\sqrt{4}=2\)

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

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

=4

√8√3 + 2√25√12 - 4√192
= √2.√4.√3 + 2.5.√4.√3 - 4.√16.√4.√3
= 2√6 + 10.√12 - 16.√12
= 2√6 - 6√12
= 2√6 - 6.√4.√3
= 2√6 - 6.2.√3
= 2√6 - 12√3

ko biết

không bt làm thì đừng có mak trả lời

\(\left(\sqrt{33+8\sqrt{2}}\right)-\left(\sqrt{33-8\sqrt{2}}\right)\)

\(=\left[\sqrt{33+8\sqrt{2}}-\left(\sqrt{33-8\sqrt{2}}\right)\right]^2\)

\(=33+8\sqrt{2}+33-8\sqrt{2}-2\sqrt{\left(33+8\sqrt{2}\right)\left(33-8\sqrt{2}\right)}\)

\(=66-62\)

\(=4\)

a) Ta có: \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right)\cdot\sqrt{3}\)

\(=\left(7\cdot4\sqrt{3}+3\cdot3\sqrt{3}-2\cdot2\sqrt{3}\right)\cdot\sqrt{3}\)

\(=33\sqrt{3}\cdot\sqrt{3}\)

=99

b) Ta có: \(\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

\(=\left(12\cdot5\sqrt{2}-8\cdot10\sqrt{2}+7\cdot15\sqrt{2}\right):\sqrt{10}\)

\(=\dfrac{85\sqrt{2}}{\sqrt{10}}=\dfrac{85}{\sqrt{5}}=17\sqrt{5}\)

c) Ta có: \(\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}-\dfrac{1}{4}\cdot2\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=\left(2\sqrt{6}-4\sqrt{3}+3\sqrt{2}\right)\cdot3\sqrt{6}\)

\(=36-36\sqrt{2}+18\sqrt{3}\)

d) Ta có: \(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

\(=3\cdot\sqrt{75\sqrt{2}}+5\cdot\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

\(=3\cdot5\sqrt{2}\cdot\sqrt{\sqrt{2}}+4\sqrt{3}\sqrt{\sqrt{2}}\)

\(=15\sqrt{\sqrt{8}}+4\sqrt{\sqrt{18}}\)

a,=\(\left(28\sqrt{3}+9\sqrt{3}-4\sqrt{3}\right).\sqrt{3}\)

   \(=28.3+9.3-4.3=99\)

b,\(=\left(60\sqrt{2}-80\sqrt{2}+175\sqrt{2}\right):\sqrt{10}\)

  \(=155\sqrt{2}:\sqrt{10}=\dfrac{155}{\sqrt{5}}\)

d,Ta có:\(3\sqrt{15\sqrt{50}}+5\sqrt{24\sqrt{8}}-4\sqrt{12\sqrt{32}}\)

        \(=3\sqrt{75\sqrt{2}}+5\sqrt{48\sqrt{2}}-4\sqrt{48\sqrt{2}}\)

        \(=15\sqrt{3\sqrt{2}}+20\sqrt{3\sqrt{2}}-16\sqrt{3\sqrt{2}}\)

        \(=19\sqrt{3\sqrt{2}}\)

a: \(x=4+\sqrt{3}+4-\sqrt{3}=8\)

Khi x=8 thì \(A=\dfrac{2-5\cdot2\sqrt{2}}{2\sqrt{2}+1}=\dfrac{2-10\sqrt{2}}{2\sqrt{2}+1}=-6+2\sqrt{2}\)