Question

Các bạn giải giúp mik bài toán này trong hôm nay nhé!:)))

\(\dfrac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}-\dfrac{6}{\sqrt{2}}-\dfrac{3}{\sqrt{2}+1}\)

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

Cảm ơn các bạn nhiềuuuuuuuuuu lắmmmmmmmmmmmmmm!:)))))))))))))))))))

Answer 1

bài này trước, bài trên để coi lại đã

Đặt \(A=\sqrt{2-\sqrt{3}}-\sqrt{\dfrac{3}{2}}\Rightarrow A\sqrt{2}=\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{3}-1-\sqrt{3}=-1\Rightarrow A=-\dfrac{1}{\sqrt{2}}=-\dfrac{\sqrt{2}}{2}\)

Answer 2

đặt \(A=\dfrac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}-\dfrac{6}{\sqrt{2}}-\dfrac{3}{\sqrt{2}+1}\)

\(\Rightarrow A\sqrt{2}=\dfrac{\sqrt{12-2\sqrt{11}}}{\sqrt{22}-\sqrt{2}}-\dfrac{6\sqrt{2}}{\sqrt{2}}-\dfrac{3\sqrt{2}}{\sqrt{2}+1}=\dfrac{\sqrt{11}-1}{\sqrt{2}\left(\sqrt{11}-1\right)}-\dfrac{6\sqrt{2}}{\sqrt{2}}-\dfrac{3\sqrt{2}}{\sqrt{2}+1}=\dfrac{1}{\sqrt{2}}-\dfrac{6\sqrt{2}}{\sqrt{2}}-\dfrac{3\sqrt{2}}{\sqrt{2}+1}=\dfrac{\sqrt{2}+1-12-6\sqrt{2}-6}{\sqrt{2}\left(\sqrt{2}+1\right)}=\dfrac{-17-5\sqrt{2}}{\sqrt{2}\left(\sqrt{2}+1\right)}\)

\(\Rightarrow A=\dfrac{-17-5\sqrt{2}}{\sqrt{2}\left(\sqrt{2}+1\right)}\cdot\dfrac{1}{\sqrt{2}}=\dfrac{-17-5\sqrt{2}}{2\left(\sqrt{2}+1\right)}=\dfrac{\left(-17-5\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2}=\dfrac{7-12\sqrt{2}}{2}\)

Kl: \(A=\dfrac{7-12\sqrt{2}}{2}\)

Answer 3

\(1.\dfrac{\sqrt{6-\sqrt{11}}}{\sqrt{22}-\sqrt{2}}-\dfrac{6}{\sqrt{2}}-\dfrac{3}{\sqrt{2}+1}=\dfrac{\sqrt{12-2\sqrt{11}}}{\sqrt{4.11}-2}-\dfrac{6\sqrt{2}}{2}-\dfrac{3\left(\sqrt{2}-1\right)}{2-1}=\dfrac{\sqrt{\left(\sqrt{11}-1\right)^2}}{2\sqrt{11}-2}-3\sqrt{2}-3\sqrt{2}+3=\dfrac{\sqrt{11}-1}{2\left(\sqrt{11}-1\right)}+3=\dfrac{1}{2}+3=\dfrac{7}{2}\)

\(2.\sqrt{2-\sqrt{3}}-\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{\dfrac{6}{2}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}}{\sqrt{2}}=\dfrac{\text{ |}\sqrt{3}-1\text{ |}-\sqrt{3}}{\sqrt{2}}=\dfrac{-1}{\sqrt{2}}=\dfrac{-\sqrt{2}}{2}\)