Đáp án:
`{2\sqrt{8}-\sqrt{12}}/{\sqrt{18}-\sqrt{48}}-{\sqrt{5}+\sqrt{27}}/{\sqrt{30}+\sqrt{162}}={-\sqrt{6}}/2`
Giải thích các bước giải:
Ta có:
`\qquad {2\sqrt{8}-\sqrt{12}}/{\sqrt{18}-\sqrt{48}}-{\sqrt{5}+\sqrt{27}}/{\sqrt{30}+\sqrt{162}}`
`={2\sqrt{8}-\sqrt{2^2 . 3}}/{\sqrt{6.3}-\sqrt{6.8}}-{\sqrt{5}+\sqrt{27}}/{\sqrt{6.5}+\sqrt{6.27}}`
`={2\sqrt{8}-2\sqrt{3}}/{\sqrt{6}.(\sqrt{3}-\sqrt{8})}-{\sqrt{5}+\sqrt{27}}/{\sqrt{6}.(\sqrt{5}+\sqrt{27})}`
`={2(\sqrt{8}-\sqrt{3})}/{-\sqrt{6}.(\sqrt{8}-\sqrt{3})}-1/\sqrt{6}`
`={-2}/\sqrt{6}-1/\sqrt{6}=-3/\sqrt{6}`
`={-3\sqrt{6}}/6={-\sqrt{6}}/2`
Vậy: `{2\sqrt{8}-\sqrt{12}}/{\sqrt{18}-\sqrt{48}}-{\sqrt{5}+\sqrt{27}}/{\sqrt{30}+\sqrt{162}}={-\sqrt{6}}/2`
