Bài 1: Tính a) (3 √2 + 2 √3 ) √2 – √24 b) 9/ √3 – √2 – 3 √27 – 4 √8 22/11/2021 Bởi Rylee Bài 1: Tính a) (3 √2 + 2 √3 ) √2 – √24 b) 9/ √3 – √2 – 3 √27 – 4 √8
$ a) (3\sqrt[]{2}+2\sqrt[]{3}).\sqrt[]{2}-\sqrt[]{24}$ `=6+2\sqrt{6}-2\sqrt{6}` `=6` $ b) \dfrac{9}{\sqrt[]{3}}-\sqrt[]{2}-3\sqrt[]{27}-4\sqrt[]{8}$ `=3\sqrt{3}-\sqrt{2}-9\sqrt{3}-8\sqrt{2}` `=-6\sqrt{3}-9\sqrt{2}` Bình luận
a,$(3√2+2√3)√2-√24$ $=3√2.√2+2√3.√2-2√6$ $=3.2+2√6-2√6$ $=6$ b,$9/√3 -√2-3√27-4√8$ $=3√3-√2-9√3-8√2$ $=-6√3-9√2$ Bình luận
$ a) (3\sqrt[]{2}+2\sqrt[]{3}).\sqrt[]{2}-\sqrt[]{24}$
`=6+2\sqrt{6}-2\sqrt{6}`
`=6`
$ b) \dfrac{9}{\sqrt[]{3}}-\sqrt[]{2}-3\sqrt[]{27}-4\sqrt[]{8}$
`=3\sqrt{3}-\sqrt{2}-9\sqrt{3}-8\sqrt{2}`
`=-6\sqrt{3}-9\sqrt{2}`
a,$(3√2+2√3)√2-√24$
$=3√2.√2+2√3.√2-2√6$
$=3.2+2√6-2√6$
$=6$
b,$9/√3 -√2-3√27-4√8$
$=3√3-√2-9√3-8√2$
$=-6√3-9√2$