Bài 1: Tính
Ta có: \(\left(\sqrt{18}+\sqrt{32}-\sqrt{50}\right)\cdot\sqrt{2}\)
\(=\sqrt{36}+\sqrt{64}-\sqrt{100}\)
\(=6+8-10=14-10=4\)
\(\left(\sqrt{18}+\sqrt{32}-\sqrt{50}\right).\sqrt{2}\)
\(=\sqrt{2}.\left(\sqrt{9}+\sqrt{16}-\sqrt{25}\right).\sqrt{2}\)
\(=2.\left(3+4-5\right)=2.2=4\)
Bài giải
\(\left(\sqrt{18}+\sqrt{32}-\sqrt{50}\right)\cdot\sqrt{2}\)
\(=\sqrt{36}+\sqrt{64}-\sqrt{100}\)
\(=6+8-10\)
\(=4\)