\(13-\sqrt{\left(\sqrt{2}-3\right)^2}\)
=13-/\(\sqrt{2}\)-3/
=13-(3-\(\sqrt{2}\))
=13-3+\(\sqrt{2}\)
=10\(\sqrt{2}\)
\(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{2-2\sqrt{2}+1}-\sqrt{4-2.2\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{2}-1-2+\sqrt{2}\)
\(=2\sqrt{2}-3\)
1) \(13-\sqrt{\left(\sqrt{2}-3\right)^2}=13-\left|\sqrt{2}-3\right|=13+\sqrt{2}-3=10+\sqrt{2}\)
2) \(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}=\left|\sqrt{2}-1\right|-\left|2-\sqrt{2}\right|=\sqrt{2}-1-2+\sqrt{2}=2\sqrt{2}-3\)
1) 13−√(√2−3)2=13−∣∣√2−3∣∣=13+√2−3=10+√213−(2−3)2=13−|2−3|=13+2−3=10+2
2) √3−2√2−√6−4√2=√(√2−1)2−√(2−√2)2=∣∣√2−1∣∣−∣∣2−√2∣∣=√2−1−2+√2=2√2−3
a: Ta có: \(13-\sqrt{\left(\sqrt{2}-3\right)^2}\)
\(=13-3+\sqrt{2}\)
\(=10+\sqrt{2}\)
b: Ta có: \(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{2}-1-2+\sqrt{2}\)
\(=2\sqrt{2}-3\)
