tính A=căn(4+2căn3) +căn(4-2căn3) Question tính A=căn(4+2căn3) +căn(4-2căn3) in progress 0 Môn Toán Huy Gia 6 years 2020-10-15T06:01:35+00:00 2020-10-15T06:01:35+00:00 2 Answers 208 views 0
Answers ( )
$A=\sqrt[]{4+2\sqrt[]{3}}+\sqrt[]{4-2\sqrt[]{3}}$
$=\sqrt[]{(\sqrt[]{3})^2+2\sqrt[]{3}.1+1^2}+\sqrt[]{(\sqrt[]{3})^2-2.\sqrt[]{3}.1+1^2}$
$=\sqrt[]{(\sqrt[]{3}+1)^2}+\sqrt[]{(\sqrt[]{3}-1)^2}$
$=|\sqrt[]{3}+1|+|\sqrt[]{3}-1|$
$=\sqrt[]{3}+1+\sqrt[]{3}-1$ (vì √3-1>0)
$=2\sqrt[]{3}$
√( 4 + 2√3 ) + √( 4 – 2√3 )
= √( 3 +2√3 + 1 ) + √( 3 – 2√3 + 1 )
= √( √3 + 1 )² + √( √3 – 1 )²
= | √3 + 1 | + | √3 – 1 |
= √3 + 1 + √3 – 1
= 2√3