\(\sqrt{8-2\sqrt{7}}-\sqrt{23-8\sqrt{7}}=\) \(\sqrt{1-2\sqrt{7}+7}-\sqrt{7-2.4.\sqrt{7}+16}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-4\right)^2}\)
\(=\sqrt{7}-1-\left(-\sqrt{7}+4\right)\)
\(=\sqrt{7}-1+\sqrt{7}-4\)\(=2\sqrt{7}-5\)
chúc bn học tốt
=\(\sqrt{\left(\sqrt{7}-1\right)^2}\)- \(\sqrt{\left(4-\sqrt{7}\right)^2}\)
= \(\sqrt{7}\)- 1 - 4 + \(\sqrt{7}\)
= \(2\sqrt{7}\)-5
đ/á ra hơi kì
#mã mã#
mình nghĩ là hai bn làm đúng đó
\(\sqrt{8-2\sqrt{7}}-\sqrt{23-8\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{16-8\sqrt{7}+7}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(4-\sqrt{7}\right)^2}\)
\(=\sqrt{7}-1-4+\sqrt{7}\)
\(=2\sqrt{7}-5\)
Học tốt
Bài làm :
Ta có :
\(\sqrt{8-2\sqrt{7}}-\sqrt{23-8\sqrt{7}}\)
\(=\sqrt{7-2\sqrt{7}+1}-\sqrt{16-8\sqrt{7}+7}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(4-\sqrt{7}\right)^2}\)
\(=\sqrt{7}-1-4+\sqrt{7}\)
\(=2\sqrt{7}-5\)
