Đặt A = \(\sqrt{10+2\sqrt{24}}-\sqrt{10-2\sqrt{24}}\) ( A > 0 )
\(A^2=10+2\sqrt{24}+10-2\sqrt{24}-2\sqrt{\left(10+2\sqrt{24}\right)\left(10-2\sqrt{24}\right)}\)
\(A^2=20-2\sqrt{4}=20-4=16\Rightarrow A=4\left(A>0\right)\)
=> ĐPcm
k liên quan 1 tí nha!!
\(A=\sqrt{a+2\sqrt{a-1}}+\sqrt{a-2\sqrt{a-1}}\)
\(A=\sqrt{a-1+2\sqrt{a-1}+1}+\sqrt{a-1-2\sqrt{a-1}+1}\)
\(A=\sqrt{\left(\sqrt{a-1}+1\right)^2}+\sqrt{\left(\sqrt{a-1}-1\right)^2}\)
trường hợp 1: \(\sqrt{a-1}-1\ge0\)
<=>\(\sqrt{a-1}\ge1\)
<=> \(a\ge2\)
thì trỉ tuyệt đối của \(\left(\sqrt{a-1}-1\right)=\sqrt{a-1}-1\)
\(A=\sqrt{a-1}+1+\sqrt{a-1}-1=2\sqrt{a-1}\)
trường hợp 2 :\(\sqrt{a-1}-1
