Ta có: \(VT=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}\right)^2-2\sqrt{7}\cdot1+1}-\sqrt{\left(\sqrt{7}\right)^2+2\sqrt{7}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2=VP\) (đpcm)
b)
√8-2√7 -√8+2√7 =-2
1)chứng minh
a)\(11+6\sqrt{2}=\left(3+\sqrt{2}\right)^2\)
b)\(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}=6\)
2)chứng minh
a)\(8-2\sqrt{7}=\left(\sqrt{7}-1\right)^2\)
b)\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=2\)
Rút gọn các biểu thức:
\(a,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(b,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
rút gọn bt
C=\(\sqrt{8+2\sqrt{7}}+\sqrt{8-2\sqrt{7}}\)
1.tính
a) \(\sqrt{3-2\sqrt{2}}\)
b)\(\sqrt{28+10\sqrt{3}}\)
c)\(\sqrt{14+6\sqrt{5}}\)
d)\(\sqrt{13-4\sqrt{3}}\)
2.tính
a)\(\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
b)\(\sqrt{18+8\sqrt{2}}-\sqrt{18-8\sqrt{2}}\)
Chứng minh :
a) \(9+4\sqrt{5}=\left(\sqrt{5}+2\right)^2\)
b) \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
c) \(\left(4-\sqrt{7}\right)^2=23-8\sqrt{7}\)
d) \(\sqrt{23+8\sqrt{7}}-\sqrt{7}=4\)
CMR:
a) \(9+4\sqrt{5}=\left(\sqrt{5}+2\right)^2\)
b) \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
c) \(23-8\sqrt{7}=\left(4-\sqrt{7}\right)^2\)
d) \(\sqrt{17-12\sqrt{2}}+2\sqrt{2}=3\)
Chứng minh rằng : 23-8√7=(4-√7) 2 2 là mũ hai
