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a. Sửa đề: \(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
biến đổi vế trái :
ta có :\(\left(3+\sqrt{5}\right)\left(\sqrt{10}+\sqrt{2}\right)\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3+\sqrt{5}}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right).\sqrt{3-\sqrt{5}}\)
=\(\sqrt{3^2-\left(\sqrt{5}\right)^2}.\sqrt{3+\sqrt{5}}.\left(\sqrt{10}-\sqrt{2}\right)\)
=2(\(\sqrt{30+10\sqrt{5}}-\sqrt{6+2\sqrt{5}}\))
=2(\(\sqrt{5}+5-\sqrt{5}-1\))
=2.4=8=VP
=> đpcm
b. Đặt vế trái là A
ta có \(A^2=\sqrt{2}+1-2\sqrt{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\sqrt{2}-1\)
=\(2\sqrt{2}-2\)
=2\(\left(\sqrt{2}-1\right)\)
=> A=\(\sqrt{2\left(\sqrt{2}-1\right)}\)
vậy VT=VP =>đpcm
a) \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)
Ta có : VT = \(2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)
\(\Leftrightarrow VT=9\) \(=VP\)
Vậy.........
b) \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}=\sqrt{6}\)
<=> \(\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2=6\)
Ta có : VT = \(2+\sqrt{3}+2-\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
= \(4+2\sqrt{4-3}=4+2=6\)
=> VT = VP
Vậy.....
c) \(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
Ta có : VT = \(\dfrac{\sqrt{4}}{\sqrt{\left(2-\sqrt{5}\right)^2}}-\dfrac{\sqrt{4}}{\sqrt{\left(2+\sqrt{5}\right)^2}}\)
= \(\dfrac{2}{\sqrt{5}-2}-\dfrac{2}{2+\sqrt{5}}=\dfrac{4+2\sqrt{5}-2\sqrt{5}+4}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
= \(\dfrac{8}{5-4}=8\)
=> VT = VP
Vậy....
a) Biến đổi vế trái ta có:
VT= \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
= \(2\sqrt{6}-4\sqrt{2}+1+4\sqrt{2}+8-2\sqrt{6}\)
= 9 = VP
Vậy đẳng thức đc chứng minh
b) Đặt vế trái = A = \(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(A^2=\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2\)
\(A^2=2+\sqrt{3}+2-\sqrt{3}+2.\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(A^2=4+2.\sqrt{4-3}=4+2.1=6\)
\(\Rightarrow A=\sqrt{6}=VP\)
Vậy đẳng thức đc chứng minh
a) \(A=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}\)
Biến đổi vế trái :
VT = \(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{\sqrt{2}\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\left|\sqrt{3}+1\right|}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\left|\sqrt{3}-1\right|}\)
\(=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)}{\sqrt{3}+3}+\frac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}=\frac{\sqrt{2}\left(2+\sqrt{3}\right)\left(\sqrt{3}-3\right)+\sqrt{2}\left(2-\sqrt{3}\right)\left(\sqrt{3}+3\right)}{\left(\sqrt{3}+3\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{\sqrt{2}\left(6-2\sqrt{3}+3\sqrt{3}-3+6+2\sqrt{3}-3\sqrt{3}-3\right)}{9-3}=\frac{6\sqrt{2}}{6}=\sqrt{2}=VP\left(đpcm\right)\)
b) \(B=\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8\)
Biến đổi vế trái :
VT = \(\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\sqrt{5+\sqrt{21}}\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5+\sqrt{21}}\sqrt{5-\sqrt{21}}\)
\(=\sqrt{2}\sqrt{5+\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{25-21}=\sqrt{10+2\sqrt{21}}\left(\sqrt{7}-\sqrt{3}\right)\sqrt{4}=\left|\sqrt{7}+\sqrt{3}\right|\left(\sqrt{7}-\sqrt{3}\right)2\)
\(=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)2=\left(7-3\right)2=4.2=8=VP\left(đpcm\right)\)
Ta có VT: \(2\sqrt{2}.\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}\)
=\(2\sqrt{6}-4\sqrt{2}+1^2+4\sqrt{2}+\left(2\sqrt{2}\right)^2-2\sqrt{6}\)
=1+8
=9(bằng VP)
Chúc học tốt:))