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a) Đặt A=\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
<=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\)=\(\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
= \(\sqrt{7}+1-\sqrt{7}+1=2\)
=> \(A=\frac{2}{\sqrt{2}}\sqrt{2}\)
b) Ta đặt \(B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
=> \(B^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)
= \(8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}\)=\(8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)
= \(5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)
=> B=\(\sqrt{5}+1\)
c) Ta xét \(A=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}\)
=> \(\sqrt{2}\cdot A=\sqrt{8+2\sqrt{3}\cdot\sqrt{5}}+\sqrt{8-2\sqrt{3}\cdot\sqrt{5}}\)
= \(\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
= \(\sqrt{3}+\sqrt{5}+\sqrt{5}-\sqrt{3}\)= \(2\sqrt{5}\)
=> A=\(\sqrt{5}\)
Ta có : \(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
= \(A-\sqrt{6-2\sqrt{5}}\)
= \(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-\sqrt{5}+1\)=1
Trả lời:
\(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(A^2=4+\sqrt{10+2\sqrt{5}}+2.\sqrt{4+\sqrt{10+2\sqrt{5}}}.\sqrt{4-\sqrt{10+2\sqrt{5}}}+4-\sqrt{10+2\sqrt{5}}\)
\(A^2=8+2\sqrt{16-10-2\sqrt{5}}\)
\(A^2=8+2\sqrt{6-2\sqrt{5}}\)
\(A^2=8+2\sqrt{5-2\sqrt{5}+1}\)
\(A^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(A^2=8+2.\left(\sqrt{5}+1\right)\)
\(A^2=8+2\sqrt{5}-2\)
\(A^2=6+2\sqrt{5}\)
\(A^2=5+2\sqrt{5}+1\)
\(A^2=\left(\sqrt{5}+1\right)^2\)
\(A=\sqrt{5}+1\)
\(B=\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
\(\sqrt{2}B=\sqrt{2}\sqrt{4+\sqrt{15}}+\sqrt{2}\sqrt{4-\sqrt{15}}-\sqrt{2}.2\sqrt{3-\sqrt{5}}\)
\(\sqrt{2}B=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)
\(\sqrt{2}B=\sqrt{5+2\sqrt{15}+3}+\sqrt{5-2\sqrt{15}+3}-2\sqrt{5-2\sqrt{5}+1}\)
\(\sqrt{2}B=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(\sqrt{2}B=\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}-2\sqrt{5}+2\)
\(\sqrt{2}B=2\)
\(B=\sqrt{2}\)
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
\(A=\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt{15}+3}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\)
\(B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{4+\sqrt{7}}\sqrt{4-\sqrt{7}}-\sqrt{\left(4-\sqrt{7}\right)^2}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}-\left|4-\sqrt{7}\right|\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=\sqrt{16-7}-4+\sqrt{7}\)
\(\Leftrightarrow\sqrt{4-\sqrt{7}}B=3-4+\sqrt{7}=-1+\sqrt{7}\)
\(\Leftrightarrow B=\frac{-1+\sqrt{7}}{\sqrt{4-\sqrt{7}}}\)
tíck mình nha bn thanks !!!!!!!!!!
Ta có A2 = 8 + \(2\sqrt{6-2\sqrt{5}}\)= 8 + 2(\(\sqrt{5}\)- 1)
= 6 + \(2\sqrt{5}\)= (\(\sqrt{5}+1\))2
Vậy A = \(\sqrt{5}+1\)
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\\ A^2=8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{6-2\sqrt{5}}\\ A^2=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\\ A=\sqrt{5}+1\)