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Câu hỏi của Trân Vũ Mai Ngọc - Toán lớp 9 - Học toán với OnlineMath
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\)
=>\(A^2=4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}+4-\)\(\sqrt{10+2\sqrt{5}}.\)
=\(8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\left(\sqrt{5}-1\right)\)
=\(8+2\sqrt{5}-2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\)
=>\(\sqrt{5}+1\)
Đặt biểu thứa là A
Bình phương 2 vế ta dc:
\(8+2\sqrt{4+\sqrt{10+2\sqrt{5}}\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{\left(10-\left(10+2\sqrt{5}\right)\right)}\)
\(A^2=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\left(\sqrt{5}-1\right)\)
Do A>0 nên :
\(A=\sqrt{8+2\left(\sqrt{5}-1\right)}=6+2\sqrt{5}=\sqrt{5}+1\)
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Như !
\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{16-10-2\sqrt{5}}\)
\(A^2=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(A^2=8+2\left|\sqrt{5}-1\right|=8+2\left(\sqrt{5}-1\right)=8+2\sqrt{5}-2=6+2\sqrt{5}\)
\(\Rightarrow\)\(A=\sqrt{A^2}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}=\left|\sqrt{5}+1\right|=\sqrt{5}+1\)
...
\(1.B=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
\(B^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\right)^2\)
\(B^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10-2\sqrt{5}}\right)}\)
\(B^2=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\text{ |}\sqrt{5}-1\text{ |}=6+2\sqrt{5}=5+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)
\(\text{ |}B\text{ |}=\text{ |}\sqrt{5}+1\text{ |}=\sqrt{5}+1\)
\(2.C=\sqrt{5\sqrt{9}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}=\sqrt{15+5\sqrt{48-10\text{ |}\sqrt{3}+2\text{ |}}}=\sqrt{15+5\sqrt{25+2.5\sqrt{3}+3}}=\sqrt{15+5\text{ |}5+\sqrt{3}\text{ |}}=\sqrt{35+5\sqrt{3}}\)
\(3.D=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2+2.\sqrt{2}.\sqrt{x-2}+2}+\sqrt{x-2-2.\sqrt{2}.\sqrt{x-2}+2}=\text{ |}\sqrt{x-2}+\sqrt{2}\text{ |}+\text{ |}\sqrt{x-2}-\sqrt{2}\text{ |}=2\sqrt{x-2}\)
a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
nhân cả hai vế với \(\sqrt{2}\), ta được:
\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)
\(=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)
\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)
\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)
\(=\sqrt{7}-1-\sqrt{7}-1\)
\(=-2\)
\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)
\(=>M^2=4-\sqrt{10-2\sqrt{5}}+2\sqrt{\left(4-\sqrt{10-2\sqrt{5}}\right)\left(4+\sqrt{10-2\sqrt{5}}\right)}\)
\(+4+\sqrt{10-2\sqrt{5}}\)
\(M^2=8+2\)\(\sqrt{16-\left(\sqrt{10-2\sqrt{5}}\right)^2}\)\(=8+2\sqrt{16-10+2\sqrt{5}}\)
\(=>M^2=8+2\sqrt{6+2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}+1\right)^2}=8+2\sqrt{5}+2\)
\(=10+2\sqrt{5}\)
\(=>M=\sqrt{10+2\sqrt{5}}\)