rút gọn biểu thức
\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)
K
Khách
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9 tháng 10 2021
\(1,\\ a,=\sqrt{\left(3+\sqrt{7}\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}=3+\sqrt{7}-\sqrt{7}+1=4\\ b,K=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}-1}{\sqrt{2}\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{2}}=\dfrac{\sqrt{2}}{2}\\ c,=\sqrt{\left(6-2\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-4\right)^2}=6-2\sqrt{6}+2\sqrt{6}-4=2\\ e,=\sqrt{\left(2-\sqrt{2}\right)^2}-\left(\sqrt{6}-\sqrt{2}\right)=2-\sqrt{2}-\sqrt{6}+\sqrt{2}=2-\sqrt{6}\)
\(2,\\ a,A=\dfrac{x-3\sqrt{x}+3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{x+9}\\ A=\dfrac{x+9}{\left(\sqrt{x}-3\right)\left(x+9\right)}=\dfrac{1}{\sqrt{x}-3}\\ b,x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\\ \Leftrightarrow A=\dfrac{1}{\sqrt{3}+1-3}=\dfrac{1}{\sqrt{3}+2}=2-\sqrt{3}\)
29 tháng 8 2019
a/ \(\sqrt{2}+\sqrt{6}\)
b/ Sửa đề:
\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}=1\)
c/ \(1+\sqrt{2}+\sqrt{5}\)
KT
2 tháng 8 2018
a) \(\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{3}\)
\(=\sqrt{6+2\sqrt{6}+2\sqrt{3}+2\sqrt{2}}-\sqrt{3}\)
\(=\sqrt{\left(\sqrt{3}+\sqrt{2}+1\right)^2}-\sqrt{3}\)
\(=\sqrt{3}+\sqrt{2}+1-\sqrt{3}\)
\(=\sqrt{2}+1\)
b) \(\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}\)
\(=\sqrt{10+2\sqrt{15}-2\sqrt{6}-2\sqrt{10}}\)
\(=\sqrt{\left(\sqrt{5}+\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{5}+\sqrt{3}-\sqrt{2}\)
GN
1
21 tháng 12 2016
\(A=\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)
\(=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)
\(=\sqrt{2+3+5+2\left(\sqrt{2.3}+\sqrt{2.5}+\sqrt{3.5}\right)}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}+\sqrt{5}\)
ND
2
8 tháng 8 2021
Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(\Leftrightarrow A^2=8+2\sqrt{16-10-2\sqrt{5}}\\ \Leftrightarrow A^2=8+2\sqrt{6-2\sqrt{5}}\\ \Leftrightarrow A^2=8+2\left(\sqrt{5}-1\right)\\ \Leftrightarrow A^2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\\ \Leftrightarrow A=\sqrt{5}+1\)
Vậy \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)
\(=\left(2\sqrt{3}+3\sqrt{2}\right)\sqrt{5}+\sqrt{2^3\sqrt{3}}\)
Ta có √[ 5 + 2 + 3 + 2√(2×3) + 2√(2×5) + 2√(3×5)] = √[(√2 + √3 + √5)2] = √2 + √3 + √5