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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 \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)
d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)
\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)
\(=\dfrac{3}{x-y}\)
#)Giải :
Bình phương hai vế, ta được :
\(B^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{\left(16-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{6-2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\left(\sqrt{5}-1\right)\)
Do \(B>0\)nên \(B=\sqrt{8+2\left(\sqrt{5}-1\right)}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
#~Will~be~Pens~#
Bình phương hai vế, ta được:
B2=8+2√(4+√10+2√5)(4−√10+2√5)=8+2√(16−(10+2√5))B2=8+2(4+10+25)(4−10+25)=8+2(16−(10+25))
B2=8+2√6−2√5=8+2√(√5−1)2=8+2(√5−1)B2=8+26−25=8+2(5−1)2=8+2(5−1)
Do B>0B>0 nên B=√8+2(√5−1)=√6+2√5=√5+1B=8+2(5−1)=6+25=5+1
Tk mk nha
~ Hok tốt ~
Thanks m.n đã tk mk
1) \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=\left|2+\sqrt{5}\right|-\left|2-\sqrt{5}\right|\)
\(=2+\sqrt{5}+2-\sqrt{5}\)
\(=4\)
2) \(\sqrt{12-6\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=\left|3-\sqrt{3}\right|+\left|3+\sqrt{3}\right|\)
\(=3-\sqrt{3}+3+\sqrt{3}\)
\(=6\)
a: \(=2\sqrt{2}+1-3=2\sqrt{2}-2\)
b: \(=\sqrt{3}+1-2\sqrt{3}-1=-\sqrt{3}\)
c: \(=2-\sqrt{3}+\sqrt{3}-1=1\)
Đặt biểu thức trên là \(A\)
Ta có \(A^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}=8+2\sqrt{6-2\sqrt{5}}\)
\(=8+2\sqrt{5-2\sqrt{5}+1}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}\)
\(\Rightarrow A=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
cho hỏi sao ra được kết quả như vậy giải thích dùm đi
Đặt cái đấy là A
A2 = 8 + \(2\sqrt{6-2\sqrt{5}}\)
= 8 + \(2\sqrt{5}-2\)
= 6 + 2\(\sqrt{5}\)= (\(1+\sqrt{5}\))2
=> A = \(1+\sqrt{5}\)
\(\sqrt{10+\sqrt{2}-\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{10+\sqrt{2}-\sqrt{2+\sqrt{8+2.2\sqrt{2}+1}}}\)
\(=\sqrt{10+\sqrt{2}-\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\)
\(=\sqrt{10+\sqrt{2}-\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{10+\sqrt{2}-\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\sqrt{10+\sqrt{2}-\sqrt{2}-1}=\sqrt{9}=3\)