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\(x=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
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PT
0
22 tháng 9 2021
a) \(=\sqrt{\left(3\sqrt{5}-2\right)^2}+\sqrt{\left(3\sqrt{5}+2\right)^2}=3\sqrt{5}-2+3\sqrt{5}+2=6\sqrt{5}\)
b) \(=\sqrt{\left(2\sqrt{5}+3\right)^2}+\sqrt{\left(2\sqrt{5}-3\right)^2}=2\sqrt{5}+3+2\sqrt{5}-3=4\sqrt{5}\)
C
2
31 tháng 8 2021
\(a,\sqrt{29-12\sqrt{5}}=2\sqrt{5}-3\\ b,\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\\ =\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\\ =\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\\ =\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\\ =\sqrt{1}=1\)
31 tháng 8 2021
a: \(\sqrt{29-12\sqrt{5}}=2\sqrt{5}-3\)
b: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
=1
VH
22 tháng 8 2020
a) \(E=\sqrt{\left|12\sqrt{5}-29\right|}-\sqrt{12\sqrt{5}+29}\)
\(\Leftrightarrow E^2=\left|12\sqrt{5}-29\right|-12\sqrt{5}-29\)
\(\Leftrightarrow E^2=29-12\sqrt{5}-12\sqrt{5}-29\)
\(\Leftrightarrow E^2=-24\sqrt{5}\)
\(\Leftrightarrow E=-2\sqrt{6\sqrt{5}}\)
b) Đặt \(F=\sqrt{\left|40\sqrt{2}-57\right|}-\sqrt{40\sqrt{2}+57}\)
\(\Leftrightarrow F^2=\left|40\sqrt{2}-57\right|-40\sqrt{2}-57\)
\(\Leftrightarrow F^2=57-40\sqrt{2}-40\sqrt{2}-57\)
\(\Leftrightarrow F^2=-80\sqrt{2}\)
\(\Leftrightarrow F=-4\sqrt{5\sqrt{2}}\)
10 tháng 9 2021
\(x=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
=1
Thay x=1 vào B, ta được:
\(B=-\sqrt{1}\cdot\left(\sqrt{1}-1\right)=0\)
5 tháng 10 2021
\(\sqrt{5-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{5-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{6-\sqrt{5}}\)
22 tháng 7 2021
H=\(\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
H=\(\sqrt{5}-\sqrt{3-\sqrt{20-12\sqrt{5}+9}}\)
H=\(\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}\right)^2-2.3.2\sqrt{5}+3^2}}\)
H=\(\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
H=\(\sqrt{5}-\sqrt{3-\left|2\sqrt{5}-3\right|}\)
H=\(\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}\)
H=\(\sqrt{5}-\sqrt{3-2\sqrt{5}+3}\)
H=\(\sqrt{5}-\sqrt{6-2\sqrt{5}}\)
H=\(\sqrt{5}-\sqrt{5-2\sqrt{5}+1}\)
H=\(\sqrt{5}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1}\)
H=\(\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
H=\(\sqrt{5}-\left|\sqrt{5}-1\right|\)
H=\(\sqrt{5}-\left(\sqrt{5}-1\right)\)
H=\(\sqrt{5}-\sqrt{5}+1\)
H=1
23 tháng 7 2021
Ta có: \(H=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{5}-\sqrt{3-2\sqrt{5}+3}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}-\sqrt{5}+1\)
=1
7 tháng 7 2023
a: Sửa đề: \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(A=2-\sqrt{3}+\sqrt{3}-1=1\)
c: \(C=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
d: Sửa đề: \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3\)
=6
H9
HT.Phong (9A5)
CTVHS
7 tháng 7 2023
a) \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(A=\left|4-\sqrt{15}\right|+\sqrt{15}\)
\(A=4-\sqrt{15}+\sqrt{15}\)
\(A=4\)
b) \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)}\)
\(B=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)
\(B=2-\sqrt{3}-1+\sqrt{3}\)
\(B=1\)
c) \(C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}\)
\(C=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(C=\left|3\sqrt{5}-2\right|-\left|3\sqrt{5}+2\right|\)
\(C=3\sqrt{5}-2-3\sqrt{5}-2\)
\(C=-4\)
d) \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}\)
\(D=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(D=\left|2\sqrt{5}+3\right|-\left|2\sqrt{5}-3\right|\)
\(D=2\sqrt{5}+3-2\sqrt{5}+3\)
\(D=6\)
x = \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
x = \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.3+3^2}\) - \(\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+3^2}\)
x = \(\sqrt{\left(2\sqrt{5}+3\right)^2}\) - \(\sqrt{\left(2\sqrt{5}-3\right)^2}\)
x = \(|\) \(2\sqrt{5}+3\) \(|\) - \(|\) \(2\sqrt{5}-3\) \(|\)
x = \(\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)\)
x = \(2\sqrt{5}+3-2\sqrt{5}+3\) = 6
\(x=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(\Rightarrow x=\sqrt{\left(3+2\sqrt{5}\right)^2}-\sqrt{\left(3-2\sqrt{5}\right)^2}\)
\(\Rightarrow x=3+2\sqrt{5}-\left(2\sqrt{5}-3\right)\)
\(\Rightarrow x=3+2\sqrt{5}-2\sqrt{5}+3\)
\(\Rightarrow x=6\)