\(\sqrt{41-12\sqrt{5}}+\sqrt{41-12\sqrt{5}}\)
K
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AH
Akai Haruma
Giáo viên
5 tháng 7 2020
Lời giải:
\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}=\sqrt{6^2+5-2.6\sqrt{5}}-\sqrt{6^2+5+2.6\sqrt{5}}\)
\(=\sqrt{(6-\sqrt{5})^2}-\sqrt{(6+\sqrt{5})^2}=|6-\sqrt{5}|-|6+\sqrt{5}|\)
\(=(6-\sqrt{5})-(6+\sqrt{5})=-2\sqrt{5}\)
27 tháng 8 2018
\(1.\sqrt{\left(5+\sqrt{7}\right)^2}-\sqrt{8-2\sqrt{7}}=5+\sqrt{7}-\sqrt{7-2\sqrt{7}+1}=5+\sqrt{7}-\sqrt{7}+1=6\)
\(2.\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{3-2\sqrt{3}+1}=\sqrt{3}+1-\sqrt{3}+1=2\)
\(3.VT=\sqrt{11}-\sqrt{20-6\sqrt{11}}=\sqrt{11}-\sqrt{11-2.3\sqrt{11}+9}=\sqrt{11}-\sqrt{11}+3=3=VP\)
Vậy , đẳng thức được chứng minh .
\(4.VT=\sqrt{41+12\sqrt{5}}-\sqrt{41-12\sqrt{5}}=\sqrt{36+2.6\sqrt{5}+5}-\sqrt{36-2.6\sqrt{5}+5}=6+\sqrt{5}-6+\sqrt{5}=2\sqrt{5}=VP\)
Vậy , đẳng thức được chứng minh .
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 8 2021
\(\sqrt{7+4\sqrt{3}}=\sqrt{\left(2+\sqrt{3}\right)^2}=2+\sqrt{3}\)
\(\sqrt{8-2\sqrt{12}}=\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}=\left|\sqrt{6}-\sqrt{2}\right|=\sqrt{6}-\sqrt{2}\)
\(\sqrt{21+6\sqrt{6}}=\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}=\left|3\sqrt{2}-\sqrt{3}\right|=3\sqrt{2}-\sqrt{3}\)
\(\sqrt{15-6\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|=3-\sqrt{6}\)
\(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}=\left|2\sqrt{5}-3\right|=2\sqrt{5}-3\)
\(\sqrt{41+12\sqrt{5}}=\sqrt{\left(6+\sqrt{5}\right)^2}=6+\sqrt{5}\)
L
1
4 tháng 9 2023
\(\sqrt{28-10\sqrt{3}}\\ =\sqrt{3-10\sqrt{3}+25}\\ =\sqrt{\left(\sqrt{3}-5\right)^2}\\ =\left|\sqrt{3}-5\right|\\ =5-\sqrt{3}\)
\(\sqrt{41+12\sqrt{5}}\\ =\sqrt{5+12\sqrt{5}+36}\\ =\sqrt{\left(\sqrt{5}+6\right)}\\ =\left|\sqrt{5}+6\right|\\ =\sqrt{5}+6\)
\(\sqrt{32-10\sqrt{7}}\\ =\sqrt{7-10\sqrt{7}+25}\\ =\sqrt{\left(\sqrt{7}-5\right)^2}\\ =\left|\sqrt{7}-5\right|\\ =5-\sqrt{7}\)
\(\sqrt{11-4\sqrt{7}}\\ =\sqrt{7-4\sqrt{7}+4}\\ =\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|\\ =\sqrt{7}-2\)
KT
10 tháng 7 2018
\(\sqrt{\left(2\sqrt{2}-3\right)^2}+2\sqrt{2}=\left|2\sqrt{2}-3\right|+2\sqrt{2}=3-2\sqrt{2}+2\sqrt{2}=3\)
\(\sqrt{\left(\sqrt{10}-3\right)^2}+\sqrt{\left(\sqrt{10}-4\right)^2}=\left|\sqrt{10}-3\right|+\left|\sqrt{10}-4\right|\)
\(=\sqrt{10}-3+4-\sqrt{10}=1\)
\(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{3}+2\right|-\left|2-\sqrt{3}\right|=\sqrt{3}+2-2+\sqrt{3}=2\sqrt{3}\)
KT
10 tháng 7 2018
\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}=\sqrt{\left(6-\sqrt{5}\right)^2}-\sqrt{\left(6+\sqrt{5}\right)^2}\)
\(=6-\sqrt{5}-6-\sqrt{5}=-2\sqrt{5}\)
\(A=\sqrt{49a^2}+3a=7\left|a\right|+3a\)
Nếu \(a\ge0\)thì: \(A=7a+3a=10a\)
Nếu \(a< 0\)thì: \(A=-7a+3a=-4a\)
\(B=3\sqrt{9a^6}-6a^3=9\left|a^3\right|-6a^3\)
Nếu \(a\ge0\)thì: \(B=9a^3-6a^3=3a^3\)
Nếu \(a< 0\)thì: \(B=-9a^3-6a^3=-15a^3\)
18 tháng 7 2017
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{8+\sqrt{2.4}+\sqrt{5.4}+\sqrt{10.4}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{1}\right)^2+\left(\sqrt{2}\right)^2+\left(\sqrt{5}\right)^2+2.\sqrt{2}.\sqrt{1}+2\sqrt{1}.\sqrt{5}+2\sqrt{5}.\sqrt{2}}-\sqrt{2}-\sqrt{5}\)
=\(\sqrt{\left(\sqrt{1}+\sqrt{2}+\sqrt{5}\right)^2}\)
= \(\sqrt{1}+\sqrt{2}+\sqrt{5}\)
18 tháng 7 2017
phần trên mk làm thiếu \(-\sqrt{2}-\sqrt{5}\)
kết quả là 1 mới đúng
20 tháng 7 2017
a) \(\sqrt{\dfrac{2-\sqrt{3}}{2}}+\dfrac{1-\sqrt{3}}{2}\)
= \(\sqrt{\dfrac{4-2\sqrt{3}}{4}}+\dfrac{1-\sqrt{3}}{2}\)
= \(\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{2}+\dfrac{1-\sqrt{3}}{2}\)
= \(\dfrac{\sqrt{3}-1+1-\sqrt{3}}{2}\)
= 0
b) \(\sqrt{41+6\sqrt{6}-12\sqrt{10}-4\sqrt{15}}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{18+20+3+2\sqrt{54}-2\sqrt{360}-2\sqrt{60}}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{\left(\sqrt{18}-\sqrt{20}+\sqrt{3}\right)^2}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{18}-2\sqrt{5}+\sqrt{3}+2\sqrt{5}-\sqrt{3}\)
= \(\sqrt{18}\)
4 tháng 9 2020
Bài 1:
a) \(=5.|2a|-5a^2\)
b) \(=7\left(a-1\right)+5a=12a-7\)
c) \(|a-2|-5\sqrt{a+2}\)
Bài 2:
a) \(=3-\sqrt{2}+5-\sqrt{2}=8-2\sqrt{2}\)
b) \(=3+\sqrt{2}-\left(3-\sqrt{2}\right)\)
\(=2\sqrt{2}\)
c) \(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)
\(=-2\sqrt{5}\)
TT
5 tháng 9 2020
a) \(5\sqrt{4a^2}-5a^2\)
\(=5.|2a|-5a^2\)
b) \(7\sqrt{\left(a-1\right)^2}+5a\)
\(=7\left(a-1\right)+5a\)
\(=12a-7\)
c) \(\sqrt{\left(2-a\right)^2}-5\sqrt{a+2}\)
\(=|a-2|-5\sqrt{a+2}\)
bài 2:
a)\(\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-5\right)^2}\)
\(=3-\sqrt{2}+5-\sqrt{2}\)
\(=8-2\sqrt{2}\)
b) \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=3+\sqrt{2}-\left(3-\sqrt{2}\right)\)
\(=2\sqrt{2}\)
c)\(\sqrt{41-12\sqrt{5}}-\sqrt{41+12\sqrt{5}}\)
\(=6-\sqrt{5}-\left(6+\sqrt{5}\right)\)
\(=-2\sqrt{5}\)
Lời giải:
\(\sqrt{41-12\sqrt{5}}+\sqrt{41-12\sqrt{5}}=2\sqrt{41-12\sqrt{5}}=2\sqrt{6^2-2.6\sqrt{5}+5}\)
\(=2\sqrt{(6-\sqrt{5})^2}=2|6-\sqrt{5}|=2(6-\sqrt{5})\)
\(=\sqrt{41-2.6\sqrt{5}}+\sqrt{41-2.6\sqrt{5}}\)
\(=\sqrt{\left(6-\sqrt{5}\right)^2}+\sqrt{\left(6-\sqrt{5}\right)^2}=\left|6-\sqrt{5}\right|+\left|6-\sqrt{5}\right|=2\left(6-\sqrt{5}\right)=12-2\sqrt{5}\)