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NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 8 2020
\(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}-2\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{\left(\sqrt{3}-1\right)^2}\left(3-4\right)\)
\(=\left(\sqrt{3}-1\right).\left(-1\right)=1-\sqrt{3}\)
b/ \(=\sqrt{4-2\sqrt{3}}\left(\sqrt{3}+1\right)=\sqrt{\left(\sqrt{3}-1\right)^2}\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=3-1=2\)
c/ \(=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{6+2\sqrt{5}-2\sqrt{5}+3}=\sqrt{9}=3\)
d/ \(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
10 tháng 7 2018
\(\sqrt{\left(4-\sqrt{15}\right)^2}=\left|4-\sqrt{15}\right|=4-\sqrt{15}\)
\(\Rightarrow\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}=4-\sqrt{15}+\sqrt{15}=4\)
\(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)
\(\sqrt{\left(1-\sqrt{3}\right)^2}=\left|1-\sqrt{3}\right|=\sqrt{3}-1\)
\(\Rightarrow\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}=2-\sqrt{3}+\sqrt{3}-1=1\)
5 tháng 7 2018
Mình làm luôn nhé :
\(\sqrt{45-2.3\sqrt{5}+1}-\sqrt{20-2.3.2\sqrt{5}+9}\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5-\sqrt{45+2.2.\sqrt{2}.3\sqrt{5}+8}}\left(\sqrt{3}+\sqrt{5}\right).\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{7+2.\sqrt{7}.\sqrt{3}+3}\) Tới đây dễ rồi , bạn tự nhóm HĐT là ra ::v
16 tháng 7 2020
Bài 1:
a) Ta có: \(\sqrt{46-6\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{45-2\cdot\sqrt{45}\cdot1+1}-\sqrt{9-2\cdot\sqrt{9}\cdot\sqrt{20}+20}\)
\(=\sqrt{\left(\sqrt{45}-1\right)^2}-\sqrt{\left(3-\sqrt{20}\right)^2}\)
\(=\left|\sqrt{45}-1\right|-\left|3-\sqrt{20}\right|\)
\(=\sqrt{45}-1-3+\sqrt{20}\)
\(=\sqrt{45}+\sqrt{20}-4\)
\(=\sqrt{5}\left(3+2\right)-4=5\sqrt{5}-4\)
b) Ta có: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{8}+8}-\sqrt{45+2\cdot\sqrt{45}\cdot\sqrt{8}+8}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{8}\right)^2}-\sqrt{\left(\sqrt{45}+\sqrt{8}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{8}\right|-\left|\sqrt{45}+\sqrt{8}\right|\)
\(=\sqrt{8}-\sqrt{5}-\sqrt{45}-\sqrt{8}\)
\(=-\sqrt{5}-\sqrt{45}=-\sqrt{5}\left(1+\sqrt{9}\right)=-4\sqrt{5}\)
c) Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{7+4\sqrt{3}}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}\)
\(=\left(3-\sqrt{2}\right)\cdot\sqrt{\left(\sqrt{3}+2\right)^2}\)
\(=\left(3-\sqrt{2}\right)\left(\sqrt{3}+2\right)\)
\(=3\sqrt{3}+6-\sqrt{6}-2\sqrt{2}\)
d) Ta có: \(\left(\sqrt{7}-\sqrt{3}\right)\sqrt{10+2\sqrt{21}}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{7+2\cdot\sqrt{7}\cdot\sqrt{3}+3}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}\)
\(=\left(\sqrt{7}-\sqrt{3}\right)\cdot\left(\sqrt{7}+\sqrt{3}\right)\)
\(=\left(\sqrt{7}\right)^2-\left(\sqrt{3}\right)^2=7-3=4\)
23 tháng 8 2020
\(A=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{3+2\sqrt{2}}\)
\(A=\sqrt{2}-1-\sqrt{\left(\sqrt{2}+1\right)^2}\)
\(A=\sqrt{2}-1-\sqrt{2}-1=-2\)
B = \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
\(B=\sqrt{6+2\sqrt{5}-\sqrt{29-6\sqrt{20}}}\)
B = \(\sqrt{6+2\sqrt{5}-\sqrt{\left(3-\sqrt{20}\right)^2}}\)
\(B=\sqrt{6+2\sqrt{5}-2\sqrt{5}+3}\)
\(B=\sqrt{9}=3\)
27 tháng 5 2022
Bài 1:
a: \(=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|\)
\(=5-\sqrt{3}-2+\sqrt{3}=3\)
b; \(B=\dfrac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\cdot\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}\)
c: \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3+3-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=1\)
d: \(A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
29 tháng 5 2017
D = (4\(\sqrt{10}\) - 4\(\sqrt{6}\) + 5\(\sqrt{6}\) - 3\(\sqrt{10}\) )\(\sqrt{4-\sqrt{15}}\)
D = (\(\sqrt{10}\) + \(\sqrt{6}\) )\(\sqrt{4-\sqrt{15}}\)
D = \(\sqrt{\left(4-\sqrt{15}\right)10}\) + \(\sqrt{\left(4-\sqrt{15}\right)6}\)
D = \(\sqrt{40-10\sqrt{15}}\) + \(\sqrt{24-6\sqrt{15}}\)
D = \(\sqrt{\left(\sqrt{15}\right)^2-2.5.\sqrt{5}+5^2}\) + \(\sqrt{\left(\sqrt{15}\right)^2-2.3.\sqrt{15}+3^2}\)
D = \(\sqrt{\left(\sqrt{15}-5\right)^2}\) + \(\sqrt{\left(\sqrt{15}-3\right)^2}\)
D = 5 - \(\sqrt{15}\) + \(\sqrt{15}\) - 3 = 2
11 tháng 6 2015
1,\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
2, (tương tự ý 1 cũng tách thành hằng đẳng thức \(\sqrt{46-6\sqrt{5}}=\sqrt{\left(3\sqrt{5}-1\right)^2}\)và \(\sqrt{29-12\sqrt{5}}=\sqrt{\left(2\sqrt{5}-3\right)^2}\)
3,\(\left(\sqrt{2}-\sqrt{9}\right)\left(\sqrt{\left(3+\sqrt{2}\right)^2}\right)=\left(\sqrt{2}-3\right)\left(\sqrt{2}+3\right)=2-9=-7\)
4, tương tự ý 3
\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\sqrt{20.9}+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)