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AT
2
PD
16 tháng 8 2020
\(\sqrt{7+\sqrt{40}}\)
\(=\sqrt{7+2\sqrt{2}.\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{5}\)
\(\)
20 tháng 7 2017
a/ \(\sqrt{21+8\sqrt{5}}+\sqrt{9-4\sqrt{5}}\)
= \(\sqrt{16+2.4.\sqrt{5}+5}+\sqrt{5-2.2\sqrt{5}+4}\)
= \(\sqrt{\left(4+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
= \(4+\sqrt{5}+\sqrt{5}-2=2+2\sqrt{5}\)
b/ \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{10}+\sqrt{15}}{2\left(\sqrt{2}+\sqrt{3}\right)}\) = \(\dfrac{\left(\sqrt{10}+\sqrt{15}\right)\left(\sqrt{2}-\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}\)
= \(\dfrac{\sqrt{20}-\sqrt{30}+\sqrt{30}-\sqrt{45}}{2\left(2-3\right)}\) = \(\dfrac{\sqrt{20}-\sqrt{45}}{-2}\) = \(\dfrac{2\sqrt{5}-3\sqrt{5}}{-2}\)
= \(\dfrac{-\sqrt{5}}{-2}=\dfrac{\sqrt{5}}{2}\)
KK
19 tháng 7 2019
Biểu thức B ko bt có sai đề ở căn thứ 2 ko ạ
Nếu nhân B với căn 2 thì cái căn thức nhất tách đc thành hđt (a+b)2 đấy ạ nhưng cái căn thứ 2 thì ko tách đc
BH
12 tháng 8 2019
\(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}=\left(2\sqrt{5}+3\right)-\left(2\sqrt{5}-3\right)=6\)
\(\sqrt{8-2\sqrt{15}}-\sqrt{23-4\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)-\left(2\sqrt{5}-\sqrt{3}\right)=-\sqrt{5}\)
\(\sqrt{8-12\sqrt{5}}+\sqrt{48+6\sqrt{15}}=\left(\sqrt{5}-\sqrt{3}\right)+\left(3\sqrt{5}+\sqrt{3}\right)=4\sqrt{5}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\left(5-2\sqrt{6}\right)+\left(5+2\sqrt{6}\right)=10\)
\(\sqrt{15-6\sqrt{15}}+\sqrt{33-12\sqrt{6}}\) đề này sai ạ
\(\sqrt{16-6\sqrt{7}}+\sqrt{64-24\sqrt{7}}=\left(3-\sqrt{7}\right)+\left(6-2\sqrt{7}\right)=9-3\sqrt{7}\)
\(\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}=\left(3-\sqrt{5}\right)+\left(3+\sqrt{5}\right)=6\)
\(\sqrt{1-6\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\left(2\sqrt{2}+5\right)+\left(2\sqrt{2}-5\right)=4\sqrt{2}\)
\(\sqrt{46-6\sqrt{5}}+\sqrt{29-12\sqrt{5}}=\left(3\sqrt{5}-1\right)+\left(2\sqrt{5}-3\right)=5\sqrt{5}-4\)
#Học tốt ạ
21 tháng 7 2022
1: \(=\sqrt{36}=6\)
2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)
3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)
4: \(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
5: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)
TN
21 tháng 8 2018
a) \(\dfrac{2\sqrt{3}+2}{4\sqrt{3}+4}=\dfrac{2\left(\sqrt{3}+1\right)}{4\left(\sqrt{3}+1\right)}=\dfrac{1}{2}\)
b) \(\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{\sqrt{4}\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{5}}{2}\)
c) \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\\ =\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}=1+\sqrt{2}\)
d) \(\sqrt{9+\sqrt{17}}.\sqrt{9-\sqrt{17}}=\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}\\ =\sqrt{81-17}=\sqrt{64}=8\)
21 tháng 8 2018
\(a.\dfrac{2\sqrt{3}+2}{4\sqrt{3}+4}=\dfrac{2\left(\sqrt{3}+1\right)}{4\left(\sqrt{3}+1\right)}=\dfrac{2}{4}=\dfrac{1}{2}\)
\(b.\dfrac{\sqrt{10}+\sqrt{15}}{\sqrt{8}+\sqrt{12}}=\dfrac{\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)}=\dfrac{\sqrt{5}}{2}\)
\(c.\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}=\dfrac{\sqrt{2}+\sqrt{3}+2+2+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+2}=\dfrac{\sqrt{2}+\sqrt{3}+2}{\sqrt{2}+\sqrt{3}+2}+\dfrac{\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}{\sqrt{2}+\sqrt{3}+2}=1+\sqrt{2}\)
\(d.\sqrt{9+\sqrt{17}}.\sqrt{9-\sqrt{17}}=\sqrt{\left(9+\sqrt{17}\right)\left(9-\sqrt{17}\right)}=\sqrt{81-17}=8\)
\(\sqrt{8+\sqrt{15}}\)
=\(\sqrt{\frac{15}{2}+2\cdot\sqrt{\frac{15}{2}}\cdot\sqrt{\frac{1}{2}}+\frac{1}{2}}\)
=\(\sqrt{\left(\sqrt{\frac{15}{2}}+\sqrt{\frac{1}{2}}\right)^2}\)
=\(|\sqrt{\frac{15}{2}}+\sqrt{\frac{1}{2}}|\)
=\(\sqrt{\frac{15}{2}}+\sqrt{\frac{1}{2}}\)
\(\sqrt{9-\sqrt{77}}\)
=\(\sqrt{\frac{11}{2}-2\cdot\sqrt{\frac{11}{2}}\cdot\sqrt{\frac{7}{2}}+\frac{7}{2}}\)
=\(\sqrt{\left(\sqrt{\frac{11}{2}}-\sqrt{\frac{7}{2}}\right)^2}\)
=\(|\sqrt{\frac{11}{2}}-\sqrt{\frac{7}{2}}|\)
=\(\sqrt{\frac{11}{2}}-\sqrt{\frac{7}{2}}\)
\(\sqrt{10+\sqrt{99}}\)
=\(\sqrt{\frac{11}{2}+2\cdot\sqrt{\frac{11}{2}}\cdot\sqrt{\frac{9}{2}}+\frac{9}{2}}\)
=\(\sqrt{\left(\sqrt{\frac{11}{2}}+\sqrt{\frac{9}{2}}\right)^2}\)
=\(|\sqrt{\frac{11}{2}}+\sqrt{\frac{9}{2}}|\)
=\(\sqrt{\frac{11}{2}}+\sqrt{\frac{9}{2}}\)