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D
26 tháng 7 2018
\(A=\sqrt{19-3\sqrt{40}}-\sqrt{19+3\sqrt{40}}=\sqrt{19-2\sqrt{90}}-\sqrt{19+2\sqrt{90}}=\sqrt{10-2.\sqrt{10}.3+9}-\sqrt{10+2.\sqrt{10}.3+9}=\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{\left(\sqrt{10}+3\right)^2}=\sqrt{10}-3-\sqrt{10}-3=-6\)\(B=\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}=\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}+\sqrt{6+2.\sqrt{3}.\sqrt{6}+3}-\sqrt{24+12\sqrt{3}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{\sqrt{3}}\right)^2}-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}=\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}=0\)
Y
4 tháng 7 2019
\(C=\sqrt{6+2\sqrt{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\) \(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\) \(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(D=\sqrt{\frac{8+2\sqrt{15}}{2}}-\sqrt{\frac{14-6\sqrt{5}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{2}}\)
\(=\frac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\frac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
\(E=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
\(F=\sqrt{\frac{24-6\sqrt{7}}{2}}-\sqrt{\frac{24+6\sqrt{7}}{2}}\) \(=\sqrt{\frac{21-2\sqrt{21\cdot3}+3}{2}}-\sqrt{\frac{21+2\sqrt{21\cdot3}+3}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{21}-\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{21}+\sqrt{3}\right)^2}{2}}\)
\(=\frac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=\frac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
\(G=\left(3+\sqrt{3}\right)\cdot\sqrt{12-6\sqrt{3}}\) \(=\left(3+\sqrt{3}\right)\cdot\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(H=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}-2-3-\sqrt{5}=-5\)
\(I=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(=2\sqrt{2}-1-2\sqrt{3}+1=2\sqrt{2}-2\sqrt{3}\)
TN
14 tháng 8 2018
\(A=\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)
\(B=\left(\sqrt{5}+\sqrt{3}\right)\left(5-\sqrt{15}\right)=\sqrt{5}\left(5-3\right)=2\sqrt{5}\)
\(C=\left(\sqrt{45}+\sqrt{63}\right)\left(\sqrt{7}-\sqrt{5}\right)=\sqrt{9}\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)=\sqrt{9}\left(7-5\right)=2\sqrt{9}\)
\(D=\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}-1+\sqrt{3}+1}{3-1}=\dfrac{2\sqrt{3}}{2}=\sqrt{3}\)
\(E=\dfrac{5+\sqrt{5}}{5-\sqrt{5}}+\dfrac{5-\sqrt{5}}{5+\sqrt{5}}=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{5^2-\sqrt{5}^2}=\dfrac{60}{20}=3\)
AH
Akai Haruma
Giáo viên
30 tháng 6 2019
a)
\((\sqrt{3}-2\sqrt{12}+2\sqrt{4})(\sqrt{27}+\sqrt{144}-2\sqrt{16})\)
\(=(\sqrt{3}-4\sqrt{3}+4)(3\sqrt{3}+12-8)\)
\(=(-3\sqrt{3}+4)(3\sqrt{3}+4)=4^2-(3\sqrt{3})^2=16-27=-11\)
b)
\((2\sqrt{5}+2\sqrt{3})^2-4\sqrt{60}\)
\(=(2\sqrt{5})^2+2.2\sqrt{5}.2\sqrt{3}+(2\sqrt{3})^2-8\sqrt{15}\)
\(=32+8\sqrt{15}-8\sqrt{15}=32\)
c)
\(\sqrt{6}(3\sqrt{12}-4\sqrt{3}+\sqrt{48}-5\sqrt{6})\)
\(=3\sqrt{72}-4\sqrt{18}+\sqrt{6.48}-5.\sqrt{36}\)
\(=18\sqrt{2}-12\sqrt{2}+12\sqrt{2}-30=18\sqrt{2}-30\)
AH
Akai Haruma
Giáo viên
30 tháng 6 2019
d)
\((\sqrt{2}-\sqrt{3})(\sqrt{6}+\sqrt{2})(\sqrt{2}+\sqrt{3})\)
\(=(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})(\sqrt{6}+\sqrt{2})\)
\(=(2-3)(\sqrt{6}+\sqrt{2})=-(\sqrt{6}+\sqrt{2})\)
e) Biểu thức bên trong căn lớn âm nên biểu căn bậc 2 không có nghĩa
f)
\((\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}).\frac{1}{\sqrt{3}+5}\)
\(=(\frac{2\sqrt{3}+15}{3-\sqrt{3}}+\frac{3}{\sqrt{3}-2}).\frac{1}{\sqrt{3}+5}\)
\(=\frac{2\sqrt{3}+15)(\sqrt{3}-2)+3(3-\sqrt{3})}{(3-\sqrt{3})(\sqrt{3}-2)}.\frac{1}{\sqrt{3}+5}\)
\(=\frac{-15+8\sqrt{3}}{(-9+5\sqrt{3})(\sqrt{3}+5)}=\frac{-15+8\sqrt{3}}{-30+16\sqrt{3}}=\frac{-15+8\sqrt{3}}{2(-15+8\sqrt{3})}=\frac{1}{2}\)
CW
9 tháng 7 2018
a) \(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}=\sqrt{2}+1-\left(\sqrt{2}-1\right)=2\)
b) \(\dfrac{1}{\sqrt{3}-1}-\dfrac{1}{\sqrt{3}+1}=\dfrac{\sqrt{3}+1-\left(\sqrt{3}-1\right)}{3-1}=1\)
c) \(2\sqrt{5}-3\sqrt{45}+\sqrt{500}=2\sqrt{5}-9\sqrt{5}+10\sqrt{5}=3\sqrt{5}\)
d) \(\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{4}\right)}{\sqrt{5}-2}=\dfrac{1}{\sqrt{3}+\sqrt{2}}-\sqrt{3}=\dfrac{1-\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\dfrac{1-3-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-2-\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\dfrac{-\sqrt{2}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=-\sqrt{2}\)
e) \(\dfrac{1}{2+\sqrt{3}}-\dfrac{1}{2-\sqrt{3}}+5\sqrt{3}=\dfrac{2-\sqrt{3}-\left(2+\sqrt{3}\right)}{4-3}+5\sqrt{3}=-2\sqrt{3}+5\sqrt{3}=3\sqrt{3}\)
f) \(\sqrt{3}-\sqrt{4+2\sqrt{3}}=\sqrt{3}-\left(\sqrt{3}+1\right)=-1\)
g) \(\dfrac{5-\sqrt{5}}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}-\dfrac{4}{\sqrt{5}+1}=\sqrt{5}-\dfrac{4}{\sqrt{5}+1}=\dfrac{5+\sqrt{5}-4}{\sqrt{5}+1}=1\)
h)\(\sqrt{37-20\sqrt{3}+\sqrt{37+20\sqrt{3}}}=\sqrt{37-20\sqrt{3}+\left(5+2\sqrt{3}\right)}=\sqrt{42-18\sqrt{3}}=\sqrt{\left(3\sqrt{3}+3\right)^2+6}\)
AH
Akai Haruma
Giáo viên
17 tháng 5 2020
h)
\(H=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2-\sqrt{3})(2+\sqrt{3})}}=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=2\sqrt{3}\)
i)
\(I=\frac{2+\sqrt{3}}{2+\sqrt{3+1+2\sqrt{3.1}}}+\frac{2-\sqrt{3}}{2-\sqrt{3+1-2\sqrt{3.1}}}=\frac{2+\sqrt{3}}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{2-\sqrt{3}}{2-\sqrt{(\sqrt{3}-1)^2}}\)
\(=\frac{2+\sqrt{3}}{2+\sqrt{3}+1}+\frac{2-\sqrt{3}}{2-(\sqrt{3}-1)}=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}\)
\(=\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{6}{6}=1\)
AH
Akai Haruma
Giáo viên
17 tháng 5 2020
ê)
\(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}\)
\(=\sqrt{(2+5+2\sqrt{2.5})+1+2(\sqrt{2}+\sqrt{5})}\)
\(=\sqrt{(\sqrt{2}+\sqrt{5})^2+1+2(\sqrt{2}+\sqrt{5})}=\sqrt{(\sqrt{2}+\sqrt{5}+1)^2}=\sqrt{2}+\sqrt{5}+1\)
g)
\(13+\sqrt{48}=13+2\sqrt{12}=12+1+2\sqrt{12.1}=(\sqrt{12}+1)^2\)
\(\Rightarrow \sqrt{13+\sqrt{48}}=\sqrt{12}+1\)
\(\Rightarrow \sqrt{3+\sqrt{13+\sqrt{48}}}=\sqrt{4+\sqrt{12}}=\sqrt{3+1+2\sqrt{3.1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)
\(\Rightarrow 2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}=2\sqrt{2-\sqrt{3}}=\sqrt{2}.\sqrt{4-2\sqrt{3}}=\sqrt{2}.\sqrt{(\sqrt{3}-1)^2}\)
\(=\sqrt{2}(\sqrt{3}-1)=\sqrt{6}-\sqrt{2}\)
\(\Rightarrow G=1\)
10 tháng 7 2022
a: \(=2\cdot\dfrac{4}{3}\sqrt{3}-3\cdot\dfrac{1}{9}\sqrt{3}-6\cdot\dfrac{2}{15}\sqrt{3}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
trả lời:
căn bậc 3 của 2197 là 13
hok tốt
............
\(\sqrt[3]{2197}\text{ là }13\)