\(\sqrt{3}\)= ..............
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19 tháng 10 2018
mẫu rút gọn như sau:
\(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
xong cộng với cái ở ngoài lại ra 4+2 căn 3 làm tương tự
27 tháng 6 2021
a, Ta có : \(\left\{{}\begin{matrix}\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\\\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\end{matrix}\right.\)
- Thay lần lượt vào A ta được :
\(A=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\sqrt{2}-1+\sqrt{2}+1\right)=2.2\sqrt{2}=4\sqrt{2}\)
b, \(B=\sqrt{2+\sqrt{3}}\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}=\sqrt{2+\sqrt{3}}\sqrt{4-2-\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\sqrt{2+\sqrt{3}}=\sqrt{4-3}=\sqrt{1}=1\)
c, \(C=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)+\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\dfrac{2\sqrt{2}+\sqrt{6}-2\sqrt{2-\sqrt{3}}-\sqrt{3}\sqrt{2-\sqrt{3}}+2\sqrt{2}-\sqrt{6}+2\sqrt{2+\sqrt{3}}-\sqrt{3}\sqrt{2+\sqrt{3}}}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\dfrac{4\sqrt{2}-2\sqrt{3}\sqrt{2-\sqrt{3}}}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
27 tháng 6 2021
a) Ta có: \(A=\left(\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\right)\left(\sqrt{3-2\sqrt{2}}+\sqrt{3+2\sqrt{2}}\right)\)
\(=\left(\sqrt{2}+1-\sqrt{2}+1\right)\left(\sqrt{2}-1+\sqrt{2}+1\right)\)
\(=2\cdot2\sqrt{2}=4\sqrt{2}\)
AH
Akai Haruma
Giáo viên
28 tháng 7 2018
Lời giải:
Đặt \(\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}=a; \sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}=b\)
Có:
\(a^2+b^2=(2+\sqrt{3}+\sqrt{2-\sqrt{3}})+(2+\sqrt{3}-\sqrt{2-\sqrt{3}})=2(2+\sqrt{3})\)
\(=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\)
\(ab=\sqrt{(2+\sqrt{3}+\sqrt{2-\sqrt{3}})(2+\sqrt{3}-\sqrt{2-\sqrt{3}})}\)
\(=\sqrt{(2+\sqrt{3})^2-(2-\sqrt{3})}=\sqrt{5+5\sqrt{3}}\)
Như vậy:
\(\frac{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}+\frac{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}=\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\)
\(=\frac{(\sqrt{3}+1)^2}{\sqrt{5+5\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{\sqrt{5}.\sqrt{\sqrt{3}+1}}=\frac{(\sqrt{3}+1)^{1.5}}{\sqrt{5}}\)
17 tháng 8 2019
\(H=\frac{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}-\frac{\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}}\)
\(H=\frac{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)\(-\frac{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)(cái này cùng dòng với cái phía trên)
\(H=\frac{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2}{\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)}\)
\(H=\frac{\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2-\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2}{2\sqrt{3}}\)
\(H=\frac{-4}{2\sqrt{3}}\)
\(H=\frac{-2}{\sqrt{3}}\)
\(H=-\frac{2\sqrt{3}}{3}\)
NH
17 tháng 8 2019
Đặt \(A=\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(A^2=2+\sqrt{3}+2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)
\(A^2=4+2\sqrt{4+2\sqrt{3}-2\sqrt{3}-3}\)
\(A^2=4+2=6\)
\(A=\sqrt{6}\)
Đặt \(B=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
\(B^2=2+\sqrt{3}-2\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+2-\sqrt{3}\)
\(B^2=4-2\sqrt{4+2\sqrt{3}-2\sqrt{3}-3}\)
\(B^2=4-2\sqrt{1}=4-2=2\)
\(B=\sqrt{2}\)
Thay vào H
\(\Rightarrow H=\frac{\sqrt{2}}{\sqrt{6}}-\frac{\sqrt{6}}{\sqrt{2}}=\frac{1}{\sqrt{3}}-\sqrt{3}=\frac{1-3}{\sqrt{3}}=\frac{-2}{\sqrt{3}}\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
a.
\(\sqrt[3]{125}.\sqrt[3]{\frac{16}{10}}.\sqrt[3]{-0,5}=\sqrt[3]{125.\frac{16}{10}.(-0,5)}=\sqrt[3]{-100}\)
b.
\(=1+\frac{1}{\sqrt[3]{4}+\sqrt[3]{2}+1}=1+\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2}-1)(\sqrt[3]{4}+\sqrt[3]{2}+1)}=1+\frac{\sqrt[3]{2}-1}{(\sqrt[3]{2})^3-1}=1+\sqrt[3]{2}-1=\sqrt[3]{2}\)
c.
\(\sqrt{3}+\sqrt[3]{10+6\sqrt{3}}=\sqrt{3}+\sqrt[3]{(\sqrt{3}+1)^3}=\sqrt{3}+\sqrt{3}+1=2\sqrt{3}+1\)
AH
Akai Haruma
Giáo viên
6 tháng 8 2021
d.
\(\frac{4+2\sqrt{3}}{\sqrt[3]{10+6\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{\sqrt[3]{(\sqrt{3}+1)^3}}=\frac{(\sqrt{3}+1)^2}{\sqrt{3}+1}=\sqrt{3}+1\)
e.
Đặt \(\sqrt[3]{2+10\sqrt{\frac{1}{27}}}=a; \sqrt[3]{2-10\sqrt{\frac{1}{27}}}=b\)
Khi đó:
$a^3+b^3=4$
$ab=\frac{2}{3}$
$E^3=(a+b)^3=a^3+b^3+3ab(a+b)$
$E^3=4+2E$
$E^3-2E-4=0$
$E^2(E-2)+2E(E-2)+2(E-2)=0$
$(E-2)(E^2+2E+2)=0$
Dễ thấy $E^2+2E+2>0$ nên $E-2=0$
$\Leftrightarrow E=2$
17 tháng 12 2023
a: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)
\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-\sqrt{2\sqrt{3}}-3\sqrt{2\sqrt{3}}\)
=0
b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|2-\sqrt{3}\right|\)
\(=\sqrt{3}+2-\sqrt{3}\)
=2
c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)
\(=\left|\sqrt{7}-4\right|-2\sqrt{7}+3\sqrt{7}\)
\(=4-\sqrt{7}+\sqrt{7}\)
=4
d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)
\(=16\sqrt{5}\)
e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)
\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
\(=-16\sqrt{3}\)
bằng 1,732050808