tính
\(\frac{9\sqrt{5}+3\sqrt{27}}{\sqrt{5}+\sqrt{3}}\)
giúp mk
Tính:
\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
giúp mk vs mk đang cần gấp//
Sủa lại đề:
\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\)
Đặt \(\hept{\begin{cases}\sqrt{3+\sqrt{5}}=a\\\sqrt{3-\sqrt{5}}=b\end{cases}}\)
Khi đó ta có \(a^2+b^2=6\), \(ab=2\), \(a+b=\sqrt{10}\), \(a-b=\sqrt{2}\), \(a^2-b^2=2\sqrt{5}\)
\(=\frac{a^2}{\sqrt{10}+a}-\frac{b^2}{\sqrt{10}+b}\)
\(=\frac{a^2.\left(\sqrt{10}+b\right)-b^2.\left(\sqrt{10}+a\right)}{\left(\sqrt{10}+a\right).\left(\sqrt{10}+b\right)}\)
\(=\frac{\sqrt{10}a^2+a^2b-\sqrt{10}b^2-ab^2}{10+\sqrt{10}a+\sqrt{10}b+ab}\)
\(=\frac{\sqrt{10}.\left(a^2-b^2\right)+ab.\left(a-b\right)}{10+\sqrt{10}.\left(a+b\right)+ab}\)
\(=\frac{\sqrt{10}.2\sqrt{5}+\sqrt{10}.\sqrt{2}}{10+\sqrt{10}.\sqrt{10}+2}\)
\(=\frac{10\sqrt{2}+2\sqrt{2}}{10+10+2}\)
\(=\frac{12\sqrt{2}}{22}\)
\(=\frac{6\sqrt{2}}{11}\)
\(\frac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\frac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}
\)
\(=\frac{3+\sqrt{5}-3-\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
\(=\frac{0}{\sqrt{10}+\sqrt{3+\sqrt{5}}}\)
\(=0\)
Bạn chưa dổi dấu kìa
Thực hiện phép tính:
1) \(2\sqrt{27}-6\sqrt{\frac{4}{3}}+\frac{3}{5}\sqrt{75}\)
2) \(\frac{1}{\sqrt{2}-\sqrt{2+\sqrt{3}}}-\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
3) \(\frac{\sqrt{3-\sqrt{5}}.\left(3+\sqrt{5}\right)}{\sqrt{10}+\sqrt{2}}\)
4)\(\sqrt{2-\sqrt{3}}.\left(\sqrt{5}+\sqrt{2}\right)\)
5) \(\sqrt{14-8\sqrt{3}}-\sqrt{24-12\sqrt{3}}\)
6) \(\sqrt{4-\sqrt{9+4\sqrt{2}}}\)
7) \(\sqrt{5}+\sqrt{9-4\sqrt{5}}\)
GIÚP MÌNH VỚI M.N~!! MÌNH ĐANG CẦN GẤP LẮM !!!! (T_T)
\(\sqrt{27}-\sqrt{5\frac{1}{3}}\)\(+4,5\sqrt{2\frac{2}{3}}\)\(+2\sqrt{27}\)
giúp mk với
Ta có:
\(\sqrt{27}-\sqrt{5\frac{1}{3}}+4,5\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
\(=3\sqrt{3}-\sqrt{\frac{16}{3}}+4,5\sqrt{\frac{8}{3}}+6\sqrt{3}\)
\(=9\sqrt{3}+\frac{4\sqrt{3}}{3}+3\sqrt{6}\)
\(=\frac{9\sqrt{6}+31\sqrt{3}}{3}\)
\(\sqrt{27}-\sqrt{5\frac{1}{3}}+4,5\sqrt{2\frac{2}{3}}+2\sqrt{27}\)
\(=\sqrt{27}-\sqrt{16.\frac{1}{3}}+4,5.\sqrt{4.\frac{1}{3}}+2\sqrt{27}\)
\(=\sqrt{27}-4\sqrt{\frac{1}{3}}+9\sqrt{\frac{1}{3}}+2\sqrt{27}\)
\(=\sqrt{27}-4\sqrt{\frac{1}{3}}+\sqrt{27}+2\sqrt{27}\)
\(=4\sqrt{27}-4\sqrt{\frac{1}{3}}\)
\(=\sqrt{54}-\sqrt{\frac{2}{3}}\)
m.n giúp mk vs ak
mk xin cảm ơn ak
a.10x-9y =1
15x +21y=36
b.\(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)
c.2\(\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\)
d.(\(\sqrt{5}+\sqrt{2}\))(\(3\sqrt{2}-1\))
Tính Q=\(\frac{\sqrt{45+27\sqrt{2}}+\sqrt{45-27\sqrt{2}}}{\sqrt{5+3\sqrt{2}}-\sqrt{5-3\sqrt{2}}}-\frac{\sqrt{3+\sqrt{2}}+\sqrt{3-\sqrt{2}}}{\sqrt{3+\sqrt{2}}-\sqrt{3-\sqrt{2}}}\)
Thực hiện phép tính:
a,\(\frac{1}{2}\sqrt{48}-5\sqrt{27}+2\sqrt{147}-\sqrt{108}\)
b,\(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(1+\sqrt{5}\right)^2}\)
c,\(\frac{12}{3+\sqrt{3}}-\frac{6}{\sqrt{3}}+\frac{\sqrt{27}-3\sqrt{2}}{\sqrt{3}-\sqrt{2}}\)
d,\(\left(\sqrt{2+\sqrt{3}}-\sqrt{3+\sqrt{5}}\right)^2\)
Giúp mk vs huhu mk hứa sẽ tick ạ :>
1) Tính:
\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5-1}}\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\)
Giúp mình với, mình cần gấp
\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)
Câu b á bạn, chỗ \(\dfrac{4}{\sqrt{5-1}}\) là đề như vậy hay là \(\dfrac{4}{\sqrt{5}-1}\) vậy?
Chứng minh rằng:
\(\sqrt[3]{\sqrt[5]{\frac{32}{5}}-\sqrt[5]{\frac{27}{5}}}=\sqrt[5]{\frac{1}{25}}+\sqrt[5]{\frac{3}{25}}-\sqrt[5]{\frac{9}{25}}\)
Chứng minh rằng:
a)\(\left(\sqrt[3]{3+2\sqrt{2}}+\sqrt[3]{3-2\sqrt{2}}\right)^8>3^6\)
b) \(\sqrt[3]{\sqrt[5]{\frac{32}{5}}-\sqrt[5]{\frac{27}{5}}}=\sqrt[5]{\frac{1}{25}}+\sqrt[5]{\frac{3}{25}}-\sqrt[5]{\frac{9}{25}}\)