1: \(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6=2\)
2: \(=6+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
3: \(=\sqrt{3}-1-\sqrt{3}+\sqrt{3}+1=\sqrt{3}\)
không dùng máy tính , tính giá trị của các biểu thức sau
1)\(\left(1+\sqrt{2}+\sqrt{3}\right)\cdot\left(1+\sqrt{2}+\sqrt{3}\right)\)
2)\(\dfrac{1}{\sqrt{2}+1}-\dfrac{\sqrt{8}-\sqrt{10}}{2-\sqrt{5}}\)
3)\(\dfrac{2+\sqrt{3}}{\sqrt{7-4\sqrt{3}}}-\dfrac{2-\sqrt{3}}{\sqrt{7+4\sqrt{3}}}\)
4)\(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)-\dfrac{\sqrt{7-4\sqrt{3}}}{\sqrt{3}-2}\)
5)\(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
6)\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)
Rút gọn:
a) \(2\sqrt{98}-3\sqrt{18}+\dfrac{1}{2}\sqrt{32}\)
b)\(\left(5\sqrt{2}+2\sqrt{5}\right).\sqrt{5}-\sqrt{250}\)
c)\(\left(2\sqrt{3}-5\sqrt{2}\right).\sqrt{3}-\sqrt{36}\)
d)\(3\sqrt{48}+2\sqrt{27}-\dfrac{1}{3}\sqrt{243}\)
e) \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}\)
f)\(4\sqrt{\dfrac{1}{2}}-\dfrac{6}{\sqrt{2}}\dfrac{2}{\sqrt{2}+1}\)
Tính:
E=(\(\sqrt{18}-3\sqrt{6}+\sqrt{2}\)) \(\sqrt{2}+6\sqrt{3}\)
G=\(\left(2\sqrt{2}-\sqrt{5}+\sqrt{18}\right)\).\(\left(\sqrt{50}+\sqrt{5}\right)\)
H=\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}\).\(\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(\left(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\right)\times\left(3\sqrt{\dfrac{2}{3}}-\sqrt{2}-\sqrt{6}\right)\times\left(-\sqrt{6}\right)\)
a, \(A=\left(\sqrt{2}+1\right)[\left(\sqrt{2}\right)^2+1][(\sqrt{2})^4+1][\left(\sqrt{2}\right)^8+1][1\left(\sqrt{2}\right)^{16}+1]\)
b, \(B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+1\sqrt{2020}}\)
c,\(C=^3\sqrt[]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\)
rÚT GỌN: G=\(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{6}}-\sqrt{2}\)
chững minh : a) \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt[]{6}=9\)
b)\(\left(3+\sqrt{5}\right)\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3-\sqrt{5}}=8\)
c)\(\sqrt{\dfrac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\dfrac{4}{\left(2+\sqrt{5}\right)^2}}=8\)
giúp mk với tối mai mk nạp rồi
\(A=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{16}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}\)
\(B=\dfrac{2\left(\dfrac{\sqrt{2}+\sqrt{3}}{6\sqrt{2}}\right)^{^{^{-1}}}+3\left(\dfrac{\sqrt{2}+\sqrt{3}}{4\sqrt{3}}\right)^{^{^{-1}}}}{\left(\dfrac{2+\sqrt{6}}{12}\right)^{^{^{-1}}}+\left(\dfrac{3+\sqrt{6}}{12}\right)^{^{^{-1}}}}\)
Cíu em với các pro ~
P/s: Câu B em làm đc r mà k biết kết quả đúng k nữa nên up lên hỏi luôn :)))
\(\dfrac{6-\sqrt{6}}{\sqrt{6}-1}+\dfrac{6-\sqrt{6}}{\sqrt{6}}\)
\(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{3}{\sqrt{18}+2\sqrt{3}}\)
\(\left(\dfrac{15}{3-\sqrt{3}}-\dfrac{2}{1-\sqrt{3}}+\dfrac{3}{\sqrt{3}-2}\right):\sqrt{28+10\sqrt{3}}\)
Rút gọn:
1) \(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}-2\sqrt{3}\)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
2) \(\sqrt{3-2\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}\)
\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}+2}\right).\dfrac{a-4}{\sqrt{4a}}\)
\(N=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\)
\(Q=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\right)\)
Làm chi tiết giúp mình với vì mình yếu phần này lắm
