\(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}=\dfrac{1}{6}\sqrt{6}\)
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MÌNH CẦN LUÔN ẠRút gọn biểu thức:1(2+sqrt{3})(7-4sqrt{3})2)left(sqrt{5-2sqrt{6}}+sqrt{2}right)sqrt{3}3)sqrt{4+2sqrt{3}}-sqrt{5-2sqrt{6}}+sqrt{2}4)sqrt{3+2sqrt{2}}+sqrt{6-4sqrt{2}}5)2+sqrt{17-4sqrt{9+4sqrt{5}}}
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MÌNH CẦN LUÔN Ạ
Rút gọn biểu thức:
1(2+\(\sqrt{3}\))(7-4\(\sqrt{3}\))
2)\(\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\sqrt{3}\)
3)\(\sqrt{4+2\sqrt{3}}-\sqrt{5-2\sqrt{6}}+\sqrt{2}\)
4)\(\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
5)\(2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\)
a/ (√10+√2) (6-2√5)√(3+√5)
b/ √(13-√160) - √(53-4√90)
c/ √(10+√24+√40+√60)
d/ (√2+√3+√6+√8+√16)/(√2+√3+√4)
e/ [√216/3-(2√3-√6)/(√8-2)] ×1/√6
f/ 1/(√2-√3) × √[(3√2-2√3)/(3√2+2√3)]
g/ 1/(√1+√2) + 1/(√2+√3) + ......+ 1/(√2017+√2018)
rút gọn các biểu thức sau
\(\dfrac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}\)+\(\dfrac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\)
\(\dfrac{\sqrt{3}}{1-\sqrt{\sqrt{3}+1}}\)+\(\dfrac{\sqrt{3}}{1+\sqrt{\sqrt{3}+1}}\)
\(\frac{6+4\sqrt{2}}{\sqrt{2}+\sqrt{6+4\sqrt{2}}}+\frac{6-4\sqrt{2}}{\sqrt{2}-\sqrt{6-4\sqrt{2}}}\)
\(\frac{4}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-2}+\frac{6}{\sqrt{3}-3}\)
1+2+3+4+5+6+7+8+9-1-2-3-4-5-6-7-8-9*1*2*3*4*5*6*7*8*9:1:2:3:4:5:6:7:8:9=???
Chứng minh các đẳng thức sau:
e) \(\left(\dfrac{3}{2}.\sqrt{6}+2.\sqrt{\dfrac{2}{3}}-4.\sqrt{\dfrac{3}{2}}\right).\left(\dfrac{3}{2}.\sqrt{6}+2.\sqrt{\dfrac{2}{3}}+4.\sqrt{\dfrac{3}{2}}\right)=-\sqrt{2}\)
1. Tính
a) \(\sqrt[3]{(\sqrt{2}+3)(11+6\sqrt{2})}\sqrt[3]{(\sqrt{2}+-3)(11-6\sqrt{2})}\)
b) (\((\sqrt[3]{9}+\sqrt[3]{6}+\sqrt[3]{4})(\sqrt[3]{3}-\sqrt[3]{2})\)
c)\(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
CMR : 3/2 *căn 6+2* căn 2/3 -4* căn 3/2 =căn 6/6
Giải hệ pt: (gồn 4 pt, em không đánh được công thức toán)
m[1]^2+m[2]^2+m[3]^2=1
2*m[4]*m[1]+2*m[5]*m[2]+2*m[3]*m[6]=-3,
2*m[6]*m[1]+2*m[4]*m[2]+2*m[3]*m[5]=0,
2*m[1]*m[2]+2*m[1]*m[3]+2*m[2]*m[3]+m[4]^2+m[5]^2+m[6]^2=2
Tìm m1,m[2],m[3],m[4],m[5],m[6] (em không đánh được m1, mọi người thông cảm ạ.
1)\(\dfrac{\sqrt{3}+\sqrt{2}-1}{2+\sqrt{6}}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+1}\left(\dfrac{\sqrt{3}}{2-\sqrt{6}}+\dfrac{\sqrt{3}}{2+\sqrt{6}}\right)-\dfrac{1}{\sqrt{2}}\)
2)\(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}+\dfrac{12}{\sqrt{6}-3}-\sqrt{6}\)
3)\(\left(\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{6}+\sqrt{2}}\right)\left(\sqrt{3}-1\right)^2\)