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Bài 8: Rút gọn biểu thức chứa căn bậc hai
Các câu hỏi tương tự
M = \(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}\)
N = \(\frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}\)
O = \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}\)
\(\frac{3\sqrt{2}+2\sqrt{3}}{\sqrt{2}+\sqrt{3}}-\frac{5}{1+\sqrt{6}}\)
\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
A) 2sqrt{3}+sqrt{48}-sqrt{75}-sqrt{243}
b) left(frac{sqrt{7}-sqrt{14}}{1-sqrt{2}}+frac{sqrt{15}-sqrt{5}}{1-sqrt{3}}right):frac{1}{sqrt{7}+sqrt{5}}
c) left(sqrt{3}+1right)sqrt{4-2sqrt{3}}
d) 5sqrt{2}+sqrt{18}-sqrt{98}-sqrt{288}
e)left(frac{sqrt{3}-sqrt{6}}{1-sqrt{2}}+frac{sqrt{15}-sqrt{5}}{1-sqrt{3}}right):frac{1}{sqrt{3}+sqrt{5}}
g)left(sqrt{3}-1right)sqrt{4+2sqrt{3}}
Đọc tiếp
A) \(2\sqrt{3}+\sqrt{48}-\sqrt{75}-\sqrt{243}\)
b) \(\left(\frac{\sqrt{7}-\sqrt{14}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}+\sqrt{5}}\)
c) \(\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}\)
d) \(5\sqrt{2}+\sqrt{18}-\sqrt{98}-\sqrt{288}\)
e)\(\left(\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{3}+\sqrt{5}}\)
g)\(\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
Rút gọn: \(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
J left(1+frac{2+sqrt{2}}{1+sqrt{2}}right) . left(1-frac{2-sqrt{2}}{1-sqrt{2}}right)
M frac{3sqrt{2}-2sqrt{3}}{sqrt{3}-sqrt{2}}:frac{1}{sqrt{6}}
N frac{6}{1+sqrt{7}}+frac{1}{sqrt{7}}
O frac{3+2sqrt{3}}{sqrt{3}}+frac{2+sqrt{2}}{1+sqrt{2}}-frac{1}{2-sqrt{3}}
Q left(frac{sqrt{6}-sqrt{2}}{1-sqrt{3}}-frac{5}{sqrt{5}}right).left(sqrt{5}-sqrt{2}right)
S frac{2}{sqrt{5}+1}-sqrt{frac{2}{3-sqrt{5}}}
Các thầy cô, các bạn giúp em với ạ. Em cảm ơn !
Đọc tiếp
J = \(\left(1+\frac{2+\sqrt{2}}{1+\sqrt{2}}\right)\) . \(\left(1-\frac{2-\sqrt{2}}{1-\sqrt{2}}\right)\)
M = \(\frac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\frac{1}{\sqrt{6}}\)
N = \(\frac{6}{1+\sqrt{7}}+\frac{1}{\sqrt{7}}\)
O = \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{1+\sqrt{2}}-\frac{1}{2-\sqrt{3}}\)
Q = \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
S = \(\frac{2}{\sqrt{5}+1}-\sqrt{\frac{2}{3-\sqrt{5}}}\)
Các thầy cô, các bạn giúp em với ạ. Em cảm ơn !
rút gọn biểu thức:
1/ frac{sqrt{15}-sqrt{5}}{sqrt{3}-1}+frac{5-2sqrt{5}}{2sqrt{5}-4}
2/ frac{sqrt{15}-sqrt{12}}{sqrt{5}-2}+frac{6+2sqrt{6}}{sqrt{3}+sqrt{2}}
3/ frac{3+2sqrt{3}}{sqrt{3}}+frac{2+sqrt{2}}{sqrt{2}+1}-left(2+sqrt{3}right)
4/ left(1-frac{5+sqrt{5}}{1+sqrt{5}}right)left(frac{5-sqrt{5}}{1-sqrt{5}}-1right)
giúp minh với cần gấp lắm
Đọc tiếp
rút gọn biểu thức:
1/ \(\frac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}+\frac{5-2\sqrt{5}}{2\sqrt{5}-4}\)
2/ \(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{6+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
3/ \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
4/ \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
giúp minh với cần gấp lắm
chứng minh rằng
a, \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=1\)
b, \(\frac{1}{x+\sqrt{x}}+\frac{2\sqrt{x}}{x-1}-\frac{1}{x-\sqrt{x}}=\frac{2}{\sqrt[]{x}}\)
* rút gọn biểu thức:
a, sqrt{frac{3}{20}}+sqrt{frac{1}{60}}-2sqrt{frac{1}{15}}
b, left(frac{1}{sqrt{5}-sqrt{3}}+frac{1}{sqrt{5}+sqrt{3}}right).sqrt{5}
c, left(5sqrt{3}+3sqrt{5}right):sqrt{15}
d, left(2+sqrt{5}right)^2-left(2+sqrt{5}right)^2
e, frac{1}{3}sqrt{48}+3sqrt{75}-sqrt{27}-10sqrt{1frac{1}{3}}
Đọc tiếp
* rút gọn biểu thức:
a, \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)
b, \(\left(\frac{1}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{3}}\right).\sqrt{5}\)
c, \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
d, \(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)
e, \(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)
1. Rút gọn
D = \(\frac{\sqrt{1+\frac{2\sqrt{2}}{3}}+\sqrt{1-\frac{2\sqrt{2}}{3}}}{\sqrt{1+\frac{2\sqrt{2}}{3}}-\sqrt{1-\frac{2\sqrt{2}}{3}}}\)
2. Chứng minh rằng:
\(\frac{a\sqrt{b}+b}{a-b}.\sqrt{\frac{ab+b^2-2\sqrt{ab^3}}{a\left(a+2\sqrt{b}\right)+b}}\left(\sqrt{a}+\sqrt{b}\right)=b\) với ( a > b > 0 )