Bài 7: Biến đối đơn giản biểu thức chứa căn bậc hai (Tiếp theo)
Các câu hỏi tương tự
Chứng minh rằng:
\(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{79}+\sqrt{80}}>4\)
Rút gọn :
\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}\)
A = \(\dfrac{4+\sqrt{3}}{\sqrt{1}+\sqrt{3}}+\dfrac{6+\sqrt{8}}{\sqrt{3}+\sqrt{5}}+...+\dfrac{2n+\sqrt{n^2-1}}{\sqrt{n-1}+\sqrt{n+1}}+\dfrac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)
B= \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{4}}+\dfrac{1}{\sqrt{4}-\sqrt{5}}-....+\dfrac{1}{\sqrt{100}-\sqrt{101}}\)
Chứng minh đẳng thức:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{2010\sqrt{2009}+2009\sqrt{2010}}=1-\dfrac{\sqrt{2010}}{2010}\)
1. dfrac{-2}{sqrt{3}-1}2. dfrac{5}{1-sqrt{6}}3. dfrac{2+sqrt{5}}{2-sqrt{5}}4. dfrac{1}{5+2sqrt{6}}5. dfrac{sqrt{5}+2}{sqrt{5}-2}6. dfrac{5sqrt{2}-2sqrt{5}}{sqrt{2}-sqrt{5}}7. dfrac{sqrt{20}-3sqrt{10}}{3-sqrt{2}}8. dfrac{6-2sqrt{5}}{3+sqrt{5}}9. dfrac{9+4sqrt{5}}{sqrt{5}+2}
Đọc tiếp
1. \(\dfrac{-2}{\sqrt{3}-1}\)
2. \(\dfrac{5}{1-\sqrt{6}}\)
3. \(\dfrac{2+\sqrt{5}}{2-\sqrt{5}}\)
4. \(\dfrac{1}{5+2\sqrt{6}}\)
5. \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\)
6. \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
7. \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{2}}\)
8. \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}\)
9. \(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}\)
M= \(\dfrac{1}{\sqrt{2}-\sqrt{3}}-\dfrac{1}{\sqrt{3}-\sqrt{4}}+\dfrac{1}{\sqrt{4}-\sqrt{5}}-....+\dfrac{1}{\sqrt{100}-\sqrt{101}}\)
Bài 1: Tính:
dfrac{1}{sqrt{3}}+dfrac{1}{3sqrt{2}}+dfrac{1}{sqrt{3}}sqrt{dfrac{5}{12}-dfrac{1}{sqrt{6}}}
Bài 2: Rút gọn rồi tính:
a) Adfrac{a^4-4a^2+3}{a^4-12a^2+27},asqrt{3}-sqrt{2}
b) Bdfrac{1}{sqrt{h+2sqrt{h-1}}}+dfrac{1}{sqrt{h-2sqrt{h-1}}},h3
c) Cdfrac{sqrt{2x+2sqrt{x^2-4}}}{sqrt{x^2-4}x+2},x2left(sqrt{3}+1right)
d) Dleft(dfrac{3}{sqrt{1+a}}+sqrt{1-a}right):left(dfrac{3}{sqrt{1-a^2}}+1right),adfrac{sqrt{3}}{2+sqrt{3}}
Mọi người giúp em với!!!!!!!!!!!!!!
Đọc tiếp
Bài 1: Tính:
\(\dfrac{1}{\sqrt{3}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{\sqrt{3}}\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}\)
Bài 2: Rút gọn rồi tính:
a) A=\(\dfrac{a^4-4a^2+3}{a^4-12a^2+27},a=\sqrt{3}-\sqrt{2}\)
b) \(B=\dfrac{1}{\sqrt{h+2\sqrt{h-1}}}+\dfrac{1}{\sqrt{h-2\sqrt{h-1}}},h=3\)
c) \(C=\dfrac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}x+2},x=2\left(\sqrt{3}+1\right)\)
d) \(D=\left(\dfrac{3}{\sqrt{1+a}}+\sqrt{1-a}\right):\left(\dfrac{3}{\sqrt{1-a^2}}+1\right),a=\dfrac{\sqrt{3}}{2+\sqrt{3}}\)
Mọi người giúp em với!!!!!!!!!!!!!!
Rút gọn biểu thức
\(a.\dfrac{\sqrt{5}-2\sqrt{3}}{\sqrt{5}+\sqrt{3}}-\dfrac{2\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
\(b.x\sqrt{2x+2}+\left(x+1\right)\sqrt{\dfrac{2}{x+1}}-4\sqrt{\dfrac{x+1}{2}}\)
Rút gọn :
\(B=\dfrac{1}{\sqrt{1}-\sqrt{2}}+\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}+...+\dfrac{1}{\sqrt{24}-\sqrt{25}}\)