23 tháng 9 2021 lúc 20:50
Các câu hỏi tương tự
chứng minh rằng với số tự nhiên n,n lớn hơn 4 ta có:
\(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}< 1\)
Chứng minh rằng với mọi số nguyên dương n, p ta có :
\(\dfrac{1}{\left(1+1\right)\sqrt[p]{1}}+\dfrac{1}{\left(2+1\right)\sqrt[p]{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt[p]{n}}\) < p
Chứng minh rằng với mọi số nguyên dương n, p ta có :
\(\dfrac{1}{\left(1+1\right)\sqrt[p]{1}}+\dfrac{1}{\left(2+1\right)\sqrt[p]{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt[p]{n}}\) < p
\(S=\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+\dfrac{1}{4\sqrt{3}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}\)
Cho n ϵ N*. Chứng minh:
a) \(1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{\left(n-1\right)^2}+\dfrac{1}{n^2}< 2\)
b) \(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-1\right)\)
19 tháng 9 2021 lúc 21:42
Cdfrac{1}{1+sqrt{2}}+dfrac{1}{sqrt{2}+sqrt{3}}+dfrac{1}{sqrt{3}+sqrt{4}}+....dfrac{1}{sqrt{99}+sqrt{100}}Cdfrac{1left(1-sqrt{2}right)}{left(1+sqrt{2}right)left(1-sqrt{2}right)}+dfrac{1left(sqrt{2}-sqrt{3}right)}{left(sqrt{2}-sqrt{3}right)left(sqrt{2}+sqrt{3}right)}+dfrac{1left(sqrt{3}-sqrt{4}right)}{left(sqrt{3}-sqrt{4}right)left(sqrt{3}+sqrt{4}right)}+........dfrac{1left(sqrt{99}-sqrt{100}right)}{left(sqrt{99}-sqrt{100}right)left(sqrt{99}+sqrt{100}right)}Cdfrac{1-sqrt{2}}{1-2}+dfrac{sqrt{2}-sqr...
Đọc tiếp
\(C=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+....\dfrac{1}{\sqrt{99}+\sqrt{100}}\)
\(C=\dfrac{1\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}+\dfrac{1\left(\sqrt{2}-\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}+\dfrac{1\left(\sqrt{3}-\sqrt{4}\right)}{\left(\sqrt{3}-\sqrt{4}\right)\left(\sqrt{3}+\sqrt{4}\right)}+........\dfrac{1\left(\sqrt{99}-\sqrt{100}\right)}{\left(\sqrt{99}-\sqrt{100}\right)\left(\sqrt{99}+\sqrt{100}\right)}\)
\(C=\dfrac{1-\sqrt{2}}{1-2}+\dfrac{\sqrt{2}-\sqrt{3}}{2-3}+\dfrac{\sqrt{3}-\sqrt{4}}{3-4}+.....+\dfrac{\sqrt{99}-\sqrt{100}}{99-100}\)
\(C=\dfrac{1-\sqrt{2}}{-1}+\dfrac{\sqrt{2}-\sqrt{3}}{-1}+\dfrac{\sqrt{3}-\sqrt{4}}{-1}+......+\dfrac{\sqrt{99}-\sqrt{100}}{-1}\)
\(C=-\left(1-\sqrt{2}\right)-\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{4}\right)-......-\left(\sqrt{99}-\sqrt{100}\right)\)
\(C=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{4}-......-\sqrt{99}+\sqrt{100}\)
\(C=-1+\sqrt{100}\)
\(C=10-1=9\)
Tính: a, \(\left(4\sqrt{2}-\dfrac{11}{2}\sqrt{8}-\dfrac{1}{3}\sqrt{288}+\sqrt{50}\right)\left(\dfrac{1}{2}\sqrt{2}\right)\)
b, \(\left(\dfrac{4}{5}\sqrt{5}-\dfrac{1}{3}\sqrt{\dfrac{1}{5}}+3\sqrt{20}+\dfrac{1}{2}\sqrt{245}\right)\div\sqrt{5}\)
28 tháng 7 2023 lúc 20:22
\(M=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
\(N=\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a+1}}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
Rút gọn :
\(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}-\sqrt{2}\right)\)
\(\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
\(\left(\dfrac{1}{\sqrt{3}-\sqrt{2}}\right)\left(\dfrac{1}{\sqrt{3}-\sqrt{2}}\right)\)