Tính giá trị biểu thức:
A= \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
21 tháng 6 2021
`a)(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)-(2sqrtx+1)/(3-sqrtx)(x>=0,x ne 4,x ne 9)`
`=(2sqrtx-9)/(x-5sqrtx+6)-(sqrtx+3)/(sqrtx-2)+(2sqrtx+1)/(sqrtx-3)`
`=(2sqrtx-9+(sqrtx-3)(sqrtx+3)+(2sqrtx+1)(sqrtx-2))/(x-5sqrtx+6)`
`=(2sqrtx-9+x-9+2x-3sqrtx-2)/(x-5sqrtx+6)`
`=(3x-sqrtx-20)/
BL
13 tháng 8 2017
\(b,\) Ta có:
\(\dfrac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\\ =\dfrac{1}{\sqrt{n}.\sqrt{n-1}\left(\sqrt{n}+\sqrt{n-1}\right)}\\ =\dfrac{\sqrt{n}}{\sqrt{n}.\sqrt{n-1}}-\dfrac{\sqrt{n-1}}{\sqrt{n}.\sqrt{n-1}}\\ =\dfrac{1}{\sqrt{n-1}}-\dfrac{1}{\sqrt{n}}\)
Thay:
\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)
\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(...\)
\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)
30 tháng 7 2017
A = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
= \(\dfrac{1}{2}\left(\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}\right)\)
= \(\dfrac{1}{2}\left(\sqrt{99}-\sqrt{3}\right)\)
B = 35 + 335 + 3335 + ... + 3333...(99 số 3)35
= 33 + 2 + 333 + 2 + 3333 + 2 + ... + 333...33 + 2
= 2 . 99 + (33 + 333 + 3333 + ... + 333...3)
= 198 + \(\dfrac{1}{3}\)(99 + 999 + 9999 + ... + 999...99)
= 198 + \(\dfrac{1}{3}\)(102 - 1 + 103 - 1 + 104 - 1 + ... + 10100 - 1)
= \(\left(\dfrac{10^{101}-10^2}{27}\right)+165\)
HN
29 tháng 11 2018
\(2\left(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\right)\)
\(>\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}+\dfrac{1}{\sqrt{99}+\sqrt{101}}\)
\(=\dfrac{1}{2}\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)
\(=\dfrac{1}{2}\left(\sqrt{101}-\sqrt{1}\right)>\dfrac{9}{2}\)
\(\Rightarrow\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}>\dfrac{9}{4}\)
11 tháng 3 2022
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>=0\\a< >1\end{matrix}\right.\)
\(A=\dfrac{1}{2\left(\sqrt{a}+1\right)}-\dfrac{1}{2\left(\sqrt{a}-1\right)}+\dfrac{a^2+1}{a^2-1}\)
\(=\dfrac{\sqrt{a}-1-\sqrt{a}-1}{2\left(a-1\right)}+\dfrac{a^2+1}{a^2-1}\)
\(=\dfrac{-1}{a-1}+\dfrac{a^2+1}{a^2-1}\)
\(=\dfrac{-a-1+a^2+1}{\left(a-1\right)\left(a+1\right)}=\dfrac{a^2-a}{\left(a-1\right)\left(a+1\right)}=\dfrac{a}{a+1}\)
b: Để A-1/3<0 thì \(\dfrac{a}{a+1}-\dfrac{1}{3}< 0\)
=>3a-a-1<0
=>2a-1<0
hay 0<a<1/2
Nhận xét 1: từng hạng tử của A có dạng:
\(\dfrac{1}{\sqrt{x}+\sqrt{x+2}}\left(x\ge3\right)\)
Nhận xét 2:
\(\left(\sqrt{x+2}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{x+2}\right)=\left(x+2\right)-x=2\)
\(\Rightarrow\dfrac{2}{\sqrt{x}+\sqrt[]{x+2}}=-\sqrt{x}+\sqrt{x+2}\)
Áp dụng vào A:
\(2A=\dfrac{2}{\sqrt{3}+\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{7}}+...+\dfrac{2}{\sqrt{97}+\sqrt{99}}\)
\(=\left(-\sqrt{3}+\sqrt{5}\right)+\left(-\sqrt{5}+\sqrt{7}\right)+...+\left(-\sqrt{97}+\sqrt{99}\right)\)
\(=-\sqrt{3}+\sqrt{99}\Leftrightarrow A=-2\sqrt{3}+2\sqrt{99}\)
A = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
=
\(\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)}+\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)}+\dfrac{\sqrt{9}-\sqrt{7}}{\left(\sqrt{7}+\sqrt{9}\right)\cdot\left(\sqrt{9}-\sqrt{7}\right)}+...+\dfrac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{97}+\sqrt{99}\right)\cdot\left(\sqrt{99}-\sqrt{97}\right)}\)
= \(\dfrac{\sqrt{5}-\sqrt{3}}{5-3}+\dfrac{\sqrt{7}-\sqrt{5}}{7-5}+\dfrac{\sqrt{9}-\sqrt{7}}{9-7}+...+\dfrac{\sqrt{99}-\sqrt{97}}{99-97}\)
=\(\dfrac{\sqrt{5}-\sqrt{3}}{2}+\dfrac{\sqrt{7}-\sqrt{5}}{2}+\dfrac{\sqrt{9}-\sqrt{7}}{2}+...+\dfrac{\sqrt{99}-\sqrt{97}}{2}\)
=\(\dfrac{1}{2}\cdot\left(\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}\right)\)
= \(\dfrac{1}{2}\cdot\left(-\sqrt{3}+\sqrt{99}\right)\)
= \(\dfrac{3\sqrt{11}-\sqrt{3}}{2}\)