tính \(A=3\sqrt{5}-\sqrt{\dfrac{1}{5}}+\dfrac{3}{\sqrt{5}-1}\)
\(B=\sqrt{2018^2+2018^22019^2+2019^2}\)
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HN
31 tháng 7 2018
a/ Ta có:
\(\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(\sqrt{n+1}+\sqrt{n}\right)\left(\sqrt{n+1}-\sqrt{n}\right)}=\sqrt{n+1}-\sqrt{n}\)
\(\Rightarrow A=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
31 tháng 7 2018
a.\(A=\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2018}}=\dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\dfrac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\dfrac{\sqrt{2019}-\sqrt{2018}}{\left(\sqrt{2019}+\sqrt{2018}\right)\left(\sqrt{2019}-\sqrt{2018}\right)}=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+...+\sqrt{2019}-\sqrt{2018}=\sqrt{2019}-1\)
AH
Akai Haruma
Giáo viên
10 tháng 8 2018
Lời giải:
Xét số hạng tổng quát:
\(\frac{1}{(n+1)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n(n+1)}(\sqrt{n+1}+\sqrt{n})}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}(\sqrt{n+1}+\sqrt{n})(\sqrt{n+1}-\sqrt{n})}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}(n+1-n)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\)
\(=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Do đó:
\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)
\(=1-\frac{1}{\sqrt{2019}}\)
AH
Akai Haruma
Giáo viên
7 tháng 8 2021
1.
Đặt biểu thức là $A$
Ta thấy:
$\frac{1}{1+\sqrt{2}}=\frac{\sqrt{2}-1}{(1+\sqrt{2})(\sqrt{2}-1)}=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$
Tương tự với các phân số còn lại và công theo vế thì:
$A=(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+...+(\sqrt{2019}-\sqrt{2018})$
$=\sqrt{2019}-1$
AH
Akai Haruma
Giáo viên
7 tháng 8 2021
2.
$\sqrt{8-2\sqrt{15}}=\sqrt{5-2\sqrt{5.3}+3}+\sqrt{3-2\sqrt{3.1}+1}$
$=\sqrt{(\sqrt{5}-\sqrt{3})^2}+\sqrt{(\sqrt{3}-1)^2}$
$=|\sqrt{5}-\sqrt{3}|+|\sqrt{3}-1|$
$=\sqrt{5}-\sqrt{3}+\sqrt{3}-1=\sqrt{5}-1$
HT
2 tháng 8 2018
1)
DKCĐ: a>0,\(a\ne1\)
\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{a}-\dfrac{1}{a}\right)\)\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\right)\)\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{1+a+1-a+2\sqrt{\left(1+a\right)\left(1-a\right)}}{\left(1+a\right)-\left(1-a\right)}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\)\(=\dfrac{2\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)}{2a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\sqrt{\left(1+a\right)\left(1-a\right)}+1}{a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)\left(\sqrt{\left(1+a\right)\left(1-a\right)}-1\right)}{a^2}\\ =\dfrac{\left(1+a\right)\left(1-a\right)-1}{a^2}\\ =\dfrac{1-a^2-1}{a^2}\\ =\dfrac{-a^2}{a^2}\\ =-1\)
MN
4 tháng 10 2018
Bài 1:Với mọi n∈N*,ta có:
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó :
A=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
6 tháng 10 2022
Bài 2:
\(A=\left(3\sqrt{2}-3+4\sqrt{2}+2-4-2\sqrt{2}\right)\cdot\left(2\sqrt{2}+2\right)\)
\(=\left(5\sqrt{2}-5\right)\left(2\sqrt{2}+2\right)\)
=10
2 tháng 9 2022
Bài 1:
a: \(\Leftrightarrow2-3\sqrt{x}+5\sqrt{x}=8\)
=>2 căn x=6
=>căn x=3
=>x=9
b: \(\Leftrightarrow\dfrac{1}{\sqrt{x}}\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{x}}=\dfrac{2}{3}:\dfrac{2}{3}=1\)
=>x=1
AH
Akai Haruma
Giáo viên
17 tháng 7 2018
Lời giải:
Xét \(1+\frac{1}{n^2}+\frac{1}{(n+1)^2}=(1+\frac{1}{n})^2-\frac{2}{n}+\frac{1}{(n+1)^2}\)
\(=\left(\frac{n+1}{n}\right)^2+\frac{1}{(n+1)^2}-\frac{2}{n}\)
\(=\left(\frac{n+1}{n}\right)^2+\frac{1}{(n+1)^2}-2.\frac{n+1}{n}.\frac{1}{n+1}\)
\(=\left(\frac{n+1}{n}-\frac{1}{n+1}\right)^2=\left(1+\frac{1}{n(n+1)}\right)^2\)
\(\Rightarrow \sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=1+\frac{1}{n(n+1)}\)
Do đó:
\(A=1+\frac{1}{1.2}+1+\frac{1}{2.3}+...+1+\frac{1}{2018.2019}\)
\(=2018+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2018.2019}\)
\(=2018+\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{2019-2018}{2018.2019}=2018+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2018}-\frac{1}{2019}\)
\(=2019-\frac{1}{2019}\)
29 tháng 11 2018
co cong thuc \(\sqrt{\dfrac{1}{1}+\dfrac{1}{a^2}+\dfrac{1}{\left(a+1\right)^2}}=\dfrac{1}{1}+\dfrac{1}{a}-\dfrac{1}{a+1}\) ban tu chung minh nha
\(A=3\sqrt{5}-\dfrac{1}{5}\sqrt{5}+\dfrac{3\left(\sqrt{5}+1\right)}{5-1}\)
\(=\dfrac{14}{5}\sqrt{5}+\dfrac{3}{4}\sqrt{5}+\dfrac{3}{4}\)
\(=\dfrac{71}{20}\sqrt{5}+\dfrac{3}{4}\)
a: \(A=3\sqrt{5}-\dfrac{1}{5}\sqrt{5}+\dfrac{3}{4}+\dfrac{3}{4}\sqrt{5}=\dfrac{71}{20}\sqrt{5}+\dfrac{3}{4}\)
b: Đặt a=2018
\(B=\sqrt{a^2+a^2\left(a+1\right)^2+\left(a+1\right)^2}\)
\(=\sqrt{a^2+\left(a^2+a\right)^2+a^2+2a+1}\)
\(=\sqrt{2a^2+1+2a+a^4+2a^3+a^2}\)
\(=\sqrt{a^4+2a^3+3a^2+2a+1}\)
\(=\sqrt{\left(a^2+a+1\right)^2}=a^2+a+1=2018^2+2018+1\)