Căn bậc 2 của 1 là 1,của 2018 bình phương là 2018,2018 bình phương/2019 bình phương là 2018/2019 nên cái căn đó có giá trị là 1+2018+2018/2019 nha.bn lấy 2018/2019+2018/2019 nếu là số tự nhiên thì biểu thức này là STN

\(\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)

\(=\)\(\sqrt{\left(1+2.2018+2018^2\right)-2.2018+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)

\(=\)\(\sqrt{2019^2-2.2018+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)

\(=\)\(\sqrt{\left(2019-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)

\(=\)\(\left|2019-\frac{2018}{2019}\right|+\frac{2018}{2019}=2019-\frac{2018}{2019}+\frac{2018}{2019}=2019\)

\(\Rightarrow\)\(\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\) là số tự nhiên ( đpcm ) 

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:v nãy giải xong thì bị lỗi please signing gì đó...(giải rất kĩ càng,lần này ko giải kĩ nx  -_-)

Đặt a = 2018 -> 2019 = a + 1..

Gọi biểu thức trên là A.Quy đồng biểu thức trong căn và rút gọn,ta được:

\(A=\sqrt{\frac{a^4+2a^3+3a^2+2a+1}{\left(a+1\right)^2}}+\frac{a}{a+1}\)

Đặt \(B=a^4+2a^3+3a^2+2a+1\)

\(=a^2\left(a^2+2a+3+\frac{2}{a}+\frac{1}{a^2}\right)\)

\(=a^2\left[\left(a+\frac{1}{a}\right)^2+2\left(a+\frac{1}{a}\right)+1\right]\)

\(=\left[a\left(a+\frac{1}{a}+1\right)\right]^2\) (Làm tắt xíu nhé)

Suy ra \(A=\frac{\left(a+\frac{1}{a}+1\right)a}{\left(a+1\right)}+\frac{a}{a+1}=\frac{a^2+2a+1}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=2019\)

Là số tự nhiên.(đpcm)

\(B=\sqrt{\frac{2019^2}{2019^2}+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)

\(B=\sqrt{\frac{\left(2018+1\right)^2}{2019^2}+\frac{2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)

\(B=\sqrt{\frac{1}{2019^2}+\frac{2018^2+2.2018+2018^2}{2019^2}+2018^2}+\frac{2018}{2019}\)

\(B=\sqrt{\frac{1}{2019^2}+2.2018.\frac{1}{2019}+2018^2}+\frac{2018}{2019}\)

\(B=\sqrt{\left(\frac{1}{2019}+2018\right)^2}+\frac{2018}{2019}\)

\(B=\frac{1}{2019}+2018+\frac{2018}{2019}=2019\) là một số tự nhiên

\(B=\sqrt{1+2018^2+\frac{2018^2}{2019^2}}+\frac{2018}{2019}\)

\(B=\sqrt{1^2+2018^2+\left(-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)

\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2+2.\frac{2018}{2019}+2.\frac{2018^2}{2019}-2.2018}\)\(+\frac{2018}{2019}\)

\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2+2\left(\frac{2018+2018.2018-2018.2019}{2019}\right)}\)\(+\frac{2018}{2019}\)

\(B=\sqrt{\left(1+2018-\frac{2018}{2019}\right)^2}+\frac{2018}{2019}\)

\(B=1+2018-\frac{2018}{2019}+\frac{2018}{2019}=2019\)

Vậy B có giá trị là 1 số tự nhiên.

Akai Haruma Nguyễn Thanh Hằng Nguyễn Thị Ngọc Thơ

Ta có: \(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{k+1}}=\frac{\left(k+1\right)\sqrt{k}-k\sqrt{k+1}}{k\left(k+1\right)^2-k^2\left(k+1\right)}\)

\(=\frac{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)}{k^3+2k^2+k-k^3-k^2}\)

\(=\frac{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)}{k\left(k+1\right)}\)

\(=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)

Lần lượt thay k=1;2;...;2018 ta được:

\(\frac{1}{2\sqrt{1}+1\sqrt{2}}=\frac{1}{1}-\frac{1}{\sqrt{2}}\)

\(\frac{1}{3\sqrt{2}+2\sqrt{3}}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\)

...

\(\frac{1}{2019\sqrt{2018}+2018\sqrt{2019}}=\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)

Cộng vế theo vế ta được:

\(C=1-\frac{1}{\sqrt{2019}}=...\)

1)

DKCĐ: a>0,\(a\ne1\)

\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}}{a}-\dfrac{1}{a}\right)\)\(=\left(\dfrac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\dfrac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right)\left(\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\right)\)\(=\dfrac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{1+a+1-a+2\sqrt{\left(1+a\right)\left(1-a\right)}}{\left(1+a\right)-\left(1-a\right)}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\)\(=\dfrac{2\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)}{2a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\sqrt{\left(1+a\right)\left(1-a\right)}+1}{a}\cdot\dfrac{\sqrt{\left(1-a\right)\left(1+a\right)}-1}{a}\\ =\dfrac{\left(\sqrt{\left(1+a\right)\left(1-a\right)}+1\right)\left(\sqrt{\left(1+a\right)\left(1-a\right)}-1\right)}{a^2}\\ =\dfrac{\left(1+a\right)\left(1-a\right)-1}{a^2}\\ =\dfrac{1-a^2-1}{a^2}\\ =\dfrac{-a^2}{a^2}\\ =-1\)

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:

\(S=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2018}}-\frac{1}{\sqrt{2019}}\)

\(=1-\frac{1}{\sqrt{2019}}\)