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\(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{\left(1+2.2017+2017^2\right)-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{\left(1+2017\right)^2-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{2018^2-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{\left(2018-\frac{2017}{2018}\right)^2}+\frac{2017}{2018}\)
Mà \(\frac{2017}{2018}< 1\Rightarrow2018-\frac{2017}{2018}>0\)
\(\Rightarrow B=2018-\frac{2017}{2018}+\frac{2017}{2018}\)
\(B=2018\)
Vậy bt B có giá trị nguyên
Gọi vế trái BPT là A.
Xét biểu thức tổng quát:
\(\sqrt{1+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^2\left(n+1\right)^2+\left(n+1\right)^2+n^2}{\left[n\left(n+1\right)\right]^2}}\\ =\frac{\sqrt{n^2\left(n^2+2n+1\right)+n^2+2n+1+n^2}}{n\left(n+1\right)}\\ =\frac{\sqrt{n^4+2n^3+3n^2+2n+1}}{n\left(n+1\right)}\\ =\frac{\sqrt{\left(n^2+n+1\right)^2}}{n\left(n+1\right)}\\ =\frac{n^2+n+1}{n\left(n+1\right)}\\ =\frac{n\left(n+1\right)+n+1-n}{n\left(n+1\right)}\\ =1+\frac{1}{n}-\frac{1}{n+1}\)
Suy ra:
\(A=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+1+...+1\right)+\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2017}-\frac{1}{2018}\right)\) (2018 số hạng 1)
\(=2018+\frac{1}{2}-\frac{1}{2018}< 2018\)
Vậy \(A< 2018\left(đpcm\right)\).
Chúc bạn học tốt nha
.
cảm ơn bạn nhé, mình đag ko bt cách chứng minh biểu thức tổng quát ;)
Áp dụng bđt Svacxo ta có :
\(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}\ge\dfrac{\left(\sqrt{2017}+\sqrt{2018}\right)^2}{\sqrt{2018}+\sqrt{2017}}=\sqrt{2017}+\sqrt{2018}\)
Dấu bằng xảy ra khi:
\(\dfrac{2017}{\sqrt{2018}}=\dfrac{2018}{\sqrt{2017}}\Leftrightarrow2017=2018\left(vl\right)\)
Suy ra không xảy ra dấu bằng
Vậy \(\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}}>\sqrt{2017}+\sqrt{2018}\)
Đặt \(2017=a\)
=>\(2018=a+1\)
Với mọi \(a\in N\) có:\(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}=\sqrt{\frac{\left(a+1\right)^2+a^2\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}=\sqrt{\frac{a^2+2a+1+a^2\left(a^2+2a+1\right)+a^2}{\left(a+1\right)^2}}=\sqrt{\frac{2a^2+2a+1+a^4+2a^3+a^2}{\left(a+1\right)^2}}=\sqrt{\frac{\left(a^4+2a^2+1\right)+2a\left(a^2+1\right)+a^2}{\left(a+1\right)^2}}\)
=\(\sqrt{\frac{\left(a^2+1\right)^2+2a\left(a^2+1\right)+a^2}{\left(a+1\right)^2}}=\sqrt{\frac{\left(a^2+a+1\right)}{\left(a+1\right)^2}}=\left|\frac{a^2+a+1}{a+1}\right|\)(do \(a\ge0\))
=\(\frac{a\left(a+1\right)+1}{a+1}=a+\frac{1}{a+1}\)
=> \(\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}=a+\frac{1}{a+1}\)
Thay a=2017 có:
\(\sqrt{1+2017^2+\left(\frac{2017}{2018}\right)^2}=2017+\frac{1}{2017+1}=2017+\frac{1}{2018}\)
=>\(\sqrt{1+22017^2+\left(\frac{2017}{2018}\right)^2}+\frac{2017}{2018}=2017+\frac{1}{2018}+\frac{2017}{2018}\)
<=> M=2017+1=2018
Vậy M=2018
Câu b đề sai nha, bây giờ đặt \(a=\sqrt{2017},b=\sqrt{2018}\)
Ta có \(\frac{a^2}{b}+\frac{b^2}{a}< a+b\Leftrightarrow ab\left(\frac{a^2}{b}+\frac{b^2}{a}\right)< ab\left(a+b\right)\)
\(\Leftrightarrow a^3+b^3< ab\left(a+b\right)\)(1)
Mà \(ab\left(a+b\right)\le\left(a^2-ab+b^2\right)\left(a+b\right)=a^3+b^3\)(2)
Từ (1), (2) => Sai
a) Ta có:
\(\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{k+1-k}{\left(k+1\right)\sqrt{k}}=\frac{\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}\)\(< \frac{2\sqrt{k+1}\left(\sqrt{k+1}-\sqrt{k}\right)}{\left(k+1\right)\sqrt{k}}=\frac{2\left(\sqrt{k+1}-\sqrt{k}\right)}{\sqrt{k+1}\sqrt{k}}=\frac{2}{\sqrt{k}}-\frac{2}{\sqrt{k+1}}\)
Cho k=1,2,....,n rồi cộng từng vế ta có:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{\left(n+1\right)\sqrt{n}}< \left(\frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}\right)+\left(\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}\right)\)\(+\left(\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}\right)+....+\left(\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n-1}}< 2\)