cmr B là số hữu tỉ \(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
1.So sánh \(\frac{2016}{2017}+\frac{2017}{2018}\)với \(1\)( không tính kết quả )
2.So sánh: \(A=\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}\)và \(B=\frac{2015+2016+2017}{2016+2017+2018}\)
3. Với n là số nguyên dương hãy so sánh 2 phân số sau: \(\frac{n}{n+8}\)và \(\frac{n-2}{n+9}\)
1. \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)>1
2. A>B
B= \(\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
chứng minh biểu thức B có giá trị nguyên
\(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{\left(1+2.2017+2017^2\right)-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{\left(1+2017\right)^2-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{2018^2-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(B=\sqrt{\left(2018-\frac{2017}{2018}\right)^2}+\frac{2017}{2018}\)
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
Cảm ơn bạn mk vừa đăng lên thì đã thấy luôn cách giải 😂
a,Cho A=\(\frac{2+2^2+2^3+...+2^{2017}}{1-2^{2017}}\)
Tinh A.
b,Cho A=\(\frac{1}{2017}+\frac{2}{2017^2}+\frac{3}{2017^3}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\)
CMR: A<\(\frac{2017}{2017^2}\)
Mik dang can gap.Giup mik voi.Thanks nhiu^^
a) \(A=\frac{2+2^2+...+2^{2017}}{1-2^{2017}}\)
Đặt \(B=2+2^2+...+2^{2017}\)
\(\Rightarrow2B=2^2+2^3+...+2^{2018}\)
\(\Rightarrow2B-B=\left(2^2+2^3+...+2^{2018}\right)-\left(2+...+2^{2017}\right)\)
\(\Rightarrow B=2^{2018}-2\)
\(\Rightarrow A=\frac{2^{2018}-2}{1-2^{2017}}\)
\(\Rightarrow A=\frac{-2.\left(1-2^{2017}\right)}{1-2^{2017}}\)
\(\Rightarrow A=-2\)
b)Đề phải là CM: \(A< \frac{2017}{2016^2}\)
\(A=\frac{1}{2017}+\frac{2}{2017^2}+...+\frac{22017}{2017^{2017}}+\frac{2018}{2017^{2018}}\)
\(\Rightarrow2017A=1+\frac{2}{2017}+...+\frac{22017}{2017^{2016}}+\frac{2018}{2017^{2017}}\)
\(\Rightarrow2017A-A=\left(1+...+\frac{2018}{2017^{2017}}\right)-\left(\frac{1}{2017}+...+\frac{2017}{2017^{2017}}+\frac{2018}{2017^{2018}}\right)\)
\(\Rightarrow2016A=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}-\frac{2018}{2017^{2018}}\)
Đặt \(\Rightarrow S=1+\frac{1}{2017}+\frac{1}{2017^2}+...+\frac{1}{2017^{2017}}\)
\(\Rightarrow2017S=2017+1+\frac{1}{2017}+...+\frac{1}{2017^{2016}}\)
\(\Rightarrow2017S-S=\left(2017+1+...+\frac{1}{2017^{2016}}\right)-\left(1+...+\frac{1}{2017^{2017}}\right)\)
\(\Rightarrow2016S=2017-\frac{1}{2017^{2017}}< 2017\)
\(\Rightarrow2016S< 2017\)
\(\Rightarrow S< \frac{2017}{2016}\)
\(\Rightarrow2016A< \frac{2017}{2016}\)
\(\Rightarrow A< \frac{2017}{2016^2}\left(đpcm\right)\)
\(\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
Lời giải:
\(\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(=\sqrt{(1+2017)^2-2.2017+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(\sqrt{2018^2-2.2018.\frac{2017}{2018}+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
\(=\sqrt{(2018-\frac{2017}{2018})^2}+\frac{2017}{2018}=2018-\frac{2017}{2018}+\frac{2017}{2018}=2018\)
Chứng minh rằng
a) Với mọi số nguyên dương n có \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+..+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\)
b) \(\frac{2017}{\sqrt{2018}}+\frac{2018}{\sqrt{2017}}< \sqrt{2017}+\sqrt{2018}\)
Hộ mình vs
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:
\(\frac{1}{2}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+....+\frac{1}{\left(n+1\right)\sqrt{n}}< \left(\frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}\right)+\left(\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}\right)\)\(+\left(\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}\right)+....+\left(\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n-1}}< 2\)
Chứng minh rằng biểu thức B = \(\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\) có giá trj là một số tự nhiên
\(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
Đặt B = 2017 => B + 1 = 2018
Khi B bằng:
\(B=\sqrt{1+B^2+\frac{B}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left(B+1\right)^2+B^2\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{B^2\left(B+1\right)^2+2B\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left[B\left(B+1\right)+1\right]^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\frac{B^2+B+1}{B+1}+\frac{B}{B+1}\left(\text{vi}:a>0\right)\)
\(B=\frac{B^2+2B+1}{B+1}\)
\(B=\frac{\left(B+1\right)^2}{B+1}\)
\(B=B+1\left(\text{vi}:a>0\Rightarrow B+1>0\right)\)
\(B=2017+1\left(\text{vi}:B=2017\right)\)
\(\Rightarrow B=2018\)
\(\text{Chứng minh rằng:}2017< \sqrt{\frac{2}{1}}+\sqrt[3]{\frac{3}{2}}+\sqrt[4]{\frac{4}{3}}+...+\sqrt[2018]{\frac{2018}{2017}}< 2018\)
Tính \(M=\sqrt{1+2017^2+\left(\frac{2017}{2018}\right)^2}+\frac{2017}{2018}\)
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
so sánh 2 số A và B nếu
\(A=-\frac{1}{2018}-\frac{3}{2017^2}-\frac{5}{2017^3}-\frac{7}{2017^4};B=\frac{-1}{2018}-\frac{7}{2017^2}-\frac{5}{2017^3}-\frac{3}{2017^4}\)