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3 tháng 12 2017
Ta có \(A=\frac{2017-2018}{2017+2018}=\frac{\left(2017-2018\right)\left(2017+2018\right)}{\left(2017+2018\right)^2}=\frac{2017^2-2018^2}{2017^2+2018^2+2.2017.2018}< \frac{2017^2-2018^2}{2017^2+2018^2}=B\)
Vậy A<B
DN
18 tháng 2 2018
À khác cái dấu nhưng đề phải là giải phương trình chứ
Đặt 2017-x=a => x-2018=-a-1 phương trình trở thành:
\(\frac{a^2+a\left(-a-1\right)+\left(a-1\right)^2}{a^2-a\left(-a-1\right)+\left(a-1\right)^2}=\frac{19}{49}\)
\(\Leftrightarrow\frac{a^2+a+1}{3a^2+3a+1}=\frac{19}{49}\)
\(\Leftrightarrow49\left(a^2+a+1\right)=19\left(3a^2+3a+1\right)\)
\(\Leftrightarrow49a^2+49a+49=57a^2+57a+19\)
\(\Leftrightarrow8a^2+8a-30=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=\frac{3}{2}\\a=-\frac{5}{2}\end{cases}\Rightarrow\orbr{\begin{cases}x=2015,5\\x=2019,5\end{cases}}}\)
Vậy......................
7 tháng 2 2020
Ta có: \(B=\frac{1}{16}+\frac{2}{16^2}+\frac{3}{16^3}+...+\frac{2018}{16^{2018}}\)
\(\Rightarrow16B=1+\frac{2}{16}+\frac{3}{16^2}+....+\frac{2018}{16^{2017}}\)
\(\Rightarrow16B-B=15B=1+\frac{1}{16}+\frac{1}{16^2}+\frac{1}{16^3}+...+\frac{1}{16^{2017}}-\frac{2018}{16^{2018}}\)
Mà: \(A=1+\frac{1}{16}+\frac{1}{16^2}+\frac{1}{16^3}+...+\frac{1}{16^{2017}}\)
\(\Rightarrow16A=16+1+\frac{1}{16}+\frac{1}{16^2}+...+\frac{1}{16^{2016}}\)
\(\Rightarrow16A-A=16-\frac{1}{16^{2017}}\)
\(\Rightarrow A=\frac{16-\frac{1}{16^{2017}}}{15}\)
\(\Rightarrow15B=\frac{16-\frac{1}{16^{2017}}}{15}-\frac{2018}{16^{2018}}\)
\(\Rightarrow15B< \frac{16}{15}\)
\(\Rightarrow B< \frac{16}{15^2}< 1\)
\(\Rightarrow B^{2017}>B^{2018}\)
HT
1
S
5 tháng 3 2019
\(\frac{x-3}{2017}-\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2017}{3}\)
\(\Leftrightarrow\frac{x-3}{2017}-1-\frac{x-2}{2018}-1=\frac{x-2018}{2}-1+\frac{x-2017}{3}-1\)
\(\Leftrightarrow\frac{x-2020}{2017}-\frac{x-2020}{2018}=\frac{x-2020}{2}+\frac{x-2020}{3}\)
\(\Leftrightarrow\frac{x-2020}{2017}-\frac{x-2020}{2018}-\frac{x-2020}{2}-\frac{x-2020}{3}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)
31 tháng 3 2018
\(B=\sqrt{1+2017^2+\frac{2017^2}{2018^2}}+\frac{2017}{2018}\)
Đặt B = 2017 => B + 1 = 2018
Khi B bằng:
\(B=\sqrt{1+B^2+\frac{B}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left(B+1\right)^2+B^2\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{B^2\left(B+1\right)^2+2B\left(B+1\right)^2+B^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\sqrt{\frac{\left[B\left(B+1\right)+1\right]^2}{\left(B+1\right)^2}}+\frac{B}{B+1}\)
\(B=\frac{B^2+B+1}{B+1}+\frac{B}{B+1}\left(\text{vi}:a>0\right)\)
\(B=\frac{B^2+2B+1}{B+1}\)
\(B=\frac{\left(B+1\right)^2}{B+1}\)
\(B=B+1\left(\text{vi}:a>0\Rightarrow B+1>0\right)\)
\(B=2017+1\left(\text{vi}:B=2017\right)\)
\(\Rightarrow B=2018\)
VT
14 tháng 10 2019
2017.2019 = (2018-1)(2018+1) = 20182 -1 => a =1
b= 20183 +1 (???)
A=24783,14746B=49566,29188
Vậy A<B
Ta thấy \(A=\frac{2018-2017}{2018+2017}=\frac{2018^2-2017^2}{\left(2018+2017\right)^2}=\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}\)
Mà \(2018^2+2.2018.2017+2017^2>2018^2+2017^2\)
\(\Rightarrow\frac{2018^2-2017^2}{2018^2+2.2018.2017+2017^2}< \frac{2018^2-2017^2}{2018^2+2017^2}\)
Vậy A<B