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a: Trường hợp 1: x<2015

A=2015-x+2016=4031-x

Trường hợp 2: x>=2015

A=x-2015+2016=x+1

b: Trường hợp 1: x<2015

B=2015-x+2016-x=4031-2x

Trường hợp 2: 2015<=x<2016

B=x-2015+2016-x=1

Trường hợp 3:x>=2016

B=x-2015+x+2016=2x-4031

\(B=\left(-5\right)^0+\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2017}\)

\(-5B=\left(-5\right)^1+\left(-5\right)^2+\left(-5\right)^3+...+\left(-5\right)^{2017}\)

\(-6B=\left(-5\right)^{2017}-1\)

\(B=\frac{\left(-5\right)^{2017}-1}{-6}\)

Ta có : B = (-5)^0 + (-5)^1 + ......+ (-5)^2017

          (-5)B = (-5)^1 + (-5)^2 + .......+ (-5)^2018

              (-4)B = (-5)^0- (-5)^2018

           B = 1 - (-5)^2018 / (-4)

Nếu có sai sót gì mong các bạn bỏ qua

\(-5B=\left(-5\right)^1+\left(-5\right)^2+...+\left(-5\right)^{2018}\)

\(-4B=\left(-5\right)^{2018}-\left(-5\right)^0\)

\(\Rightarrow B=\frac{\left(-5\right)^{2018}-\left(-5\right)^0}{-4}\)

Bài 1:

ta có: \(B=\frac{12}{\left(2.4\right)^2}+\frac{20}{\left(4.6\right)^2}+...+\frac{388}{\left(96.98\right)^2}+\frac{396}{\left(98.100\right)^2}\)

\(B=\frac{4^2-2^2}{2^2.4^2}+\frac{6^2-4^2}{4^2.6^2}+...+\frac{98^2-96^2}{96^2.98^2}+\frac{100^2-98^2}{98^2.100^2}\)

\(B=\frac{1}{2^2}-\frac{1}{4^2}+\frac{1}{4^2}-\frac{1}{6^2}+...+\frac{1}{96^2}-\frac{1}{98^2}+\frac{1}{98^2}-\frac{1}{100^2}\)

\(B=\frac{1}{2^2}-\frac{1}{100^2}\)

\(B=\frac{1}{4}-\frac{1}{100^2}< \frac{1}{4}\)

\(\Rightarrow B< \frac{1}{4}\)

Bài 2:

ta có: \(B=\frac{2015+2016+2017}{2016+2017+2018}\)

\(B=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

mà \(\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\)

\(\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\)

\(\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\)

\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)

\(\Rightarrow A>B\)

Học tốt nhé bn !!

Ta có: \(B=4^{2017}+4^{2016}+...+4^2+4^1+4^0\)

\(\Leftrightarrow4\cdot B=4^{2018}+4^{2017}+...+4^3+4^2+4^1\)

\(\Leftrightarrow3\cdot B=4^{2018}-1\)

\(\Leftrightarrow A=165\cdot\dfrac{4^{2018}-1}{3}+55\)

\(\Leftrightarrow A=4^{2018}\)

Ta có: 1+(1+2)+(1+2+3)+...+(1+2+3+...+2017)=2017x1+2016x2+2015x3+...+2x2016+1x2017

=> K-2016=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2017\right)}{2017x1+2016x2+2015x3+...+2x2016+1x2017}\)=\(\frac{2017x1+2016x2+2015x3+...+2x2016+1x2017}{2017x1+2016x2+2015x3+...+2x2016+1x2017}=1\)

=> K=2016+1=2017

Toán tiếng anh hả bạn

Bài này thì bạn mình có thể giải được

Thank you

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