Vì 2016²⁰¹⁶ + 1 < 2016²⁰¹⁷ + 1

\(\Rightarrow B< \frac{2016^{2016}+1+2015}{2016^{2017}+1+2015}=\frac{2016^{2016}+2016}{2016^{2017}+2016}=\frac{2016\left(2016^{2015}+1\right)}{2016\left(2016^{2016}+1\right)}=\frac{2016^{2015}+1}{2016^{2016}+1}=A\)

Vậy A > B

Theo kết luận kết quả là A > B

A =2016^2016+2/2016^2016-1 và B= 2016^2016/2016^2016-3

Giải:

a)Ta có:

C=1957/2007=1957+50-50/2007

                      =2007-50/2007

                      =2007/2007-50/2007

                      =1-50/2007

D=1935/1985=1935+50-50/1985

                      =1985-50/1985

                      =1985/1985-50/1985

                      =1-50/1985

Vì 50/2007<50/1985 nên -50/2007>-50/1985

⇒C>D

b)Ta có:

A=2016²⁰¹⁶+2/2016²⁰¹⁶-1

A=2016²⁰¹⁶-1+3/2016²⁰¹⁶-1

A=2016²⁰¹⁶-1/2016²⁰¹⁶-1+3/2016²⁰¹⁶-1

A=1+3/2016²⁰¹⁶-1

Tương tự: B=2016²⁰¹⁶/2016²⁰¹⁶-3

                 B=1+3/2016²⁰¹⁶-3

Vì 2016²⁰¹⁶-1>2016²⁰¹⁶-3 nên 3/2016²⁰¹⁶-1<3/2016²⁰¹⁶-3

⇒A<B

Chúc bạn học tốt!

Làm tiếp:

c)Ta có:

M=10²⁰¹⁸+1/10²⁰¹⁹+1

10M=10.(10²⁰¹⁸+1)/20²⁰¹⁹+1

10M=10²⁰¹⁹+10/10²⁰¹⁹+1

10M=10²⁰¹⁹+1+9/10²⁰¹⁹+1

10M=10²⁰¹⁹+1/10²⁰¹⁹+1 + 9/10²⁰¹⁹+1

10M=1+9/10²⁰¹⁹+1

Tương tự:

N=10²⁰¹⁹+1/10²⁰²⁰+1

10N=1+9/10²⁰²⁰+1

Vì 9/10²⁰¹⁹+1>9/10²⁰²⁰+1 nên 10M>10N

⇒M>N

Chúc bạn học tốt!

con cặc

\(A=\dfrac{2016^{2016}+1}{2016^{2015}+1}>1\)

\(B=\dfrac{2016^{2016}+1}{2016^{2017}+1}< 1\)

\(\Rightarrow A< B\)

Vậy A < B

A và B có tử số nguyên dương bằng nhau, mà mẫu số nguyên dương A<B nên A>B( để dễ hiểu thì ví dụ đây: 1/5 bé hơn 1/6)

Làm cách nào cx được à bạn :v mình biết có mỗi 1 cách cho cái số mũ to này :v

Đặt a = 2016, xét hiệu A - B :

\(A-B=\dfrac{a^{2014}+1}{a^{2015}+1}-\dfrac{a^{2016}+1}{a^{2017}+1}=\dfrac{\left(a^{2014}+1\right)\left(a^{2017}+1\right)-\left(a^{2016}+1\right)\left(a^{2015}+1\right)}{\left(a^{2015}+1\right)\left(a^{2017}+1\right)}\)

Xét tử số : \(T=a^{4031}+a^{2014}+a^{2017}+1-\left(a^{4031}+a^{2016}+a^{2015}+1\right)\)

\(=a^{2014}+a^{2017}-a^{2016}-a^{2015}=a^{2014}\left(1+a^3-a^2-a\right)=a^{2014}\left(a+1\right)\left(a-1\right)^2>0\)

\(\Rightarrow A-B>0\Rightarrow A>B\)

\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)

\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)

\(10A=\dfrac{10^{2015}+2016+9\cdot2016}{10^{2015}+2016}=1+\dfrac{18144}{10^{2015}+2016}\)

\(10B=\dfrac{10^{2016}+9+18144}{10^{2016}+2016}=1+\dfrac{18144}{10^{2016}+2016}\)

mà \(\dfrac{18144}{10^{2015}+2016}>\dfrac{18144}{10^{2016}+2016}\)

nên A>B

TA có :\(\frac{2015.2016-1}{2015.2016}=\frac{2015.2016}{2015.2016}-\frac{1}{2015.2016}=1-\frac{1}{2015.2016}\)

Ta có:\(\frac{2016.2017-1}{2016.2017}=\frac{2016.2017}{2016.2017}-\frac{1}{2016.2017}=1-\frac{1}{2016.2017}\)

Vì \(2015.2016< 2016.2017\)

\(\Rightarrow\frac{1}{2015.2016}>\frac{1}{2016.2017}\)

\(\Rightarrow1-\frac{1}{2015.2016}< 1-\frac{1}{2016.2017}\)

\(\Rightarrow\frac{2015.2016-1}{2015.2016}< \frac{2016.2017-1}{2016.2017}\)

Vậy \(\frac{2015.2016-1}{2015.2016}< \frac{2016.2017-1}{2016.2017}\)

help me

Nếu:

\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)

Ta có:

\(x=\dfrac{2016^{2017}+1}{2016^{2016}+1}< 1\)

\(\Rightarrow x< \dfrac{2016^{2017}+1+2015}{2016^{2016}+1+2015}\Rightarrow x< \dfrac{2016^{2017}+2016}{2016^{2016}+2016}\Rightarrow x< \dfrac{2016\left(2016^{2016}+1\right)}{2016\left(2016^{2015}+1\right)}\Rightarrow x< \dfrac{2016^{2016}+1}{2016^{2015}+1}=y\)

\(\Rightarrow x< y\)