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câu 1. tìm x nguyên để \(\frac{-35}{6}\)<x<\(\frac{-18}{5}\)
<=> -4,375<x<-3,6
mà x\(\in\)Z nên x={-4}
câu 2. A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)
B=\(\frac{2015+2016}{2016+2017}\)=\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)
Vì \(\frac{2015}{2016+2017}\)<\(\frac{2015}{2016}\); \(\frac{2016}{2016+2017}\)<\(\frac{2016}{2017}\)
Vậy B<A
\(A=\frac{2015}{2016}+\frac{2016}{2017}=1-\frac{1}{2016}+1-\frac{1}{2017}>1\)
\(B=\frac{2015+2016}{2016+2017}< \frac{2016+2017}{2016+2017}=1\)
Suy ra \(A>B\).
\(\frac{2015}{2016}+\frac{2016}{2017}>\frac{\left(2015+2016\right)}{\left(2016+2017\right)}=\frac{2015}{2016+2017}+\frac{2016}{2016+2017}\)
phân tích B ta có
B = \(\frac{2014+2015}{2015+2016}=\frac{2014}{2015+2016}+\frac{2015}{2015+2016}\)
vì \(\frac{2014}{2015+2016}<\frac{2014}{2015}\) ., \(\frac{2015}{2015+2016}<\frac{2015}{2016}\)
=> B< A
A=2014/2015+2015/2016. B=(2014+2015)/(2015+2016)
A=1-1/2015+1-1/2016. B=1-2/4031
A=1+1-(2015+2016)/(2015x2016). So sánh
A=1+1-(4031)/(2015x2x1008). 1+1-[4031/(4030x1008)]>1;1-2/4031<1.
A=1+1-[4031/(4030x1008)]. Vậy 1+1-[4031/(4030x1008)]>1-2/4031.
=>A>B
Ta có:\(Q=\frac{2015+2016+2017}{2016+2017+2018}=\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
Vì \(\hept{\begin{cases}\frac{2015}{2016}>\frac{2015}{2016+2017+2018}\\\frac{2016}{2017}>\frac{2016}{2016+2017+2018}\\\frac{2017}{2018}>\frac{2017}{2016+2017+2018}\end{cases}}\)
\(\Rightarrow\frac{2015}{2016}+\frac{2016}{2017}+\frac{2017}{2018}>\frac{2015}{2016+2017+2018}+\frac{2016}{2016+2017+2018}+\frac{2017}{2016+2017+2018}\)
\(\Rightarrow P>Q\)
Vậy P > Q
\(1-\frac{2015}{2016}=\frac{1}{2016}\)
\(1-\frac{2016}{2017}=\frac{1}{2017}\)
Vì \(\frac{1}{2016}>\frac{1}{2017}\) (phần bù so với 1)
Nên \(\frac{2015}{2016}< \frac{2016}{2017}\)
sai rồi