x+1/2017+x+2/2016+x+3/2015=x+10/2008+x+11/2007+x+12/2006
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1) 9 x 10 + 10 x 11 + 11 x 12 + ....+ 2015 x 2016
=9x10x11-8x9x10+10x11x12-9x10x11+...+2015x2016x2017-2014x2015x2013
=2015x2016x2017-8x9x10
=8193537360
![](https://rs.olm.vn/images/avt/0.png?1311)
Thấy số chính phương là các số có dạng 3k hoặc 3k+1
A=1015+1=1000.....000000000001
Tổng các chữ số của A là 1+0+0+...+0+1=2
2 có dạng 3k+2
=> A có dạng 3k+2 nên A ko phải số chính phương
B chia hết cho B thì chắc chia hết cho 3
C thì
2) x2 + y2 = 3z2 => x2 + y2 chia hết cho 3
Vì x2 ; y2 là số chính phương nên x2 ; y2 chia cho 3 dư 0 hoặc 1
Nếu x2 hoặc y2 hoặc x2 và y2 chia cho 3 dư 1 => x2 + y2 chia cho 3 dư 1 hoặc 2 ( trái với đề bai)
=> x2 ; y2 đều chia hết cho 3. 3 là số nguyên tố => x; y đều chia hết cho 3
=> x2; y2 chia hết cho 9 => 3z2 chia hết cho 9 => z2 chia hết cho 3 ; 3 là số nguyên tố => z chia hết cho 3
Vậy...
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Cho x,y là các số nguyên dương, chứng minh rằng:
\(1< \frac{x}{x+y}+\frac{y}{y+z}+\frac{z}{z+x}< 2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
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\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Rightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+1010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)=\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}\right)\)
\(\Rightarrow x+2010=0\) vì \(0< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}< \frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}\)
\(\Rightarrow x=-2010\)
Bài giải
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-(\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998})=0\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
Vì \(\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)\ne0\) nên \(x+2010=0\)
\(x=0-2010=-2010\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{x+10}{2000}+\dfrac{x+11}{1999}+\dfrac{x+12}{1998}\)
\(\Rightarrow\left(\dfrac{x+1}{2009}+1\right)+\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)=\left(\dfrac{x+10}{2000}+1\right)+\left(\dfrac{x+11}{1999}+1\right)+\left(\dfrac{x+12}{1998}+1\right)\)
\(\Rightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}=\dfrac{x+2010}{2000}+\dfrac{x+2010}{1999}+\dfrac{x+2010}{1998}\)\(\Rightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}-\dfrac{x+2010}{2000}-\dfrac{x+2010}{1999}-\dfrac{x+2010}{1998}=0\)\(\Rightarrow\left(x+2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2007}-\dfrac{1}{2000}-\dfrac{1}{1999}-\dfrac{1}{1998}\right)=0\)\(\Rightarrow x+2010=0\Rightarrow x=-2010\)
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a: \(\Leftrightarrow-\dfrac{3}{2x-3}=\dfrac{2}{5}-\dfrac{3}{2}-3=\dfrac{-41}{10}\)
=>41(2x-3)=30
=>82x-123=30
=>82x=153
hay x=153/82
b: \(\Leftrightarrow\left(x-1\right)\left(7-2x\right)=0\)
=>x=1 hoặc x=7/2
c: \(\Leftrightarrow\left(\dfrac{x+1}{2018}+1\right)+\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+10}{2009}+1\right)+\left(\dfrac{x+11}{2008}+1\right)+\left(\dfrac{x+12}{2007}+1\right)\)
=>x+2019=0
hay x=-2019