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(|x|-2011)(n+2008)(n+2009)=-(23-32)2009=-(-1)2009=1=1(n+2008)(n+2009)
=>|x|-2011=1
|x|=1+2011
|x|=2012
=>x=2012 hoặc x=-2012
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\(\left(\left|x\right|-2011\right)^{\left(2+2008\right)}\cdot\left(2+2009\right)=-\left(2^3-3^2\right)^{2009}\)
\(\left(\left|x\right|-2011\right)^{2010}\cdot2011=-\left(8-9\right)^{2009}\)
\(\left(\left|x\right|-2011\right)^{2010}\cdot2011=-\left(-1\right)^{2009}\)
\(\left(\left|x\right|-2011\right)^{2010}\cdot2011=-\left(-1\right)\)
\(\left(\left|x\right|-2011\right)^{2010}\cdot2011=1\)
\(\left(\left|x\right|-2011\right)^{2010}=\dfrac{1}{2011}\)
???
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\(\dfrac{x-1}{2011}+\dfrac{x-2}{2010}-\dfrac{x-3}{2009}=\dfrac{x-4}{2008}\)
<=> \(\left(\dfrac{x-1}{2011}-1\right)+\left(\dfrac{x-2}{2010}-1\right)-\left(\dfrac{x-3}{2009}-1\right)=\left(\dfrac{x-4}{2008}-1\right)\)
<=> \(\dfrac{x-2012}{2011}+\dfrac{x-2012}{2010}-\dfrac{x-2012}{2009}-\dfrac{x-2012}{2008}=0\)
<=> \(\left(x-2012\right)\left(\dfrac{1}{2011}+\dfrac{1}{2010}-\dfrac{1}{2009}-\dfrac{1}{2008}\right)=0\)
<=> x - 2012 = 0
<=> x = 2012
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trừ 1 vào mỗi tỉ số,ta đc:
\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1-\frac{x-3}{2009}-1=\frac{x-4}{2008}-1\)
\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}-\frac{x-3-2009}{2009}=\frac{x-4-2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}=\frac{x-2012}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}-\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(mà\frac{1}{2011}<\frac{1}{2010}<\frac{1}{2009}<\frac{1}{2008}\Rightarrow\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\)
=>x-2012=0
=>x=2012
vậy x=2012
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Bỏ dấu giá trị tuyệt đối:
x \(\le\) 2008 | 2008 < x < 2009 | 2009 \(\le\) x < 2010 | 2010\(\le\)x < 2011 | x \(\ge\) 2011 | |
|x- 2008| | 2008-x | x-2008 | x-2008 | x-2008 | x-2008 |
|x-2009| | 2009-x | 2009-x | x-2009 | x-2009 | x-2009 |
|x-2010| | 2010-x | 2010 - x | 2010 - x | x - 2010 | x - 2010 |
|x-2011| | 2011 - x | 2011 - x | 2011 - x | 2011 - x | x - 2001 |
=>
+) Nếu x \(\le\) 2008 => A = 2008 - x + 2009 - x + 2010 - x + 2011 - x + 2008 = 10 046 - 4x \(\ge\) 10 046 - 4.2008 = 2014
+) Nếu 2008 < x < 2009 => A = x - 2008 + 2009 - x + 2010 - x + 2011 - x + 2008 = 6030 - 2x > 6030 - 2.2009 = 2012
+) Nếu 2009 \(\le\) x < 2010 => A = x - 2008 + x - 2009 + 2010 - x + 2011 - x + 2008 = 2012
+) Nếu 2010 \(\le\) x < 2011 => A = x - 2008 + x - 2009 + x - 2010 + 2011 - x + 2008 = 2x - 2008 \(\ge\) 2.2010 - 2008 = 2012
+) Nếu x \(\ge\) 2011 => A = x - 2008 + x - 2009 + x - 2010 + x - 2011 + 2008 = 4x - 6030 \(\ge\) 4.2011 - 6030 = 2014
Từ các trường hợp trên => A nhỏ nhất bằng 2012 khi x = 2009 ; hoặc x = 2010
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\(\frac{x-1}{2011}+\frac{x-2}{2010}=\frac{x-3}{2009}+\frac{x-4}{2008}\)
\(\Rightarrow\frac{x-1}{2011}-1+\frac{x-2}{2010}-1=\frac{x-3}{2009}-1+\frac{x-4}{2008}-1\)
\(\Rightarrow\frac{x-1-2011}{2011}+\frac{x-2-2010}{2010}=\frac{x-3-2009}{2009}+\frac{x-4-2008}{2008}\)
\(\Rightarrow\frac{x-2012}{2011}+\frac{x-2012}{2010}=\frac{x-2012}{2009}+\frac{x-2012}{2008}\)
\(\Rightarrow\left(x-2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)
\(\Rightarrow x-2012=0\)
\(\Rightarrow x=2012\)
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Ta có:
\(\frac{x+4}{2008}+1+\frac{x+3}{2009}+1=\frac{x+2}{2010}+1+\frac{x+1}{2011}+1\)
\(\frac{x+2012}{2008}+\frac{x+2012}{2009}=\frac{x+2012}{2010}+\frac{x+2012}{2011}\)
\(\left(x+2012\right)\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
\(x=-2012\)
Ta có
(|x|-2011)^(n+2008)(n+2009)=-(-355)^2009=355 ^2009
n+2008 và n+2009 là 2 số nguyên liên tiếp => (n+2009)(n+2008) là số chẵn => (|x|-2011)^(n+2008)(n+2009) là số chính phương
=> 355 ^2009 là số chính phương mà 355^2009=5^2009 x 71^2009
5,7 là số nguyên tố
=> 2009 là số lẻ (vô lý)
Không có x t/m
(Có gì mắc bạn nhắn mình giải đáp cho)