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1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
x1+x2+x3+x4+...x49+x50+x51=.
\(\Rightarrow\)(x1+x2)+(x3+x4)+...+(x49+x50)+x51=0
Theo de bai ta co:\(\Rightarrow\)1+1+...+1+x51=0
\(\Rightarrow\)1*25+x51=0
\(\Rightarrow\)x51=-25
Ma x60+x51=0\(\Rightarrow\)-50+(-25)=1\(\Rightarrow\)x50=26
Nho tich nha
-(x-6+85)=(x+51)-54
-x+6-85=x+51-54
-x-x=-6+85+51-54
-2x=76
x=-38
(x1 + x2) + (x3 + x4) +..... + (x49 + x50) + x51
= 1 + 1+ 1 + .... + 1 + x51 = 0
= 25 + x51 = 0
x51 = 0 - 25
x51 = -25
x1+x2+x3+...+x50+x51=0
=>(x1+x2)+(x3+x4)+...+(x49+x50)+x51=0
=>1+1+1+...+1+x51=0
=>25+x1=0
=>x51=-25
=>x50=1-(-25)=26
x1+x2+x3+...+x49+x50+x51=0
(x1+x2)+(X3+x4)+ ...+ (x49+x50) + x51=0
1.25+x51=0
=> x51= -25 ( vì -25+25=0 nhá ^^)
mà x50+ (-25)(tức là x51 ý ) =1
=> x50=26