59-x/41+57-x/43+55-x/45+53-x/47+51-x/49=-5
giải phương trình: (59-x/41) + (57-x/43) + (55-x/45) + (53-x/47) + (51-x/49) = -5
ta có: (59-x)/41 +(57-x)/43 +(55-x)/45 +(53-x)/47 +(51-x)/49 =-5
<=>[(59-x)/41 +1 ] +[(57-x)/43 +1] +[(55-x)/45 +1] +[(53-x)/47 +1] +[(51-x)/49 +1] =0
<=>(59-x-41)/41 + (57-x-43)/43 +(55-x-45)/45 +(53-x-47)/47 +(51-x-49)/49 =0
<=>(100-x)/41 + (100-x)/43 + (100-x)/45 +(100-x)/47 + (100-x)/49 =0
<=>(100-x).( 1/41 + 1/43 + 1/45 + 1/47 + 1/49 ) =0
mà (1/41 + 1/43 + 1/45 + 1/47 + 1/49) khác 0 nên 100-x =0 <=>x=100
vậy nghiệm của pt là x=100
giải phương trình 59-x/41+57-x/43+55-x/45+53-x/47+51-x/49=-5
\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
giải phương trình trên
Tìm x E Z biết
59-x/41 +57-x/43 +55-x/45 + 53-x/47 +51-x/49=-5
\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Leftrightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
Mà \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\)
\(\Leftrightarrow100-x=0\Leftrightarrow x=100\)
Vậy x = 100
Tìm x biết 59-x \41 + 57-x\ 43 + 55-x\45 + 53-x\47 + 51-x\49 = -5 . Giúp mình mình like cho . Ok. <3
\(\frac{59-x}{41}+\frac{57-x}{43}+\frac{55-x}{45}+\frac{53-x}{47}+\frac{51-x}{49}=-5\)
Câu 2: (2 điểm) Tim x, biết:
1) (x−1)/2009+(x−2)/2008=(x−3)/2007+(x−4)/2006
2) (59−x)/41+(57−x)/43+(55−x)/45+(53−x)/47+(51−x)/49=−5
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 \(100-x=0\Leftrightarrow x=100\)
giải phương trình
\(\frac{59-x}{41}+\frac{57-x}{43}+\frac{55-x}{45}+\frac{53-x}{47}+\frac{51-x}{49}=-5\)
\(\frac{59-x}{41}+\frac{57-x}{43}+\frac{55-x}{45}+\frac{53-x}{47}+\frac{51-x}{49}=-5\)
\(\Rightarrow\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{53-x}{47}+1+\frac{51-x}{49}+1\)\(=-5+5\)
\(\Rightarrow\frac{59-x+49}{41}+\frac{57-x+43}{43}+\frac{55-x+45}{45}+\frac{53-x+47}{47}\)\(+\frac{51-x+49}{49}=0\)
\(\Rightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)
\(\Rightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)
Vì \(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\ne0\)
\(\Rightarrow100-x=0\)
\(\Rightarrow x=100\)
\(=\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{53-x}{47}+1+\)
\(\frac{51-x}{49}+1=-5+5\)
đoạn này có 5 là do mik mượn 5 con 1 khi đó nha
\(=\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\)
\(\frac{100-x}{49}=0\)
\(=\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)
mà \(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}< 0\)
nên 100-x=0
còn lại bn từ lm
\(\frac{59-x}{41}\)+ \(\frac{57-x}{43}\)+ \(\frac{55-x}{45}\)+ \(\frac{53-x}{47}\)+ \(\frac{51-x}{49}\)= -5