\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
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K
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K
14 tháng 1 2019
a, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Leftrightarrow\left(\dfrac{59-x}{49}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{45}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Leftrightarrow\left(100-x\right).\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
Mà \(\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)\ne0\)
\(\Rightarrow100-x=0\)
\(\Rightarrow x=100\)
Vậy \(S=\left\{100\right\}\)
K
14 tháng 1 2019
b, \(6x^2-5x+3=2x-3x\left(3-2x\right)\)
\(\Leftrightarrow6x^2-5x+3=2x-9x+6x^2\)
\(\Leftrightarrow6x^2-5x+3=-7x+6x^2\)
\(\Leftrightarrow6x^2-5x+3+7x-6x^2=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
\(\Leftrightarrow x=\dfrac{-3}{2}\)
Vậy \(S=\left\{\dfrac{-3}{2}\right\}\)
FT
10 tháng 11 2017
\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Rightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)\(\Rightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\)
\(\Rightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\)
\(\Rightarrow100-x=0\Rightarrow x=100\)
NH
1
20 tháng 7 2017
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
NH
5 tháng 3 2023
\(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)
\(< =>\dfrac{x+1}{59}+1+\dfrac{x+3}{57}+1+\dfrac{x+5}{55}+1=\dfrac{x+7}{53}+1+\dfrac{x+9}{51}+1+\dfrac{x+11}{49}+1\)
\(< =>\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}=\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)
\(< =>\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)=0\\ < =>x+60=0\\ < =>x=-60\)
NT
Nguyễn thành Đạt
CTVHS
5 tháng 3 2023
Ta có : \(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)
\(\Leftrightarrow\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}+3\text{=}\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}+3\)
\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)
\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)
\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}\text{=}\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)
\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}-\dfrac{x+60}{53}-\dfrac{x+60}{51}-\dfrac{x-60}{49}\text{=}0\)
\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)\text{=}0\)
\(Do\) \(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\ne0\)
\(\Leftrightarrow\left(x+60\right)\text{=}0\)
\(x\text{=}-60\)
\(Vậy...\)
12 tháng 3 2023
\(\dfrac{x-45}{55}+\dfrac{x-47}{53}=\dfrac{x-55}{45}+\dfrac{x-53}{47}\)
\(\Leftrightarrow\left(\dfrac{x-45}{55}-1\right)+\left(\dfrac{x-47}{53}-1\right)=\left(\dfrac{x-55}{45}-1\right)+\left(\dfrac{x-53}{47}-1\right)\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}-\dfrac{x-100}{45}-\dfrac{x-100}{47}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\right)=0\)
Do \(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\) nên x - 100 = 0 <=> x = 100
PT
0
6 tháng 5 2019
\(\frac{59-x}{41}+\frac{57-x}{43}+\frac{55-x}{45}+\frac{53-x}{47}+\frac{51-x}{49}=-5\)
\(\Rightarrow\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{53-x}{47}+1+\frac{51-x}{49}+1\)\(=-5+5\)
\(\Rightarrow\frac{59-x+49}{41}+\frac{57-x+43}{43}+\frac{55-x+45}{45}+\frac{53-x+47}{47}\)\(+\frac{51-x+49}{49}=0\)
\(\Rightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)
\(\Rightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)
Vì \(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\ne0\)
\(\Rightarrow100-x=0\)
\(\Rightarrow x=100\)
K
6 tháng 5 2019
\(=\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{53-x}{47}+1+\)
\(\frac{51-x}{49}+1=-5+5\)
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
9 tháng 2 2021
Ta có : \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{49}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)
\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{49}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)
\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{49}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)
<=> x - 100 = 0
<=> x = 100
Vậy ..
9 tháng 2 2021
Ta có: \(\dfrac{x-50}{50}+\dfrac{x-51}{49}+\dfrac{x-52}{48}+\dfrac{x-53}{47}+\dfrac{x-200}{25}=0\)
\(\Leftrightarrow\dfrac{x-50}{50}-1+\dfrac{x-51}{49}-1+\dfrac{x-52}{48}-1+\dfrac{x-53}{47}-1+\dfrac{x-200}{25}+4=0\)
\(\Leftrightarrow\dfrac{x-100}{50}+\dfrac{x-100}{49}+\dfrac{x-100}{48}+\dfrac{x-100}{47}+\dfrac{x-100}{25}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)
mà \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}>0\)
nên x-100=0
hay x=100
Vậy: S={100}
10 tháng 4 2022
\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\Leftrightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)
=>100-x=0
hay x=100
