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9 tháng 2 2022

bn châu sai rồi =7/3 mới đúng

~HT~

K cho mình nha cảm ơn các bn

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\(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)

\(\Leftrightarrow\dfrac{59-x}{41}+1+\dfrac{57-x}{43}+1+\dfrac{55-x}{45}+1+\dfrac{53-x}{47}+1+\dfrac{51-x}{49}+1=0\)

=>100-x=0

hay x=100

26 tháng 8 2015

x=20     

26 tháng 8 2015

Lí do = 0 nè:

\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-5}{39}+\frac{x-69}{49}\)

=> \(\frac{x-39}{19}+\frac{x-49}{29}=0+\left(\frac{x-5}{39}+\frac{x-69}{49}\right)\)

=> \(\frac{x-39}{19}+\frac{x-49}{29}-\left(\frac{x-5}{39}+\frac{x-69}{49}\right)=0\)

=> \(\frac{x-39}{19}+\frac{x-49}{29}-\frac{x-5}{39}-\frac{x-69}{49}=0\)

56x59+49x50+52x50

=3304+2450+2600

=5754+2600

=8304

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

HQ
Hà Quang Minh
Giáo viên
29 tháng 11 2023

5 tháng 3 2023

\(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)

\(< =>\dfrac{x+1}{59}+1+\dfrac{x+3}{57}+1+\dfrac{x+5}{55}+1=\dfrac{x+7}{53}+1+\dfrac{x+9}{51}+1+\dfrac{x+11}{49}+1\)

\(< =>\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}=\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)

\(< =>\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)=0\\ < =>x+60=0\\ < =>x=-60\)

 

 

5 tháng 3 2023

Ta có : \(\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}=\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}\)

\(\Leftrightarrow\dfrac{x+1}{59}+\dfrac{x+3}{57}+\dfrac{x+5}{55}+3\text{=}\dfrac{x+7}{53}+\dfrac{x+9}{51}+\dfrac{x+11}{49}+3\)

\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)

\(\Leftrightarrow\left(\dfrac{x+1}{59}+1\right)+\left(\dfrac{x+3}{57}+1\right)+\left(\dfrac{x+5}{55}+1\right)\text{=}\left(\dfrac{x+7}{53}+1\right)+\left(\dfrac{x+9}{51}+1\right)+\left(\dfrac{x+11}{49}+1\right)\)

\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}\text{=}\dfrac{x+60}{53}+\dfrac{x+60}{51}+\dfrac{x+60}{49}\)

\(\Leftrightarrow\dfrac{x+60}{59}+\dfrac{x+60}{57}+\dfrac{x+60}{55}-\dfrac{x+60}{53}-\dfrac{x+60}{51}-\dfrac{x-60}{49}\text{=}0\)

\(\Leftrightarrow\left(x+60\right)\left(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\right)\text{=}0\)

\(Do\) \(\dfrac{1}{59}+\dfrac{1}{57}+\dfrac{1}{55}-\dfrac{1}{53}-\dfrac{1}{51}-\dfrac{1}{49}\ne0\)

\(\Leftrightarrow\left(x+60\right)\text{=}0\)

\(x\text{=}-60\)

\(Vậy...\)