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AH
Akai Haruma
Giáo viên
8 tháng 5 2023

Đề thiếu rồi. Bạn xem lại.

11 tháng 5 2023

\(\dfrac{55-x}{1963}\) + \(\dfrac{50-x}{1968}\) + \(\dfrac{45-x}{1973}\) + \(\dfrac{40-x}{1978}\) + 4 = 0

(1 + \(\dfrac{55-x}{1963}\) ) + (  1 + \(\dfrac{50-x}{1968}\)) + (1+ \(\dfrac{45-x}{1973}\))+ (1 + \(\dfrac{40-x}{1978}\)) = 0

\(\dfrac{1963+55-x}{1963}\) + \(\dfrac{1968+50-x}{1968}\)+\(\dfrac{1973+45-x}{1973}\)+\(\dfrac{1978+40-x}{1978}\)=0

\(\dfrac{2018-x}{1963}\)+\(\dfrac{2018-x}{1968}\)+\(\dfrac{2018-x}{1973}\)+\(\dfrac{2018-x}{1973}\)+\(\dfrac{2018-x}{1978}\)=0

(2018 - \(x\))\(\times\)\(\dfrac{1}{1963}\)+\(\dfrac{1}{1986}\)+\(\dfrac{1}{1973}\)+) =0

                              2018 \(-x\) = 0

                              \(x\) = 2018

 

                           

17 tháng 5 2023

\(\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+4=0\)

\(\Rightarrow\text{ }\dfrac{55-x}{1963}+\dfrac{50-x}{1968}+\dfrac{45-x}{1973}+\dfrac{40-x}{1978}+1+1+1+1=0\)

\(\Rightarrow\text{ }\left(\dfrac{55-x}{1963}+1\right)+\left(\dfrac{50-x}{1968}+1\right)+\left(\dfrac{45-x}{1973}+1\right)+\left(\dfrac{40-x}{1978}+1\right)=0\)

\(\Rightarrow\text{ }\dfrac{2018-x}{1963}+\dfrac{2018-x}{1968}+\dfrac{2018-x}{1973}+\dfrac{2018-x}{1978}=0\)

\(\Rightarrow\text{ }\left(2018-x\right)\left(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\right)=0\)

Mà \(\dfrac{1}{1963}+\dfrac{1}{1968}+\dfrac{1}{1973}+\dfrac{1}{1978}\ne0\)

\(\Rightarrow\text{ }2018-x=0\)

\(\Rightarrow\text{ }x=2018-0\)

\(\Rightarrow\text{ }x=2018\)

Vậy, \(x=2018.\)

19 tháng 6 2020

\(\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0\)

\(\Leftrightarrow\left(\frac{55-x}{1963}+1\right)+\left(\frac{50-x}{1968}+1\right)+\left(\frac{45-x}{1973}+1\right)+\left(\frac{40-x}{1978}+1\right)=0\)

\(\Leftrightarrow\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0\)

\(\Leftrightarrow\left(2018-x\right).\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

\(\Leftrightarrow2018-x=0\)

\(\Leftrightarrow x=2018\)

Vậy \(x=2018\)

VC
19 tháng 6 2020

Dễ dàng :v

Có \(\frac{55-x}{1963}+\frac{50-x}{1968}+\frac{45-x}{1973}+\frac{40-x}{1978}+4=0\)

\(\Rightarrow\left(\frac{55-x}{1963}+1\right)+\left(\frac{50-x}{1968}+1\right)+\left(\frac{45-x}{1973}+1\right)+\left(\frac{40-x}{1978}+1\right)=0\)

\(\Rightarrow\frac{2018-x}{1963}+\frac{2018-x}{1968}+\frac{2018-x}{1973}+\frac{2018-x}{1978}=0\)

\(\Rightarrow\left(2018-x\right)\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)=0\)

Mà \(\Rightarrow\left(\frac{1}{1963}+\frac{1}{1968}+\frac{1}{1973}+\frac{1}{1978}\right)>0\Rightarrow2018-x=0\)

\(\Rightarrow x=2018-8=2018\)

Vậy x = 2018

a: =1250:25=50

b: =234(45+55)+234=23400+234=23634

c: =1000x1000=1000000

5 tháng 4 2015

 

45%.x + 55%.x = 20%

x(45% + 55%)   = 20%

x.100%             = 20 %

x.1                   = 20%

x                      = 20% 

x                      = 0,2 

x - 55 = 45

x = 45 + 55

x = 100

12 tháng 7 2016

x - 55 = 45 

     x = 55 + 45

      x  = 100

4 tháng 11 2017

\(\left(3x-1964\right).1967=\left(2004-1964\right).1967\)

\(\left(3x-1964\right).1967-\left(2004-1964\right).1967=0\)

\(1967.\left(3x-1964-2004+1964\right)=0\)

\(3x-2004=0\)

\(3x=2004\)

\(x=668\)

vậy \(x=668\)

4 tháng 11 2017

( 3 * x - 1964 ) * 1964 = ( 2004 - 1964 ) * 1967

Vì cả hay vế đều nhân với 1967 nên 3 * x - 1964 = 2004 - 1964

x = 2004 : 3

x = 668