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21 tháng 1 2016

x-5/1990+x-15/1980+x-25/1970=x-1990/5+x-1980/15+x-1970/25

<=> (x-5/1990-1)+(x-15/1980-1)+(x-25/1970-1)=(x-1990/5-1)+(x-1980/15-1)+(x-1970/25-1)

<=> x-1995/1990+x-1995/1980+x-1995/1970=x-1995/5+x-1995/15+x-1995/25

<=> (x-1995)(1/1990+1/1980+1/1970-1/5-1/15-1/25)=0

<=> x-1995=0 

<=> x=1995

\(\Leftrightarrow\left(\dfrac{x-5}{1990}-1\right)+\left(\dfrac{x-15}{1980}-1\right)+\left(\dfrac{x-25}{1970}-1\right)\\ +\left(\dfrac{x-1990}{5}-1\right)+\left(\dfrac{x-1980}{15}-1\right)+\left(\dfrac{x-1970}{25}-1\right)=0\\ \Leftrightarrow\dfrac{x-1995}{1990}+\dfrac{x-1995}{1980}+\dfrac{x-1995}{1970}+\dfrac{x-1995}{5}\\ +\dfrac{n-1995}{15}+\dfrac{n-1995}{25}=0\\ \Rightarrow\left(x-1995\right)\left(\dfrac{1}{1990}+\dfrac{1}{1980}+\dfrac{1}{1970}+\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}\right)=0\) 

\(\Rightarrow x-1995=0\\ \Rightarrow x=1995\)

16 tháng 2 2022

trừ 1 riết rồi không bù vào cho nó à :>

19 tháng 1 2018

xong r nhé. thanks m.n

12 tháng 2 2020

Ta có: \(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}\)

\(\Leftrightarrow\)\(\frac{x-5}{1990}+\frac{x-15}{1980}+\frac{x-25}{1970}-3=\frac{x-1990}{5}+\frac{x-1980}{15}+\frac{x-1970}{25}-3\)

\(\Leftrightarrow\)\(\frac{x-5}{1990}-1+\frac{x-15}{1980}-1+\frac{x-25}{1970}-1=\frac{x-1990}{5}-1+\frac{x-1980}{15}-1+\frac{x-1970}{25}-1\)\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}=\frac{x-1995}{5}+\frac{x-1995}{15}+\frac{x-1995}{25}\)

\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}+\frac{x-1995}{1970}-\frac{x-1995}{5}-\frac{x-1995}{15}-\frac{x-1995}{25}=0\)

\(\Leftrightarrow\)\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}-\frac{1}{15}-\frac{1}{25}\right)=0\)

\(\Leftrightarrow\)\(x-1995=0\)

\(\Leftrightarrow\)\(x=1995\)

13 tháng 3 2018

pt <=> (x-5/1990 -  1) + (x-15/1980 - 1) = (x-1980/15 - 1) + (x-1990/5 - 1)

<=> x-1995/1990 + x-1995/1980 = x-1995/15 + x-1995/5

<=> x-1995/15 + x-1995/5 - x-1995/1990 - x-1995/1980 = 0

<=> (x-1995).(1/5+1/15-1/1990-1/1980) = 0

<=> x-1995 = 0 ( vì 1/5 + 1/15 - 1/1990 - 1/1980 > 0 )

<=> x = 1995

Vậy S={1995}

Tk mk nha

13 tháng 3 2018

Ta có : 

\(\frac{x-5}{1990}+\frac{x-15}{1980}=\frac{x-1980}{15}+\frac{x-1990}{5}\)

\(\Leftrightarrow\)\(\left(\frac{x-5}{1990}-1\right)+\left(\frac{x-15}{1980}-1\right)=\left(\frac{x-1980}{15}-1\right)+\left(\frac{x-1990}{5}-1\right)\)

\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}=\frac{x-1995}{15}+\frac{x-1995}{5}\)

\(\Leftrightarrow\)\(\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)

\(\Leftrightarrow\)\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{15}+\frac{1}{5}\right)=0\)

Vì \(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{15}+\frac{1}{5}\ne0\)

Nên \(x-1995=0\)

\(\Rightarrow\)\(x=1995\)

Vậy \(x=1995\)

Chúc bạn học tốt ~

10 tháng 1 2016

Ta có: \(\frac{x-5}{1990}+\frac{x-15}{1980}=\frac{x-1980}{15}+\frac{x-1990}{5}\)

=> \(\left(\frac{x-5}{1990}-1\right)+\left(\frac{x-15}{1980}-1\right)=\left(\frac{x-1980}{15}-1\right)+\left(\frac{x-1990}{5}-1\right)\)

=> \(\frac{x-5-1990}{1990}+\frac{x-15-1980}{1980}=\frac{x-1980-15}{15}+\frac{x-1990-5}{5}\)

=> \(\frac{x-1995}{1990}+\frac{x-1995}{1980}=\frac{x-1995}{15}+\frac{x-1995}{5}\)

=> \(\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)

=> \(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\right)=0\)

Vì \(\frac{1}{1990}+\frac{1}{1980}\ne\frac{1}{15}+\frac{1}{5}\)           =>   \(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\ne0\)

=> x - 1995 = 0

=> x = 1995

10 tháng 1 2016

\(\frac{x-5}{1990}+\frac{x-15}{1980}=\frac{x-1980}{15}+\frac{x-1990}{5}\)

\(\Leftrightarrow\frac{x-5}{1990}-1+\frac{x-15}{1980}-1-\frac{x-1980}{15}+1-\frac{x-1990}{5}+1=0\)

\(\Leftrightarrow\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)

\(\Leftrightarrow\left(x-1995\right).\left(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\right)=0\)

<=>x=1995 

:(

17 tháng 1 2018

\(\frac{x-5}{1990}-1+\frac{x-15}{1980}-1=\frac{x-1980}{15}-1+\frac{x-1990}{5}-1\)

\(\frac{x-1995}{1990}+\frac{x-1995}{1980}-\frac{x-1995}{15}-\frac{x-1995}{5}=0\)

\(\left(x-1995\right)\left(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\right)=0\)

Mà \(\frac{1}{1990}+\frac{1}{1980}-\frac{1}{15}-\frac{1}{5}\ne0\)

Nên \(x-1995=0\Leftrightarrow x=1995\)

28 tháng 2 2019

\(\frac{x+10}{2000}+\frac{x+20}{1990}+\frac{x+30}{1980}+\frac{x+40}{1970}=-4\)

\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)

\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)

Vì  \(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}>0\)

\(\Rightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

28 tháng 2 2019

\(\Leftrightarrow\frac{x+10}{2000}+1+\frac{x+20}{1990}+1+\frac{x+30}{1980}+1+\frac{x+40}{1970}+1=0\)

\(\Leftrightarrow\frac{x+2010}{2000}+\frac{x+2010}{1990}+\frac{x+2010}{1980}+\frac{x+2010}{1970}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)=0\)

\(\left(\frac{1}{2000}+\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}\right)\ne0\Rightarrow\left(x+2010\right)=0\\ \Rightarrow x=-2010\)

5 tháng 3 2020

1/Bạn cộng tất cả các phân số ở 2 vế với 1, tất cả các phân số sẽ có chung tử, cậu nhóm tử đó lại thành PT tích..với mẫu =0 tìm đc x

2/Trừ 1 vào từng phân thức đc

\(\frac{x-b-c}{a}-1+\frac{x-a-c}{b}-1+\frac{x-a-b}{c}-1=0\)

\(\Leftrightarrow\frac{x-\left(a+b+c\right)}{a}+\frac{x-\left(a+b+c\right)}{b}+\frac{x-\left(a+b+c\right)}{c}=0\)

\(\Leftrightarrow\left(x-\left(a+b+c\right)\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

\(\Rightarrow x=a+b+c\)