K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

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

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

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

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

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

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

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

nên x-1995=0

hay x=1995

Vậy: S={1995}

6 tháng 6 2020

còn câu a?

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

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

16 tháng 2 2022

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

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

7 tháng 2 2018

1) Ta có:

\(\left(3x+5\right)\left(11+3m\right)-7\left(x+2\right)=115\) có nghiệm x=1

Thay x = 1 vào pt ta được:

\(\left(3.1+5\right)\left(11+3m\right)-7\left(1+2\right)=115\)

\(\Leftrightarrow8\left(11+3m\right)-7.3=115\)

\(\Leftrightarrow88+24m-21=115\)

\(\Leftrightarrow88+24m=136\)

\(\Leftrightarrow24m=48\)

\(\Leftrightarrow m=2\)

Vậy để pt nhận x=1 làm nghiệm thì m = 2

7 tháng 2 2018

2) Ta có:

\(2\left(x+n\right)\left(x+2\right)-3\left(x-1\right)\left(x^2+1\right)=15\) có nghiệm x = -1

Thay x = -1 vào pt ta được:

\(2\left(-1+n\right)\left(-1+2\right)-3\left(-1-1\right)\left[\left(-1\right)^2+1\right]=15\)

\(\Leftrightarrow\left(-2+2n\right).1+6.2=15\)

\(\Leftrightarrow-2+2n+12=15\)

\(\Leftrightarrow2n+10=15\)

\(\Leftrightarrow n=2,5\)

19 tháng 1 2018

xong r nhé. thanks m.n

12 tháng 2 2020

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

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

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

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

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

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

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

DT
23 tháng 1 2023

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

19 tháng 6 2016

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

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

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

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

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

\(\Leftrightarrow x-1995=0\).Do \(\frac{1}{1990}+\frac{1}{1980}+\frac{1}{1970}-\frac{1}{5}+\frac{1}{15}+\frac{1}{25}\ne0\)

\(\Leftrightarrow x=1995\)

5 tháng 7 2018

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

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

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

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

\(\Leftrightarrow x-1995=0\)

\(\Leftrightarrow x=1995\)

5 tháng 7 2018
\(Giải\): \(\dfrac{x-5}{1990}\)+\(\dfrac{x-15}{1990}\)=\(\dfrac{x-1980}{15}\)+\(\dfrac{x-1990}{5}\) ⇔(\(\dfrac{x-5}{1990}\)- 1) + (\(\dfrac{x-15}{1980}\)- 1) = (\(\dfrac{x-1980}{15}\)-1) +\(\dfrac{x-1990}{5}\) - 1) ⇔ \(\dfrac{x-1995}{1990}\)+\(\dfrac{x-1995}{1980}\)-\(\dfrac{x-1995}{15}\)-\(\dfrac{x-1995}{5}\)= 0 ⇔ (\(x-1995\)) (\(\dfrac{1}{1990}\)+\(\dfrac{1}{1980}\)-\(\dfrac{1}{15}\)-\(\dfrac{1}{5}\)) = 0 ⇔\(x-1995=0\)\(x=1995\)
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 ~