Tìm x :
392-x/32 +390-x/34+388-x/36+386-x/38+384-x/40 = -5
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
PB
1
MN
22 tháng 1 2020
Ta có :
\(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
\(\Leftrightarrow\left(\frac{392-x}{32}+1\right)+\left(\frac{390-x}{34}+1\right)+\left(\frac{388-x}{36}+1\right)+\left(\frac{386-x}{38}+1\right)+\left(\frac{384-x}{40}\right)=0\)
\(\Leftrightarrow\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
\(\Leftrightarrow\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
Mà : \(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\ne0\)
\(\Leftrightarrow424-x=0\)
\(\Leftrightarrow x=424\)
Vậy x = 424
30 tháng 4 2017
ta có : \(\dfrac{392-x}{32}+\dfrac{390-x}{34}+\dfrac{388-x}{36}+\dfrac{386-x}{38}\)+\(\dfrac{384-x}{40}=-5\)
\(\Leftrightarrow\)\(\dfrac{392-x}{32}+1+\dfrac{390-x}{34}+1+\dfrac{388-x}{36}+1\)+\(\dfrac{384-x}{40}+1=0\)
\(\Leftrightarrow\)\(\dfrac{424-x}{32}+\dfrac{424-x}{34}+\dfrac{424-x}{36}+\dfrac{424-x}{38}+\dfrac{424-x}{40}=0\)\(\Leftrightarrow\left(424-x\right)\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\right)=0\)
\(\Leftrightarrow x=424\)(vì \(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\ne0\))
Vậy tập nghiệm của phương trình là s=\(\left\{424\right\}\)
24 tháng 2 2020
a, Ta có : \(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
=> \(\frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}+1=-5+5=0\)
=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{38}+\frac{424-x}{40}=0\)
=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{38}+\frac{1}{40}\right)=0\)
=> \(424-x=0\)
=> \(x=424\)
Vậy phương trình có nghiệm là x = 424 .
b, Ta có : \(\frac{x+1}{2014}+\frac{x+3}{2012}=\frac{x+5}{2010}+\frac{x+6}{2009}\)
=> \(\frac{x+1}{2014}+1+\frac{x+3}{2012}+1=\frac{x+5}{2010}+1+\frac{x+6}{2009}+1\)
=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}=\frac{x+2015}{2010}+\frac{x+2015}{2009}\)
=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}-\frac{x+2015}{2010}-\frac{x+2015}{2009}=0\)
=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
=> \(x+2015=0\)
=> \(x=-2015\)
Vậy phương trình có nghiệm là x = -2015 .
24 tháng 2 2020
a) \(\frac{392-x}{32}+\frac{390-x}{34}+\frac{388-x}{36}+\frac{386-x}{38}+\frac{384-x}{40}=-5\)
<=> \(\frac{392-x}{32}+1+\frac{390-x}{34}+1+\frac{388-x}{36}+1+\frac{386-x}{38}+1+\frac{384-x}{40}=0\)
<=> \(\frac{424-x}{32}+\frac{424-x}{34}+\frac{424-x}{36}+\frac{424-x}{40}=0\)
<=> \(\left(424-x\right)\left(\frac{1}{32}+\frac{1}{34}+\frac{1}{36}+\frac{1}{40}\right)=0\)
<=> 424 - x = 0
<=> x = 424
Vậy S = {424}
b) \(\frac{x+1}{2014}+\frac{x+3}{2012}=\frac{x+5}{2010}+\frac{x+6}{2009}\)
<=> \(\left(\frac{x+1}{2014}+1\right)+\left(\frac{x+3}{2012}+1\right)=\left(\frac{x+5}{2010}+1\right)+\left(\frac{x+6}{2009}+1\right)\)
<=> \(\frac{x+2015}{2014}+\frac{x+2015}{2012}=\frac{x+2015}{2010}+\frac{x+2015}{2009}\)
<=> \(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2012}-\frac{1}{2010}-\frac{1}{2009}\right)=0\)
<=> x + 2015 = 0
<=> x= -2015
Vậy S = {-2015}
6 tháng 2 2023
Các bước giải chi tiết
a) \(\dfrac{392-x}{32}\) + \(\dfrac{390-x}{34}\) + \(\dfrac{388-x}{36}\) = -3
⇔ \(\dfrac{392-x}{32}\)+1+\(\dfrac{390-x}{34}\)+1+\(\dfrac{388-x}{36}\)+1 = 0
⇔\(\dfrac{424-x}{32}\)+\(\dfrac{424-x}{34}\)+\(\dfrac{424-x}{36}\)=0
⇔\(\left(424-x\right)\)\(\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}\right)\)=0
⇔\(424-x\) = 0\(\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}\ne\forall x\right)\)
⇔\(x=424\)
b) \(\dfrac{x-3}{3}\)- \(x\) = \(5-\dfrac{x+1}{4}\)
⇔\(\dfrac{x-3-3x}{3}\) = 5 + \(\dfrac{-\left(x+1\right)}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = 5 + \(\dfrac{-x-1}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = \(\dfrac{20-x-1}{4}\)
⇔\(\dfrac{-2x-3}{3}\) = \(\dfrac{-x+19}{4}\)
⇔ \(4\left(-2x-3\right)\) = \(3\left(-x+19\right)\)
⇔\(-8x-12\) = \(-3x+57\)
⇔\(-8x\) = \(-3x+69\)
⇔\(-5x=69\)
⇔ \(x=-\dfrac{69}{5}\)
6 tháng 2 2023
(392-x)/32+(390-x)/34+(388-x)/36=-3
=>392-x)/32 +1 + (390-x)/34 +1 +(388-x)/36 +1=0
=>(424-x)/32+(424-x)/34+(424-x)/36=0
=>424-x=0(vì 1/32+1/34+1/36 khác 0)
=>x=424
27 tháng 12 2016
Câu x ) là bằng - 5 nhé mấy bạn. Làm giúp mình tất cả nhé ! Mình cảm ơn nhiều lắm !
NK
4
12 tháng 2 2020
(x-99) (1/30 + 1/32 + 1/34 - 1/36 - 1/38) = 0
SUy ra x - 99 = 0
VẬy x =99
\(\dfrac{392-x}{32}+\dfrac{390-x}{34}+\dfrac{388-x}{36}+\dfrac{386-x}{38}+\dfrac{384-x}{40}=-5\\ =>\left(\dfrac{392-x}{32}+1\right)+\left(\dfrac{390-x}{34}+1\right)+\left(\dfrac{388-x}{36}+1\right)+\left(\dfrac{386-x}{38}+1\right)+\left(\dfrac{384-x}{40}+1\right)=-5+5\\ =>\dfrac{424-x}{32}+\dfrac{424-x}{34}+\dfrac{424-x}{36}+\dfrac{424-x}{38}+\dfrac{424-x}{40}=0\\ =>\left(424-x\right)\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\right)=0\\ =>424-x=0\\ =>x=424\)
\(\dfrac{392-x}{32}+\dfrac{390-x}{34}+\dfrac{388-x}{36}+\dfrac{386-x}{38}+\dfrac{384-x}{40}=-5\\\Rightarrow\dfrac{392-x}{32}+1+\dfrac{390-x}{34}+1+\dfrac{388-x}{36}+1+\dfrac{386-x}{38}+1+\dfrac{384-x}{40}+1=0\\ \Rightarrow\dfrac{424-x}{32}+\dfrac{424-x}{34}+\dfrac{424-x}{36}+\dfrac{424-x}{38}+\dfrac{424-x}{40}=0\\ \Rightarrow\left(424-x\right)\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\right)=0\\ \Rightarrow424-x=0\\ \Rightarrow x=424\)