x-97/3 + x-92/4 + x-85/5 =6
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
TG
0
20 tháng 2 2023
\(\Leftrightarrow\left(\dfrac{x-11}{111}+1\right)+\left(\dfrac{x-12}{112}+1\right)=\left(\dfrac{x-23}{123}+1\right)+\left(\dfrac{x-24}{124}+1\right)\)
=>x+100=0
=>x=-100
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 4 2020
\(\Leftrightarrow\frac{x+8}{92}+1+\frac{x+7}{93}+1+\frac{x+6}{94}+1\ge\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Leftrightarrow\frac{x+100}{92}+\frac{x+100}{93}+\frac{x+100}{94}\ge\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{92}-\frac{1}{98}+\frac{1}{93}-\frac{1}{97}+\frac{1}{94}-\frac{1}{96}\right)\ge0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{6}{92.98}+\frac{4}{93.97}+\frac{2}{94.96}\right)\ge0\)
\(\Leftrightarrow x+100\ge0\Rightarrow x\ge-100\)
GL
2
TT
8 tháng 2 2021
\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
=> x = - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
8 tháng 2 2021
Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
6 tháng 10 2016
a) 1 - 2 + 3 - 4 + 5 - 6 + ... + 97 - 98 + 99 - 100 + 101 (có 101 số; 101 chia 2 dư 1)
= (1 - 2) + (3 - 4) + (5 - 6) + ... + (97 - 98) + (99 - 100) + 101 (có 50 nhóm và dư 1 số)
= -1 + (-1) + (-1) + ... + (-1) + (-1) + 101
= -1.50 + 101
= -50 + 101
= 51
2 câu c` lại cx chia thành nhóm r` lm tương tự
HB
4 tháng 2 2019
1 - 2 + 3 - 4 + ........ + 99 - 100 + 101
= ( 1 - 2) + ( 3 - 4 ) + ..... + ( 99 -100 ) + 101 ( 50 nhóm )
= - 1 + ( - 1 ) + ...... + ( -1) + 101 ( 50 số -1 )
= - 1 . 50 + 101
= 50 + 101
= 151
TT
0
24 tháng 1 2018
Ta có\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)
<=> \(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+9}{91}+1\right)=\left(\frac{x+91}{9}+1\right)+\left(\frac{x+92}{8}+1\right)+\left(\frac{x+61}{39}+1\right)\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}=\frac{x+100}{9}+\frac{x+100}{8}+\frac{x+100}{39}\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}-\frac{x+100}{9}-\frac{x+100}{8}-\frac{x+100}{39}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\right)=0\)
Do \(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\ne0\)
Nên x+100=0 => x=-100
NH
1
KN
4 tháng 3 2020
\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(\Rightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=0\)
\(\Rightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
Dễ thấy \(\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)>0\)nên x + 2004 = 0
Vậy x = -2004
\(\frac{x-97}{3}+\frac{x-92}{4}+\frac{x-85}{5}=6\)
\(\Rightarrow\frac{20x-1940+15x-1380+12x-1020}{60}=6\)
\(\Rightarrow47x-4310=360\)
\(\Rightarrow47x=4670\)
\(x=\frac{4670}{47}\)