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22 tháng 6 2022
b: \(\Leftrightarrow\left(\dfrac{29-x}{21}+1\right)+\left(\dfrac{27-x}{23}+1\right)+\left(\dfrac{25-x}{25}+1\right)+\left(\dfrac{23-x}{27}+1\right)+\left(\dfrac{21-x}{29}+1\right)=0\)
=>50-x=0
hay x=50
c: \(\Leftrightarrow\dfrac{x-2}{2001}+1=\dfrac{x-1}{2002}+\dfrac{x}{2003}\)
\(\Leftrightarrow\left(\dfrac{x-2}{2001}-1\right)=\left(\dfrac{x-1}{2002}-1\right)+\left(\dfrac{x}{2003}-1\right)\)
=>x-2003=0
hay x=2003
17 tháng 3 2019
Ta có: \(\frac{x-29}{1970}+\frac{x-27}{1972}+\frac{x-25}{1974}+\frac{x-23}{1976}+\frac{x-21}{1978}+\frac{x-19}{1980}\)\(=\frac{x-1970}{29}+\frac{x-1972}{27}+\frac{x-1974}{25}+\frac{x-1976}{23}+\frac{x-1978}{21}+\frac{x-1980}{19}\)
\(\Leftrightarrow\left(\frac{x-29}{1970}-1\right)+\left(\frac{x-27}{1972}-1\right)+\left(\frac{x-25}{1974}-1\right)+\left(\frac{x-23}{1976}-1\right)+\left(\frac{x-21}{1978}-1\right)+\left(\frac{x-19}{1980}-1\right)\)\(=\left(\frac{x-1970}{29}-1\right)+\left(\frac{x-1972}{27}-1\right)+\left(\frac{x-1974}{25}-1\right)+\left(\frac{x-1976}{23}-1\right)+\left(\frac{x-1978}{21}-1\right)+\left(\frac{x-1980}{19}-1\right)\)
\(\Leftrightarrow\frac{x-1999}{1970}+\frac{x-1999}{1972}+\frac{x-1999}{1974}+\frac{x-1999}{1976}+\frac{x-1999}{1978}+\frac{x-1999}{1980}\)\(=\frac{x-1999}{29}+\frac{x-1999}{27}+\frac{x-1999}{25}+\frac{x-1999}{24}+\frac{x-1999}{21}+\frac{x-1999}{19}\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}\right)\)\(=\left(x-1999\right)\left(\frac{1}{29}+\frac{1}{27}+\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)\)
\(\Leftrightarrow\left(x-1999\right)\left(\frac{1}{1970}+\frac{1}{1972}+\frac{1}{1974}+\frac{1}{1976}+\frac{1}{1978}+\frac{1}{1980}-\frac{1}{29}-\frac{1}{27}-\frac{1}{25}-\frac{1}{23}-\frac{1}{21}-\frac{1}{19}\right)=0\)\(\Leftrightarrow\) \(x-1999=0\) (Vì ...khác 0)
\(\Leftrightarrow x=1999\)(thỏa mãn)
Vậy \(x=1999\)
L
3
26 tháng 7 2017
148-x/25-1 + 169-x/23-2 + 186-x/21-3 + 199-x/19-4
123-x/25 + 123-x/23 + 123-x/21 + 123-x/19 =0
123-x=0 => x=123
18 tháng 2 2020
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)
\(\left(\frac{148-x}{25}-1\right)+\left(\frac{169-x}{23}-2\right)+\left(\frac{186-x}{21}-3\right)+\left(\frac{199-x}{19}-4\right)=0\)
=> \(\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\)
=> \(\left(123-x\right)\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
=> 123 - x = 0
=> x = 123
5 tháng 1 2018
a) \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}-1+\dfrac{x-4}{101}-1+\dfrac{x-3}{102}-1=\dfrac{x-100}{5}-1+\dfrac{x-101}{4}-1+\dfrac{x-102}{3}-1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow\left(x-105\right)=0;\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)\ne0\)
\(\Leftrightarrow x=105\)
b) \(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)=0\)
\(\Leftrightarrow50-x=0;\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)\ne0\)
\(\Leftrightarrow x=50\)
19 tháng 1 2019
\(a.\dfrac{x-2}{2000}+\dfrac{x-3}{1999}=\dfrac{x-4}{1998}+\dfrac{x-5}{1997}\\ \Leftrightarrow\dfrac{x-2}{2000}-1+\dfrac{x-3}{1999}-1=\dfrac{x-4}{1998}-1+\dfrac{x-5}{1997}-1\\ \Leftrightarrow\dfrac{x-2}{2000}-\dfrac{2000}{2000}+\dfrac{x-3}{1999}-\dfrac{1999}{1999}=\dfrac{x-4}{1998}-\dfrac{1998}{1998}+\dfrac{x-5}{1997}-\dfrac{1997}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}=\dfrac{x-2002}{1998}+\dfrac{x-2002}{1997}\\ \Leftrightarrow\dfrac{x-2002}{2000}+\dfrac{x-2002}{1999}-\dfrac{x-2002}{1998}-\dfrac{x-2002}{1997}=0\\ \Leftrightarrow\left(x-2002\right)\left(\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\right)=0\\ \)
\(Do:\dfrac{1}{2000}+\dfrac{1}{1999}-\dfrac{1}{1998}-\dfrac{1}{1997}\ne0\\ \Rightarrow x-2002=0\\ \Leftrightarrow x=2002\\ Vậy:S=\left\{2002\right\}\)
Mấy câu khác tương tự :v
2 tháng 1 2023
b: \(\Leftrightarrow\left(\dfrac{148-x}{25}-1\right)+\left(\dfrac{169-x}{23}-2\right)+\left(\dfrac{186-x}{21}-3\right)+\left(\dfrac{199-x}{19}-4\right)=0\)
=>123-x=0
=>x=123
c: \(\Leftrightarrow\dfrac{x-2}{2017}+1=\dfrac{x-1}{2018}+\dfrac{x}{2019}\)
\(\Leftrightarrow\left(\dfrac{x-2}{2017}-1\right)=\left(\dfrac{x-1}{2018}-1\right)+\left(\dfrac{x}{2019}-1\right)\)
=>x-2019=0
=>x=2019
H
19 tháng 1 2019
a, \(\Leftrightarrow3x^2-3+5=3x^2+2x-3x-2\)
\(\Leftrightarrow3x^2-3x-2x+3x=-2+3-5\)
<=>x=-4
b, \(\Leftrightarrow\dfrac{x+4}{5}-\dfrac{5x}{5}+\dfrac{20}{5}=\dfrac{2x}{6}-\dfrac{3\left(x-2\right)}{6}\)
\(\Leftrightarrow\dfrac{x+4-5x+20}{5}=\dfrac{2x-3x+6}{6}\)
\(\Leftrightarrow\dfrac{6\left(-4x+24\right)}{30}=\dfrac{5\left(-x+6\right)}{30}\)
<=>-24x+144=-5x+30
<=>-5x+24x=144-30
<=>19x=114
<=>x=6
16 tháng 6 2022
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
17 tháng 6 2022
d: \(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8=12x^2-12x-8\)
\(\Leftrightarrow12x^2+16=12x^2-12x-8\)
=>-12x=24
hay x=-2
e: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(x-5\right)^2\)
\(\Leftrightarrow x^2+7x+10-12x+9=x^2-10x+25\)
=>-5x+19=-10x+25
=>5x=6
hay x=6/5
f: \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
=>x-105=0
hay x=105
2 tháng 7 2022
c: =>|x-2|+3=-5 hoặc |x-2|+3=5
=>|x-2|=2
=>x-2=2 hoặc x-2=-2
=>x=4 hoặc x=0