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x + 1/100 + x + 2/101 = x + 3/102 - 1
<=> x + 1/100 - 1 + x + 2/101 - 1 = x + 3/102 - 1 - 2
<=> x - 99/100 + x - 99/101 = x - 99/102 - 2
<=> x - 99/100 + x - 99/101 - x - 99/102 = -2
<=> (x - 99)(1/100 + 1/101 - 1/102) = -2
<=> x - 99 = -2/1/100 + 1/101 - 1/102
<=> x = -2/1/100 + 1/101 - 1/102 + 99
Bạn chịu khó bấm máy hộ mình, số to quá
\(< =>\left(\dfrac{x-5}{100}-1\right)+\left(\dfrac{x-4}{101}-1\right)+\left(\dfrac{x-3}{102}-1\right)+3=\left(\dfrac{x-100}{5}-1\right)+\left(\dfrac{x-101}{4}-1\right)+\left(\dfrac{x-102}{3}-1\right)+3\)\(< =>\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}\)
\(< =>\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)\) = 0
<=> x - 105 = 0
<=> x = 105
Vậy tập nghiệm của phương trình là S = \(\left\{105\right\}\)
\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)
\(\Leftrightarrow\dfrac{x-5}{100}-1+\dfrac{x-4}{101}-1+\dfrac{x-3}{102}-1=\dfrac{x-100}{5}-1+\dfrac{x-101}{4}-1+\dfrac{x-102}{3}-1\)
\(\Leftrightarrow\dfrac{x-5-100}{100}+\dfrac{x-4-101}{101}+\dfrac{x-3-102}{102}-\dfrac{x-100-5}{5}-\dfrac{x-101-4}{4}-\dfrac{x-102-3}{3}=0\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)=0\)
\(\Leftrightarrow x-105=0\) ( vì \(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\ne0\) )
\(\Leftrightarrow x=105\)
Vậy tập nghiệm của pt là S ={ 105 }
a, \(\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow\dfrac{2-x}{2001}-1+2=\dfrac{1-x}{2002}-\dfrac{x}{2003}+2\)
\(\Leftrightarrow\dfrac{2-x}{2001}+1=\left(\dfrac{1-x}{2002}+1\right)+\left(\dfrac{-x}{2003}+1\right)\)
\(\Leftrightarrow\dfrac{2003-x}{2001}=\dfrac{2003-x}{2002}+\dfrac{2003-x}{2003}\)
\(\Leftrightarrow\dfrac{2003-x}{2001}-\dfrac{2003-x}{2002}-\dfrac{2003-x}{2003}=0\)
\(\Leftrightarrow\left(2003-x\right)\left(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Vì \(\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)
\(\Rightarrow2003-x=0\)
\(\Rightarrow x=2003\)
Vậy : \(s=\left\{2003\right\}\)
b, \(\dfrac{x-5}{100}+\dfrac{x-4}{101}=\dfrac{x-100}{5}+\dfrac{x-101}{4}\)
\(\Leftrightarrow\dfrac{x-5}{100}+\dfrac{x-4}{101}-2=\dfrac{x-100}{5}+\dfrac{x-101}{4}-2\)
\(\Leftrightarrow\left(\dfrac{x-5}{100}-1\right)+\left(\dfrac{x-4}{101}-1\right)=\left(\dfrac{x-100}{5}-1\right)+\left(\dfrac{x-101}{4}-1\right)\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}=\dfrac{x-105}{5}+\dfrac{x-105}{4}\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}-\dfrac{x-105}{5}-\dfrac{x-105}{4}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)
Vì \(\dfrac{1}{100}+\dfrac{1}{101}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)
\(\Rightarrow x-105=0\)
\(\Rightarrow x=105\)
Vậy : \(s=\left\{105\right\}\)
\(a,\dfrac{2-x}{2001}-1=\dfrac{1-x}{2002}-\dfrac{x}{2003}\)
\(\Leftrightarrow\)haizzz bạn cộng mỗi hạng tử ở mỗi vế cho một. Chuyển vế và giải ra x=2003
b, Tương tự bạn -1 cho mỗi vế. GIải phương trình đc x=105
\(\dfrac{x-5}{100}+\dfrac{x-4}{101}=\dfrac{x-100}{5}+\dfrac{x-101}{4}\)
\(\Leftrightarrow\dfrac{x-5}{100}-1+\dfrac{x-4}{101}-1=\dfrac{x-100}{5}-1+\dfrac{x-101}{4}-1\)
\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}-\dfrac{x-105}{5}-\dfrac{x-105}{4}=0\)
\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\Leftrightarrow x=105\)
Câu 2:
\(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
Có \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge0\)
do đó phương trình ban đầu tương đương với:
\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)
\(\Leftrightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)
\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)
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) \(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)
\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)
\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)
\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)=0\)
\(\Leftrightarrow50-x=0;\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)\ne0\)
\(\Leftrightarrow x=50\)
a: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)
=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)
=>2x-2018<0
=>x<2019
b: \(\Leftrightarrow\left(\dfrac{3-x}{100}+\dfrac{4-x}{101}\right)>\dfrac{5-x}{102}+\dfrac{6-x}{103}\)
=>\(\dfrac{x-3}{100}+\dfrac{x-4}{101}-\dfrac{x-5}{102}-\dfrac{x-6}{103}< 0\)
=>\(x+97< 0\)
=>x<-97
Vì \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|>0\forall x\)
mà \(\left|x+\frac{1}{101}\right|+\left|x+\frac{2}{101}\right|+\left|x+\frac{3}{101}\right|+...+\left|x+\frac{100}{101}\right|=101x\)
nên x>0
Với x>0, ta được:
\(x+\frac{1}{101}+x+\frac{2}{101}+x+\frac{3}{101}+...+x+\frac{100}{101}=101x\)
\(\Leftrightarrow100x-101x+\frac{5050}{101}=0\)
\(\Leftrightarrow-x+50=0\)
hay x=50
Vậy: S={50}