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a) |2x-3|+x=21
|2x-3|=21-x
\(\Rightarrow\)\(\orbr{\begin{cases}2x-3=21-x\\2x-3=-\left(21-x\right)\end{cases}}\)
TH1: 2x-3=21-x
2x-x=21+3
x=24
TH2: 2x-3=-(21-x)
2x-3 = -21+x
2x-x=-21+3
x=-18
Vậy x \(\varepsilon\){-18;24}
ta có 605x > 0 suy ra x>0
ta có x+1+x+2+x+...+100=605x
100x +5050=605x
505x=5050
x=10
|x-1|+|x-2|+...|x-100|=2500
+ xét trường hợp x\geq 0
\Rightarrow |x-1|+|x-2|+...|x-100|=2500
hay x-1+x-2+.................+x-100=2500
\Rightarrow 100x-5050=2500
\Rightarrow x=755
+ xét trường hợp x<0
\Rightarrow |x-1|+|x-2|+...|x-100|=2500
hay 1-x+2-x+.............+100-x=2500
\Rightarrow 5050-100x=2500
\Rightarrow x=255
\(\frac{x+1}{2015}+\frac{x+2}{2014}=\frac{x+3}{2013}+\frac{x+4}{2012}\)
\(=>\frac{x+1}{2015}+1+\frac{x+2}{2014}+1=\frac{x+3}{2013}+1+\frac{x+4}{2012}+1\)
\(=>\frac{x+2016}{2015}+\frac{x+2016}{2014}=\frac{x+2016}{2013}+\frac{x+2016}{2012}\)
\(=>\left(\frac{x+2016}{2015}+\frac{x+2016}{2014}\right)-\left(\frac{x+2016}{2013}+\frac{x+2016}{2012}\right)=0\)
\(=>\left(x+2016\right).\left[\left(\frac{1}{2015}+\frac{1}{2014}\right)-\left(\frac{1}{2013}+\frac{1}{2012}\right)\right]=0\)
\(=>\orbr{\begin{cases}x+2016=0\\\left(\frac{1}{2015}+\frac{1}{2014}\right)-\left(\frac{1}{2013}+\frac{1}{2012}\right)=0\end{cases}}\)
Do 1/2015 + 1/2014 < 1/2013 + 1/2012
=> (1/2015 + 1/2014) - (1/2013 + 1/2012) khác 0
=> x - 2016 = 0
=> x = 2016
Vậy x = 2016
Ủng hộ mk nha ^_-
Ta có:
\(\left(\frac{1}{5}+\frac{1}{3}+\frac{3}{10}\right)+-\frac{1}{2}=\frac{1}{5}+\frac{1}{3}+\frac{3}{10}\)\(-\frac{1}{2}\)
=\(\frac{6}{30}+\frac{10}{30}+\frac{9}{30}-\frac{15}{30}=\frac{6+10+9-15}{30}=\frac{10}{30}=\frac{1}{3}\)
Đề:........
<=> (24)x < (27)4
<=> 24x < 228
<=> 4x < 28
<=> x < 7
Vậy x = {0; 1; 2; 3; 4; 5; 6}
Theo đề bài, ta có:
\(\frac{2}{3}x^2-2=\frac{2}{3}\)
\(\frac{2}{3}x^2=\frac{2}{3}+2\)
\(\frac{2}{3}x^2=\frac{8}{3}\)
\(x^2=\frac{8}{3}\div\frac{2}{3}\)
\(x^2=4\)
\(x=\text{±}2\)