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![](https://rs.olm.vn/images/avt/0.png?1311)
\(-1+\left|x+4,5\right|=-6,2.\)
\(\left|x+4,5\right|=-6,2+1\)
\(\left|x+4,5\right|=-5,2\)( loại ).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-1+\left|x+4,5\right|=-6,2\)
\(\Leftrightarrow\left|x+4,5\right|=-5,2\)
Vì dấu GTTT không mang giá trị âm -> vô nghiệm
\(\Rightarrow x\in\varnothing\)
\(-1+\left|x+4.5\right|=-6.2\)
\(-1+\left|x+20\right|=-12\)
\(\left|x+20\right|=-12-\left(-1\right)\)
\(\left|x+20\right|=-11\)
\(\Rightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giá trị tuyệt đối của 1 số luôn lớn hơn 0
=>/2Y+7,4/+6,2+/-X+2,1/\(\ge\) 0+6,2+0=6,2
GTNN của A là 6,2 khi 2Y+7,4=0 ( Y=3,7) và /-X+2,1/=0 ( X=2,1)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{3.4}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{3}-\frac{1}{x+1}\)
\(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x\left(x+1\right)}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+.....+\frac{1}{x}-\frac{1}{x+1}\)
\(=\frac{1}{3}-\frac{1}{x+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1/3.4+1/4.5+1/5.6+1/6.7+....+1/x(x+1)=3/10
<=> \(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{\left(x+1\right)x}=\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
<=> \(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}=\frac{1}{30}\)=> x+1=30=>x=29
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{3}-\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{30}\)
\(\Rightarrow x+1=30\)
\(x=30-1=29\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a/|x|-2,5=27,5
=>|x|=27,5+2,5=30
=>x=30 hoặc x=-30
b/\(\dfrac{3}{4}+\dfrac{2}{5}.x=\dfrac{29}{60}\)
=>\(\dfrac{2}{5}.x\)=\(\dfrac{29}{60}-\dfrac{3}{4}\)=\(\dfrac{-4}{15}\)
=>x=\(\dfrac{-4}{15}:\dfrac{2}{5}\)=\(\dfrac{-2}{3}\)
c/(x-1)\(^5\)=-32
=>x-1=-2 vì (-2)\(^5\)=-32
=>x=-2+1=-1
d/\(\dfrac{4}{5}.x+0,5=4.5\)
=>\(\dfrac{4}{5}.x+0,5=20\)
=>\(\dfrac{4}{5}.x=20-0,5=19,5\)
=>\(x=19,5:\dfrac{4}{5}\)=\(\dfrac{195}{8}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|x-3,5\right|+\left|4,5-x\right|=0\)
\(\Rightarrow\orbr{\begin{cases}x-3,5=0\\4,5-x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3,5\\x=4,5\end{cases}}\)
Vậy, ...
~ Học tốt ~
|x-3.5|+|4.5-x|=0
\(\orbr{\begin{cases}x-3.5=0\\4.5-x=0\end{cases}}\)
\(\orbr{\begin{cases}x=3.5\\x=4.5\end{cases}}\)
học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
-1+|x+4.5|=-6.2
-1+|x+4.5|=-6.2=-12
|x+4.5|=-12-(-1)=13
=>x+4.5=13=>13:5=2,6=>2,6-4=-2,6
x+4.5=-13=>-13:5=-2,6=>-2,6-4=-6,6
Vậy x \(\in\)\([\)-2,6;-6,6\(]\)