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a) \(\left|1,5x\right|-\left|0,6\right|=\left(-2\right)+\left|0,4\right|\)
\(\left|1,5x\right|-0,6=\left(-2\right)+0,4\)
\(\left|1,5x\right|-0,6=-1,6\)
\(\left|1,5x\right|=-1,6+0,6\)
\(\left|1,5x\right|=-1\)
\(\Rightarrow\) x không tồn tại
b) \(\left|3x+1\right|=2-\left|-\frac{4}{5}\right|\)
\(\left|3x+1\right|=2-\frac{4}{5}\)
\(\left|3x+1\right|=\frac{6}{5}\)
\(\Rightarrow3x+1\in\left\{\frac{6}{5};-\frac{6}{5}\right\}\)
\(\Rightarrow3x\in\left\{\frac{1}{5};\frac{-11}{5}\right\}\)
\(\Rightarrow x\in\left\{\frac{1}{15};\frac{-11}{15}\right\}\)
c)\(\left|1-2x\right|+4=\left|-10\frac{1}{2}\right|-\frac{1}{2}\)
\(\left|1-2x\right|+4=10\frac{1}{2}-\frac{1}{2}\)
\(\left|1-2x\right|+4=10\)
\(\left|1-2x\right|=10-4\)
\(\left|1-2x\right|=6\)
\(\Rightarrow1-2x\in\left\{6;-6\right\}\)
\(\Rightarrow2x\in\left\{-5;7\right\}\)
\(\Rightarrow2x\in\left\{-\frac{5}{2};\frac{7}{2}\right\}\)
d)\(\left|x+\frac{7}{3}\right|-\left|\frac{1}{3}\right|=0\)
\(\left|x+\frac{7}{3}\right|-\frac{1}{3}=0\)
\(\Rightarrow\left|x+\frac{7}{3}\right|=\frac{1}{3}\)
\(\Rightarrow x+\frac{7}{3}\in\left\{\frac{1}{3};\frac{-1}{3}\right\}\)
\(\Rightarrow x\in\left\{-2;\frac{-8}{3}\right\}\)
|1,5x|-|0,6|=(-2)+|0,4|
|1,5x|-0=(-2)+0,4
|1,5x|=-2+(0,4+0,6)
|1,5x|=-2+1=-1
mà GTTT của 1 số luôn luôn là số nguyên dương =>x\(\in\phi\)
b)|3x+1|=2-|4/5|
|3x+1|=2-4/5
|3x+1|=6/5
=>3x+1=6/5 hay 3x+1=-6/5
3x=6/5-1 hay 3x=-6/5-1
3x=1/5 hay 3x=-11/5
x=1/5:3 hay x=-11/5:3
x=1/15 hay x=-11/15
c)|1-2x|+4=|-10/1/2|-1/2
|1-2x|=-21/2-1/2-4
|1-2x|=-15
=>1-2x=15 hay 1-2x=-15
2x=1-15 hay 2x=1-(-15)
2x=-14 hay 2x=16
x=14/2 hay x=16/2
x=7 hay x=8
d)|x+7/3|-|1/3|=0
|x+7/3|-1/3=0
|x+7/3|=0+1/3
|x+7/3|=1/3
=>x+7/3=1/3 hay x+7/3=-1/3
x=1/3-7/3 hay x=-1/3-7/3
x=-6/3=-2 hay x=-8/3
Bài giải
\(\left|\sqrt{x+1}-0,5\right|-0,6=\sqrt{\left(-3\right)^2}+0,4\)
\(\left|\sqrt{x+1}-0,5\right|-0,6=3+0,4\)
\(\left|\sqrt{x+1}-0,5\right|-0,6=3,4\)
\(\left|\sqrt{x+1}-0,5\right|=3,4+0,6\)
\(\left|\sqrt{x+1}-0,5\right|=4\)
\(\Rightarrow\orbr{\begin{cases}\sqrt{x+1}-0,5=-4\\\sqrt{x+1}-0,5=4\end{cases}\Rightarrow}\orbr{\begin{cases}\sqrt{x+1}=-3,5\text{ ( loại ) }\\\sqrt{x+1}=4,5\end{cases}}\Rightarrow\text{ }x+1=20,25\text{ }\Rightarrow\text{ }x=19,25\)
\(\Rightarrow\text{ }x=19,25\)
Ta có: \(|\sqrt{x+1}-0,5|=4\)\(\left(ĐK:x\ge-1\right)\)
<=> \(\orbr{\begin{cases}\sqrt{x+1}-0,5=4\\\sqrt{x+1}-0,5=-4\end{cases}}\)
<=> \(\orbr{\begin{cases}x=19,25\\x\in\varnothing\end{cases}}\)
Ta có:
\(0,4\left(3\right)=\frac{43-4}{90}=\frac{39}{90}=\frac{13}{30}.\)
\(0,6\left(2\right)=\frac{62-6}{90}=\frac{56}{90}=\frac{28}{45}.\)
\(0,6\left(8\right)=\frac{68-6}{90}=\frac{62}{90}=\frac{31}{45}.\)
Vậy:
\(\frac{13}{30}+\frac{28}{45}.\frac{5}{2}-\frac{\frac{5}{6}}{\frac{31}{45}}.\frac{53}{50}\)
\(=\frac{13}{30}+\frac{14}{9}-\frac{75}{62}.\frac{53}{50}\)
\(=\frac{13}{30}+\frac{14}{9}-\frac{159}{124}\)
\(=\frac{179}{90}-\frac{159}{124}\)
\(=\frac{3943}{5580}.\)
Chúc bạn học tốt!
a: =>|3/2x|=-2+0,4+0,6=-1(vô lý)
b: =>|x+7/3|=1/3
=>x+7/3=1/3 hoặc x+7/3=-1/3
=>x=-2 hoặc x=-8/3
\(0,4\left(3\right)=\frac{43-4}{90}=\frac{39}{90}=\frac{13}{30}\)đó Michiel Girl Mít ướt
Đó là công thức đưa 1 số thập phân vô hạn tuần hoàn sang phân số đó Michiel Girl Mít ướt
a)\(\left( { - 0,4} \right) + \frac{3}{8} + \left( { - 0,6} \right) = \left[ {\left( { - 0,4} \right) + \left( { - 0,6} \right)} \right] + \frac{3}{8} = - 1 + \frac{3}{8} = \frac{{ - 5}}{8}\).
b)
\(\frac{4}{5} - 1,8 + 0,375 + \frac{5}{8} = (0,8 - 1,8) + (0,375 + 0,625) = ( - 1) + 1 = 0\)
Ta có: \(0,4\left(3\right)=\dfrac{43-4}{90}=\dfrac{39}{90};\) \(0,6\left(2\right)=\dfrac{62-6}{90}=\dfrac{56}{90}\)
\(0,5\left(8\right)=\dfrac{58-5}{90}=\dfrac{53}{90}\)
Vậy biểu thức M có giá trị:
\(\dfrac{39}{90}+\dfrac{56}{90}.\dfrac{5}{2}-\dfrac{\dfrac{5}{6}}{\dfrac{53}{90}}.\dfrac{53}{30}=\dfrac{13}{30}+\dfrac{14}{9}-\dfrac{5}{6}.\dfrac{90}{53}.\dfrac{53}{50}=\dfrac{13}{30}+\dfrac{14}{9}-\dfrac{3}{2}\)
\(=\dfrac{13.9+14.30-3.135}{270}=\dfrac{402}{270}=\dfrac{67}{45}\)
a: \(=\left\{\left[\left(20-\dfrac{1}{4}\right)\cdot0.2\right]+\dfrac{3}{20}\right\}\cdot5:\left[\left(2+\dfrac{25}{11}\cdot\dfrac{22}{100}\cdot10\right)\cdot\dfrac{1}{33}\right]\)
\(=\left\{\left[\dfrac{79}{20}+\dfrac{3}{20}\right]\right\}\cdot5:\left[\dfrac{356}{55}\cdot\dfrac{1}{33}\right]\)
\(=\dfrac{82}{20}\cdot5:\dfrac{3856}{1815}\simeq104,516\)
b: \(=\dfrac{13}{30}+\dfrac{28}{45}\cdot\dfrac{5}{2}\cdot\left[\dfrac{5}{6}:\dfrac{53}{90}\right]\cdot\dfrac{53}{50}\)
\(=\dfrac{13}{30}+\dfrac{14}{9}\cdot\dfrac{3}{2}=\dfrac{83}{30}\)
\(|1,5x|-0,6=0,4-2\) \(\Leftrightarrow|1,5x|=-1\left(ktm\right)\)