/x/ + 0,75=2
/x+1/3/-4=1
/x/+2,5=0
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3
18 tháng 6 2019
\(a,\left[0,75x+\frac{5}{2}\right]-\frac{4}{7}-\left[-\frac{1}{3}\right]=-\frac{5}{6}\)
\(\Leftrightarrow\left[0,75x+\frac{5}{2}\right]-\frac{4}{7}+\frac{1}{3}=-\frac{5}{6}\)
\(\Leftrightarrow\left[0,75x+\frac{5}{2}\right]=-\frac{5}{6}-\frac{1}{3}+\frac{4}{7}\)
\(\Leftrightarrow0,75x+\frac{5}{2}=-\frac{25}{42}\)
\(\Leftrightarrow0,75x=-\frac{65}{21}\Leftrightarrow x=-\frac{260}{63}\)
18 tháng 6 2019
\(b,\left[4x-9\right]\left[2,5+-\frac{7}{3}x\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x-9=0\\2,5+-\frac{7}{3}x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)
Vậy \(x\in\left\{\frac{9}{4};\frac{15}{14}\right\}\)
Tham khảo nhé phamthiminhanh
7 tháng 2 2021
a) Ta có: \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow x^2=\left(-15\right)\cdot\left(-60\right)=900\)
hay \(x\in\left\{30;-30\right\}\)
Vậy: \(x\in\left\{30;-30\right\}\)
b) Ta có: \(\left|x\right|+0.573=2\)
\(\Leftrightarrow\left|x\right|=1.427\)
hay \(x\in\left\{1.427;-1.427\right\}\)
Vậy: \(x\in\left\{1.427;-1.427\right\}\)
c) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=-1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=3\\x+\dfrac{1}{3}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{8}{3}\\x=-\dfrac{10}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{8}{3};-\dfrac{10}{3}\right\}\)
d) Ta có: \(0.01:2.5=\left(0.75x\right):0.75\)
\(\Leftrightarrow\dfrac{0.75\cdot x}{0.75}=\dfrac{0.01}{2.5}\)
\(\Leftrightarrow x=\dfrac{1}{250}\)
Vậy: \(x=\dfrac{1}{250}\)
HN
0
24 tháng 8 2018
a) 0,01 : 25 = ( 0,75 . x ) : 0,75
\(\Rightarrow\frac{1}{2500}=0,75\cdot x:0,75\)
\(\Rightarrow x=\frac{1}{2500}\)
b) \(\left(3,8\right):\left(2x\right)=\frac{1}{4}:\frac{8}{3}\)
\(\Rightarrow3,8:\left(2x\right)=\frac{3}{32}\)
\(\Rightarrow2x=3,8:\frac{3}{32}=\frac{608}{15}\)
\(\Rightarrow x=\frac{608}{15}:2=\frac{304}{15}\)
17 tháng 10 2020
b) (5/2-3x)=25/9
3x = 5/2-25/9
3x =-5/18
x =-5/18:3
x=-5/54
17 tháng 10 2020
\(e.\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=\left(-2\right)^5\)
\(x-1=-2\)
\(x\) \(=-2+1\)
\(x\) \(=-1\)
Vậy \(x=-1\)
VH
0
a, \(\left|x\right|+0,75=2\Rightarrow\left|x\right|=1,25\Rightarrow x=\pm1,25\)
b, \(\left|x+\dfrac{1}{3}\right|-4=1\Rightarrow\left|x+\dfrac{1}{3}\right|=5\Rightarrow\left|x\right|=\dfrac{14}{3}\Rightarrow x=\pm\dfrac{14}{3}\)
c,
\(\left|x\right|+2,5=0\Rightarrow\left|x\right|=-0,25\) (vô lí)
Vì \(\left|x\right|\ge0\) => x vô nghiệm
a) \(\left|x\right|+0,75=2\)
\(\Leftrightarrow\left|x\right|=2-0,75=1,25\)
\(\Rightarrow\left[{}\begin{matrix}x=-1,25\\x=1,25\end{matrix}\right.\)
b) \(\left|x+\dfrac{1}{3}\right|-4=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=1+4=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=5\\x+\dfrac{1}{3}=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{3}\\x=-\dfrac{16}{3}\end{matrix}\right.\)
c) \(\left|x\right|+2,5=0\)
\(\Leftrightarrow\left|x\right|=0-2,5=-2,5\)
Vì \(\left|x\right|\ge0\) mà \(-2,5< 0\) nên \(x\in\varnothing\)