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\(\dfrac{3}{2}\)(\(x\) - \(\dfrac{5}{3}\)) - \(\dfrac{4}{5}\) = \(x\) + 1
\(\dfrac{3}{2}\) \(x\) - \(\dfrac{15}{6}\) - \(\dfrac{4}{5}\) = \(x\) + 1
\(\dfrac{3}{2}\)\(x\) - \(x\) = 1 + \(\dfrac{15}{6}\) + \(\dfrac{4}{5}\)
\(\dfrac{1}{2}\)\(x\) =\(\dfrac{43}{10}\)
\(x\) = \(\dfrac{43}{10}\) \(\times\) 2
\(x\) = \(\dfrac{43}{5}\)
\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3.\left(x-\dfrac{5}{3}\right)}{2}-\dfrac{4}{5}=x+1\\ \Rightarrow\dfrac{3x-5}{2}-\dfrac{4}{5}=x+1\Rightarrow\dfrac{5\left(3x-5\right)}{10}-\dfrac{8}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1\\ \Rightarrow\dfrac{15x-33}{10}-x=x+1\\ \Rightarrow\dfrac{15x-33}{10}=x+1-x\\ \Rightarrow5x-33=10\\ \Rightarrow5x=10+33\\\Rightarrow5x=43\\ \Rightarrow x=\dfrac{43}{5} \)
a) x - 1/2 = 3/5
x = 3/5 + 1/2
x = 11/10
b) x - 1/2 = -2/3
x = -2/3 + 1/2
x = -1/6
c) 2/5 - x = 0,25
x = 2/5 - 0,25
x = 2/5 - 1/4
x = 3/20
`@` `\text {Ans}`
`\downarrow`
\(\left(\dfrac{x}{3}+\dfrac{1}{2}\right)\left(75\%-1\dfrac{1}{2}x\right)=0\)
`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}+\dfrac{1}{2}=0\\\dfrac{75}{100}-\dfrac{3}{2}x=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}\dfrac{x}{3}=-\dfrac{1}{2}\\\dfrac{3}{2}x=\dfrac{75}{100}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=-1\cdot3\\x=\dfrac{75}{100}\div\dfrac{3}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy, `x={-3/2; 1/2}.`
Lời giải:
$(x-15)-x.13=0$
$x-15-x.13=0$
$(x-x.13)-15=0$
$x(1-13)-15=0$
$x.(-12)-15=0$
$x.(-12)=15$
$x=15:(-12)=\frac{-5}{4}$
-7|x+4|= -7
|x+4|= -7 : -7
|x+4|= 1
=> x+4=1 hay x+4=-1
x= -3 x=3
-7.|x+4| = 21 : (-3)
-7.|x+4| = -7
|x+4| = -7 : - 7
|x+4| = 1
x = 1 - 4
x = - 3
\(a.x+\left(x+10\right)+\left(x+14\right)+....+\left(x+162\right)=3434\\ x+\left(x+x+.....+x\right)+\left(10+14+18+.....+162\right)=3434\\ x+x+39+3354=3434\\ x\cdot40=80\\ x=2\)
| x | + | 2x - 3 | = 0 (1)
Ta có \(\hept{\begin{cases}\left|x\right|\ge0\\\left|2x-3\right|\ge0\end{cases}}\forall x\)
\(\Rightarrow\left|x\right|+\left|2x-3\right|\ge0\forall x\) (2)
Từ (1) và (2) => (1) \(\Leftrightarrow\) \(\hept{\begin{cases}\left|x\right|=0\\\left|2x-3\right|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\2x-3=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\2x=3\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)
\(\Leftrightarrow x\in\varnothing\)
Vậy \(x\in\varnothing\)
@@ Học tốt
!!! K chắc