|x-1,5|-1/2=0
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3 tháng 7 2021
a) Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow-\left(x-2\right)-2\left(x-2\right)^3=0\)
\(\Leftrightarrow\left(x-2\right)\left[1+2\left(x-2\right)^2\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
b) Ta có: \(8x^3-72x=0\)
\(\Leftrightarrow8x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy: S={0;3;-3}
c) Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^2=0\)
\(\Leftrightarrow\left(x-1.5\right)^2\left[\left(x-1.5\right)^4+2\right]=0\)
\(\Leftrightarrow x-1.5=0\)
hay x=1,5
d) Ta có: \(2x^3+3x^2+3+2x=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
\(\Leftrightarrow2x=-3\)
hay \(x=-\dfrac{3}{2}\)
e) Ta có: \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x-1\right)+x\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-2\end{matrix}\right.\)
Vậy: S={0;1;-2}
f) Ta có: \(x^3-4x-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)-14x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=12\end{matrix}\right.\)
Vậy: S={0;2;12}
27 tháng 6 2019
1) \(|5x-3|=|7-x|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=7-x\\5x-3=x-7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x=10\\4x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
Vậy...
27 tháng 6 2019
2) \(2.|3x-1|-3x=7\)
\(\Leftrightarrow2.|3x-1|=7+3x\)
\(\Leftrightarrow\orbr{\begin{cases}2.\left(3x-1\right)=7+3x\\2.\left(3x-1\right)=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}6x-2=7+3x\\6x-2=-7-3x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=9\\9x=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-5}{9}\end{cases}}\)
Vậy...
17 tháng 4 2021
Ta có: \(\left(3x-1.5\right)\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1.5=0\\\dfrac{-1}{2}x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1.5\\\dfrac{-1}{2}x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=10\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{1}{2};10\right\}\)
17 tháng 4 2021
Giải:
(3.x-1.5).(-1/2.x+5)=0
=> 3x-1.5=0 hoặc -1/2x+5=0
x=1/2 hoặc x=10
Vậy x ∈{1/2;10}
Chúc bạn học tốt!
TN
2
16 tháng 9 2021
\(a,\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{3}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\\ b,\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(1,5+\dfrac{-3}{x}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\-\dfrac{3}{x}=-1,5=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=2\end{matrix}\right.\)
16 tháng 9 2021
a: \(\left(x-\dfrac{1}{3}\right)\left(x+\dfrac{2}{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: \(\left(\dfrac{3}{4}x-\dfrac{9}{16}\right)\left(\dfrac{1}{5}+\left(-3\right):x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{9}{16}\\\left(-3\right):x=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{16}:\dfrac{3}{4}=\dfrac{9}{16}\cdot\dfrac{4}{3}=\dfrac{3}{4}\\x=\left(-3\right):\dfrac{-1}{5}=15\end{matrix}\right.\)
11 tháng 8 2019
\(\left(x-1,5\right)^6+2.\left(1,5-x\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\1,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow x=1,5\)
Vậy : \(x=1,5\)
R
27 tháng 6 2020
\(\left(3x-1,5\right).\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow-\frac{3}{2}x^2+15x+\frac{3}{4}x-\frac{15}{2}=0\)
\(\Leftrightarrow-\frac{3}{2}x^2+\frac{63}{4}x-\frac{15}{2}=0\)
\(\Leftrightarrow-6x^2+63x-30=0\)
\(\Leftrightarrow2x^2-21x+10=0\)
\(\Leftrightarrow2x^2-20x-x+10=0\)
\(\Leftrightarrow2x\left(x-10\right)-\left(x-10\right)=0\)
\(\Leftrightarrow\left(x-10\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-10=0\\2x-1=0\end{cases}}\) \(\orbr{\begin{cases}x=10\\x=\frac{1}{2}\end{cases}}\)
#Chúc bạn học tốt!
`|x-1,5|-1/2=0`
`=> |x-1,5| = 1/2`
`=> |x-1,5| = 0,5`
`=> x - 1,5 = 0,5` hoặc `x - 1,5 = -0,5`
`=> x = 0,5 + 1,5 ` hoặc `x = -0,5 + 1,5`
`=> x = 2` hoặc `x = 1`
Vậy ...
=3991