tìm X
(-0.6X-1/2)x3/4-(-1)=1/3
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HV
27 tháng 8 2021
a,x(2x-1)-(x-1)^2-x^2=0
<=>x(2x-1-x)-(x-1)^2=0
<=>x(x-1)-(x-1)^2=0
<=>(x-x+1)(x-1)=0
<=>x-1=0
<=>x=1
b,(x+2)^3-x^3-6x^2=4
<=>x^3+6x^2+12x+8-x^3-6x^2=4
<=>12x+8=4
<=>x=-1/3
tick mik nha
27 tháng 8 2021
`a)x(2x-1)-(x-1)^2-x^2=0`
`<=>2x^2-x-x^2+2x-1-x^2=0`
`<=>x-1=0`
`<=>x=1`
Vậy `x=1.`
`b)(x+2)^3-x^3-6x^2=4`
`<=>x^3+6x^2+12x+8-x^3-6x^2=4`
`<=>12x+8=4`
`<=>12x=-4`
`<=>x=-1/3`
Vậy `x=-1/3.`
18 tháng 12 2022
3: =>x(x+1)=0
=>x=0 hoặc x=-1
4: =>(2x-3)(x+2)=0
=>x=3/2 hoặc x=-2
6: =>6x=7 hoặc 6x=-7
=>x=7/6 hoặc x==7/6
17 tháng 10 2021
\(a,\Leftrightarrow x^2+2x+1-x^2+3x-2x=3\\ \Leftrightarrow3x=2\Leftrightarrow x=\dfrac{3}{2}\\ b,\Leftrightarrow x^2-x-6-x^2+6x-9=15\\ \Leftrightarrow5x=30\Leftrightarrow x=6\\ c,\Leftrightarrow x^3+3x^2+3x+1-x^3-3x^2-2x+3=0\\ \Leftrightarrow x=-4\)
17 tháng 10 2021
a) \(\left(x+1\right)^2-x\left(x-3\right)=2x+3\Rightarrow x^2+2x+1-x^2+3x=2x+3\)
\(\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
AL
4
AC
2
28 tháng 6 2018
a/ \(\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=180\)
\(\Leftrightarrow x\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)=180\)
\(\Leftrightarrow\frac{13}{12}x=180\)
\(\Leftrightarrow x=\frac{2160}{13}\)
b/ \(\frac{1}{x.2}+\frac{2}{x.3}+\frac{3}{x.4}=120\)
\(\Leftrightarrow\frac{1}{x}\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}\right)=120\)
\(\Leftrightarrow\frac{1}{x}.\frac{23}{12}=120\)
\(\Leftrightarrow\frac{1}{x}=\frac{1440}{23}\)
\(\Leftrightarrow x=\frac{23}{1440}\)
6 tháng 7 2021
\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)
\(\Leftrightarrow-5x-18=0\)
\(\Leftrightarrow x=-\dfrac{18}{5}\)
Vậy ...
\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Leftrightarrow12x+6=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy ...
\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)
\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)
Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)
\(\Rightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)
Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy ...
26 tháng 10 2021
a: \(\Leftrightarrow x\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
c: \(\Leftrightarrow\left(x-1\right)\left(3x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
26 tháng 10 2021
a) \(x^2-6x=0\\ \Leftrightarrow x\left(x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
b) \(\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c) \(9x^2\left(x-1\right)=x-1\\ \Leftrightarrow\left(9x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(3x+1\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\\x=1\end{matrix}\right.\)
d) \(x^2-4=\left(x-2\right)^2\\ \Leftrightarrow\left(x-2\right)\left(x+2-x+2\right)=0\\ \Leftrightarrow4\left(x-2\right)=0\\ \Leftrightarrow x=2\)
e) \(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
f) \(x^3-0,36=0\\ \Leftrightarrow x\left(x^2-0,36\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{3}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
g) \(\Leftrightarrow\left(5x-1\right)\left(x-2018\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2018\end{matrix}\right.\)
h) \(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(\left(-0,6x-\frac{1}{2}\right)\times\frac{3}{4}-\left(-1\right)=\frac{1}{3}\)
\(\left(-0,6x-\frac{1}{2}\right)\times\frac{3}{4}=\frac{1}{3}+\left(-1\right)\)
\(\left(-0,6x-\frac{1}{2}\right)\times\frac{3}{4}=-\frac{2}{3}\)
\(-0,6x-\frac{1}{2}=-\frac{2}{3}\div\frac{3}{4}\)
\(-0,6x-\frac{1}{2}=-\frac{8}{9}\)
\(-0,6x=-\frac{8}{9}+\frac{1}{2}\)
\(-0,6x=-\frac{7}{18}\)
\(x=-\frac{7}{18}\div\left(-0,6\right)\)
\(x=\frac{35}{54}\)