tìm x, biết:
x3-13x=0
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DP
2
PD
9 tháng 10 2021
Pt $\Leftrightarrow (x-4)^3=0\\\Leftrightarrow x-4=0\\\Leftrightarrow x=4$
NT
1
22 tháng 7 2021
b) 5x(x-2000)-x+2000=0
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
CM
29 tháng 7 2019
x3 = 13x
⇔ x3 – 13x = 0
⇔ x.x2 – x.13 = 0
(Có nhân tử chung x)
⇔ x(x2 – 13) = 0
⇔ x = 0 hoặc x2 – 13 = 0
+ x2 – 13 = 0 ⇔ x2 = 13 ⇔ x = √13 hoặc x = –√13
Vậy có ba giá trị của x thỏa mãn là x = 0, x = √13 và x = –√13.
27 tháng 10 2016
\(x^3+6x^2-13x-42=0\)
\(\Leftrightarrow\left(x^3-3x^2\right)+\left(9x^2-27x\right)+\left(14x-42\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)+9x\left(x-3\right)+14\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)+\left(x^2+9x+14\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+7x+2x+14\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[x\left(x+7\right)+2\left(x+7\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x+7\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\x+2=0\\x+7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-2\\x=-7\end{array}\right.\)
TH
27 tháng 10 2016
x3 + 6x2 - 13x - 42 = 0
=> x3 - 3x2 + 9x2 - 27x + 14x - 42 = 0
=> x2 ( x - 3 ) + 9x ( x - 3 ) + 14 ( x - 3 ) = 0
=> ( x - 3 ) ( x2 + 9x + 14) = 0
=> ( x - 3 ) ( x2 + 2x + 7x + 14 ) = 0
=> ( x - 3 ) [ x ( x + 2 ) + 7 ( x + 2 ) ] = 0
=> ( x - 3 ) ( x + 2 ) ( x + 7 ) = 0
=> x - 3 = 0 => x = 3
=> x + 2 = 0 => x = -2
=> x + 7 = 0 => x = -7
PT
1
25 tháng 11 2016
a)\(6x^2+5x-6=0\)
\(\Leftrightarrow6x^2-4x+9x-6=0\)
\(\Leftrightarrow2x\left(3x-2\right)+3\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)
b)\(6x^2-13x+6=0\)
\(\Leftrightarrow6x^2-4x-9x+6=0\)
\(\Leftrightarrow2x\left(3x-2\right)-3\left(3x-2\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=\frac{2}{3}\end{array}\right.\)
c)\(10x^2-13x-3=0\)
\(\Leftrightarrow10x^2-15x+2x-3=0\)
\(\Leftrightarrow5x\left(2x-3\right)+\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=0\\5x+1=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{1}{5}\end{array}\right.\)
d)\(20x^2+19x-3=0\)
\(\Delta=19^2-\left(-4\left(20.3\right)\right)=601\)
\(\Rightarrow x_{1,2}=\frac{-19\pm\sqrt{601}}{40}\)
e)\(3x^2-x+6=0\)
\(\Delta=\left(-1\right)^2-4\left(3.6\right)=-71< 0\)
Suy ra vô nghiệm
\(\Leftrightarrow x\cdot\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)
hay \(x\in\left\{0;\sqrt{13};-\sqrt{13}\right\}\)
\(x\left(x^2-13\right)=0\)
\(\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x=\sqrt{13}\\x=-\sqrt{13}\end{matrix}\right.\end{matrix}\right.\)