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NN
1
26 tháng 8 2016
a ) \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}5x-1=0\\x-2000=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=2000\end{array}\right.\)
b ) \(x^3-13x=0\)
\(\Leftrightarrow x\left(x^2-13\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x^2-13=0\Rightarrow\left[\begin{array}{nghiempt}x=\sqrt{13}\\x=-\sqrt{13}\end{array}\right.\end{array}\right.\)
SG
6
20 tháng 4 2017
Bài giải:
a) 5x(x -2000) - x + 2000 = 0
5x(x -2000) - (x - 2000) = 0
(x - 2000)(5x - 1) = 0
Hoặc 5x - 1 = 0 => 5x = 1 => x = 1515
Vậy x = 1515; x = 2000
b) x3 – 13x = 0
x(x2 - 13) = 0
Hoặc x = 0
Hoặc x2 - 13 = 0 => x2 = 13 => x = ±√13
Vậy x = 0; x = ±√13
29 tháng 5 2017
a) 5x(x-2000)-x+2000=0
5x(x-2000)-(x-2000)=0
(x-2000)(5x-1)=0
\(\Leftrightarrow\) x-2000=0 hoặc 5x-1=0
\(\Leftrightarrow\) x=2000 hoặc x=\(\dfrac{1}{5}\)
b) \(x^3-13x=0\)
\(x\left(x^2-13\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(x^2-13=0\)
\(\Leftrightarrow x=0\) hoặc \(x=13\) hoặc \(x=-13\)
NT
1
DN
20 tháng 8 2018
a,\(5x\left(x-2000\right)-x+2000=0\)
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
b,\(x^3-13x=0\)
\(\Rightarrow x.x^2-13x=0\)
\(\Rightarrow x\left(x^2-13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2=13\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\end{matrix}\right.\)
11 tháng 12 2018
a, 3x 3 - 3x = 0
=> 3x ( x 2 - 1 ) = 0
=> \(\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}\Rightarrow[}\begin{cases}x=0\\x=1\\x=-1\end{cases}}\)
b, x ( x - 2 ) + ( x - 2 ) = 0
=> ( x - 2 ) ( x + 1 ) = 0
=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
c, 5x ( x - 2000 ) - x + 2000 = 0
=> ( x - 2000 ) ( 5x - 1 ) = 0
=> \(\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}}\)
11 tháng 12 2018
a) \(3x^3-3x=0\)
\(\Rightarrow3x\left(x^2-1\right)=0\)
\(\Rightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{0;\pm1\right\}\)
b) \(x\left(x-2\right)+x-2=0\)
\(\Rightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{-1;2\right\}\)
c) \(5x\left(x-2000\right)-x+2000=0\)
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-2000=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2000\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{5};2000\right\}\)
6 tháng 4 2020
a) \(2x^3+5x^2-3x=0\Leftrightarrow x\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x^2+5x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=0\\x=-3\\x=\frac{1}{2}\end{matrix}\right.\)
b) \(2x^3+6x^2=x^2+3x\Leftrightarrow2x^3+5x^2-3x=0\)
Vậy $\orpt{\begin{matrix}x=0\\x=-3\\x=\frac{1}{2}\end{matrix}}$ (Giải câu a)
c) \(x^3-12=13x\Leftrightarrow x^3-13x-12=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x-12\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\\x=-3\end{matrix}\right.\)
Vậy $\orpt{\begin{matrix}x=-1\\x=4\\x=-3\end{matrix}}$
d) \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=\frac{3}{4}\end{matrix}\right.\)
17 tháng 7 2018
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
G
9 tháng 2 2018
Giải:
a) \(8\left(3x-2\right)-13x=5\left(12-3x\right)+7x\)
\(\Leftrightarrow24x-16-13x=60-15x+7x\)
\(\Leftrightarrow24x-13x+15x-7x=60+16\)
\(\Leftrightarrow19x=76\)
\(\Leftrightarrow x=\dfrac{76}{19}=4\)
Vậy ...
b) \(\dfrac{5x}{x+2}-\dfrac{3}{x-2}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\) (1)
ĐKXĐ: \(x\ne\pm2\)
\(\left(1\right)\Leftrightarrow\dfrac{5x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow5x\left(x-2\right)-3\left(x+2\right)+3x^2+6=0\)
\(\Leftrightarrow5x^2-10x-3x-6+3x^2+6=0\)
\(\Leftrightarrow8x^2-13x=0\)
\(\Leftrightarrow x\left(8x-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\8x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{13}{8}\left(TM\right)\end{matrix}\right.\)
Vậy ...
c) \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) (2)
ĐKXĐ: \(x\ne-1;x\ne3\)
\(\left(2\right)\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
\(\Leftrightarrow x\left(x+1+x-3\right)=4x\)
\(\Leftrightarrow x\left(2x-2\right)=4x\)
\(\Leftrightarrow2x-2=4\)
\(\Leftrightarrow x=3\)
Vậy ...
a) 5x(x - 2000) - (x - 2000) = 0
tương đương (x - 2000)(5x - 1) = 0
tương đương x = 2000 hoặc x = 1/5
b) x(x^2 -13) = 0
\(x\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)
tương đương x = 0 hoặ x = \(\sqrt{13}\)hoặc x = \(-\sqrt{13}\)