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21 tháng 9 2017
a ) \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x=2000\) và \(x=\dfrac{1}{5}\)
b ) \(x^3-13x=0\)
\(\Leftrightarrow x\left(x^2-13\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=\sqrt{13}\end{matrix}\right.\)
Vậy \(x=0\) và \(x=\sqrt{13}\)
c ) \(x+5x^2=0\)
\(\Leftrightarrow x\left(1+5x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\1+5x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy \(x=0\) và \(x=-\dfrac{1}{5}\)
d ) \(\left(x+1\right)=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left[1-\left(x+1\right)\right]=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy \(x=0\) và \(x=-1\)
e ) \(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\\left(loại\right)\end{matrix}\right.\)
Vậy \(x=0\)
21 tháng 9 2017
a, \(5x\left(x-2000\right)-x+2000=0\)
\(\Leftrightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2000\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\\x-2000=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2000\end{matrix}\right.\)
b,\(x^3-13x=0\)
\(\Leftrightarrow x\left(x ^2-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{13}\end{matrix}\right.\)
c,\(x+5x^2=0\)
\(\Leftrightarrow x\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\5x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{5}\end{matrix}\right.\)
d,\(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow\left(x+1\right)-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow-x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
e,\(x^3+x=0\)
\(\Leftrightarrow x\left(x^2+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
CHÚC BẠN HỌC TỐT........
HT
2
DN
28 tháng 6 2015
a, 5x(x-2000)-x+2000=0
<=>5x(x-2000)-(x-2000)=0
<=>(5x-1)(x-2000)=0
<=>5x-1=0 hoặc x-2000=0
<=>x=1/5 hoặc x=2000
b, x3-13x=0
<=>x(x2-13)=0
<=>x=0 hoặc x2-13=0
<=>x=0 hoặc x=\(\sqrt{13}\) hoặc x=\(-\sqrt{13}\)
28 tháng 6 2015
a,5x(x-2000)-x+2000=0
=>5x(x-2000)-(x-2000)=0
=>(5x-1)(x-2000)=0
=>x-2000=0 hoặc 5x-1=0
=>x=2000 hoặc x=1/5
vậy x=1/5;2000
b,x3-13x=0
=>(x2-13)x=0
=>x2-13=0 hoặc x=0
=>x=0 hoặc x=\(\sqrt{13}\)
vậy x=0;\(\sqrt{13}\)
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 \(x=-13\)
22 tháng 7 2021
b) 5x(x-2000)-x+2000=0
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
NT
1
19 tháng 8 2018
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 = \(-\sqrt{13}\)
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
25 tháng 8 2017
a) 5x ( x - 2000 ) - x + 2000 = 0
5x ( x - 2000 ) - ( x - 2000 ) = 0
5x ( x - 2000 ) = 0
\(\Rightarrow\orbr{\begin{cases}5x=0\\x-2000=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2000\end{cases}}\)
Vậy ....
b) x3 - 13x = 0
x ( x2 - 13 ) = 0
x ( x - \(\sqrt{13}\)) - ( x + \(\sqrt{13}\)) = 0
\(\Rightarrow\hept{\begin{cases}x=0\\x-\sqrt{13}\\x+\sqrt{13}\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=\sqrt{13}\\x=\sqrt{-13}\end{cases}}\)
Vậy ....
a) x2 + 6 + 9
= x2 + 2 . 3 . x + 32
= ( x + 3 )2
b) 10x - 25 - x2
= - ( x2 - 10x + 25 )
= - ( x - 5 )2
c) 8x3 - 1/8
= ( 2x )3 - ( 1/2 )3
= ( 2x - 1/2 ) ( 4x2 + x + 1/4 )
d) 1/25 x2 - 64x2
= ( 1/5x )2 - ( 8x )2
= ( 1/5x + 8x ) ( 1/5 - 8x )
30 tháng 9 2018
\(x^3-13x=0\)
<=> \(x\left(x^2-13\right)=0\)
<=> \(x\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)
<=> \(x=0\)
hoặc \(x-\sqrt{13}=0\)
hoặc \(x+\sqrt{13}=0\)
<=> .....
12 tháng 8 2019
\(a,4x\left(x+1\right)=8\left(x+1\right)\)
\(\Leftrightarrow4x^2+4x-8x-8=0\)
\(\Leftrightarrow4x^2-4x-8=0\)
\(\Leftrightarrow4\left(x^2-x-2\right)=0\)
\(\text{⇔}4\left(x^2-2x+x-2\right)=0\)
\(\text{⇔}4\left(x-2\right)\left(x+1\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
\(c,2x\left(x-2\right)-\left(2-x\right)^2=0\)
\(\text{⇔}2x\left(x-2\right)-\left(x-2\right)^2=0\)
\(\text{⇔}\left(x-2\right)\left(2x-x+2\right)=0\)
\(\text{⇔}\left(x-2\right)\left(x+2\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(d,\left(x-3\right)^3+\left(3-x\right)=0\)
\(\text{⇔}\left(x-3\right)^3-\left(x-3\right)=0\)
\(\text{⇔}\left(x-3\right)\left(x^2-6x+9-1\right)=0\)
\(\text{⇔}\left(x-3\right)\left(x^2-6x+8\right)=0\)
\(\text{⇔}\left(x-3\right)\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=4\end{matrix}\right.\)
\(g,5x\left(x-2000\right)-x+2000=0\)
\(\text{⇔}\left(x-2000\right)\left(5x-1\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=2000\\x=\frac{1}{5}\end{matrix}\right.\)
\(n,\left(x+1\right)^2-1+x=0\)
\(\text{⇔}x^2+2x+1-1+x=0\)
\(\text{⇔}x^2+3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
\(k,\left(1-x\right)^2-1+x=0\)
\(\text{⇔}\left(1-x\right)^2-\left(1-x\right)=0\)
\(\text{⇔}\left(1-x\right)\left(1-x-1\right)=0\)
\(\text{⇔}\left(1-x\right).\left(-x\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
\(m,x+6x^2=0\)
\(\text{⇔}x\left(1+6x\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=0\\x=-\frac{1}{6}\end{matrix}\right.\)
\(h,x^2-4x=0\)
\(\text{⇔}x\left(x-4\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
a) 5x(x - 2000) - x + 2000 = 0
=> 5x(x - 2000) - (x - 2000) = 0
=> (x - 2000).(5x - 1) = 0
=> x - 2000 = 0 hoặc 5x - 1 = 0
=> x = 2000 hoặc 5x = 1
=> x = 2000 hoặc x = 1/5
b) x3 - 13x = 0
=> x.(x2 - 13) = 0
=> x = 0 hoặc x2 - 13 = 0
=> x = 0 hoặc x2 = 13, vô lí
=> x = 0
a) 5x(x-2000)-(x-2000)=(5x-1)(x-2000)=0 nên x=1/5 hoặc x=2000
b)\(x^3-13x=x\left(x^2-13\right)=0\)\(\Rightarrow\)x=0 hoặc x^2=13 hay x=\(\sqrt{13}\)