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XO
18 tháng 8 2020
1) x2 - 7x = 0
=> x(x - 7) = 0
=> \(\orbr{\begin{cases}x=0\\x=7\end{cases}}\)
2) -3x2 + 5x = 0
=> x(-3x + 5) = 0
=> \(\orbr{\begin{cases}x=0\\-3x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}\)
3) x2 - 19x - 20 = 0
=> x2 - 20x + x - 20 = 0
=> x(x - 20) + (x - 20) = 0
=> (x + 1)(x - 20) = 0
=> \(\orbr{\begin{cases}x+1=0\\x-20=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=20\end{cases}}\)
4) x2 - 5x - 24 = 0
=> x2 - 8x + 3x - 24 = 0
=> x(x - 8) + 3(x - 8) = 0
=> (x + 3)(x - 8) = 0
=> \(\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
20 tháng 8 2020
1) x2 - 7x = 0
<=> x( x - 7 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x=7\end{cases}}\)
2) -3x2 + 5x = 0
<=> x( -3x + 5 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x=\frac{5}{3}\end{cases}}\)
3) x2 - 19x - 20 = 0
<=> x2 + x - 20x - 20 = 0
<=> x( x + 1 ) - 20( x + 1 ) = 0
<=> ( x - 20 )( x + 1 ) = 0
<=> \(\orbr{\begin{cases}x=20\\x=-1\end{cases}}\)
4) x2 - 5x - 24 = 0
<=> x2 + 3x - 8x - 24 = 0
<=> x( x + 3 ) - 8( x + 3 ) = 0
<=> ( x - 8 )( x + 3 ) = 0
<=> \(\orbr{\begin{cases}x=8\\x=-3\end{cases}}\)
DA
3
SC
18 tháng 9 2018
a,x2+6x-7=0
=>x2+7x-x-7=0
=>(x^2+7x)-(x+7)=0
=>x(x+7)-(x+7)=0 =>(x+7)(x-1)=0
=>\(\orbr{\begin{cases}x+7=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-7\\x=1\end{cases}}}\)
b, x^3-2x^2-5x+6=0
=>x(x^2-2x-5+6)=0
=>x(x^2-2x+1)=0\(^{\orbr{\begin{cases}x=0\\\left(x-1^2\right)=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
c, 2x^2-5x+3=0
=>2x^2-2x-3x+3=0
D
18 tháng 9 2018
\(x^3-19x-30=0\)
\(\Rightarrow x^3+5x^2+6x-5x^2-25x-30=0\)
\(\Rightarrow\left(x-5\right)\left(x^2+5x+6\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x^2+2x+3x+6\right)=0\)
\(\Rightarrow\left(x-5\right)[x\left(x+2\right)+3\left(x+2\right)]=0\)
\(\Rightarrow\left(x-5\right)\left(x+3\right)\left(x+2\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-5=0\\x+3=0\\x+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=-3\\x=-2\end{cases}}\)
SG
1
23 tháng 5 2017
a, x2- 2x -3 = 0
\(\Leftrightarrow\) x2 + x - 3x - 3 =0 \(\Leftrightarrow\) x(x+1) - 3(x+1) = 0
\(\Leftrightarrow\) (x+1)(x-3) = 0
\(\Leftrightarrow\) x+1 = 0 hoặc x - 3 =0
1, x+1 = 0 \(\Leftrightarrow\) x = -1 2, x-3 = 0 \(\Leftrightarrow\) x = 3
b, \(2x^2+5x-3=0\)
\(\Leftrightarrow\)\(2x^2-x+6x-3=0\)
\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\) 2x - 1 = 0 hoặc x + 3 = 0
1, 2x -1 = 0 \(\Leftrightarrow x=\dfrac{1}{2}\) 2, x + 3 = 0 \(\Leftrightarrow x=-3\)
VD
27 tháng 8 2017
\(a,\)\(x^4-4x^3+4x^2=0\)
\(\Leftrightarrow x^2.\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow x^2.\left(x^2-2.x.2+2^2\right)=0\)
\(\Leftrightarrow x^2.\left(x-2\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\\left(x-2\right)^2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
\(b,\)\(x^2+5x+4=0\)
\(\Leftrightarrow x^2+x+4x+4=0\)
\(\Leftrightarrow x.\left(x+1\right)+4.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+4=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-4\end{cases}}\)
\(c,\)\(9x-6x^2-3=0\)
\(\Leftrightarrow-3.\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow2x^2-2x-x+1=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right).\left(2x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\2x=1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(d,\)\(2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+4x+x+2=0\)
\(\Leftrightarrow2x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\2x=-1\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}\)
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
=> \(\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}}\)
7 tháng 5 2017
a) \(x+5x^2=0\)
<=>\(x\left(1+5x\right)=0\)
+) \(x=0\) (TM)
+)\(1+5x=0\)
<=>\(5x=-1\)
<=>\(x=\dfrac{-1}{5}\) (TM)
Vậy \(x\) có 2 giá trị: \(x=\dfrac{-1}{5}\); \(x=0\)
b)\(x+1=\left(x+1\right)^2\)
<=>\(x+1-\left(x+1\right)^2=0\)
<=>\(\left(x+1\right)\left(1-x-1\right)=0\)
<=>\(\left(x+1\right)\left(-x\right)=0\)
+)\(x+1=0\)
<=>\(x=-1\) (TM)
+)\(-x=0\)
<=>\(x=0\) (TM)
Vậy \(x\) có 2 giá trị : \(x=-1\); \(x=0\)
c) \(x^3+x=0\)
<=> \(x\left(x^2+1\right)=0\)
+) \(x=0\) (TM)
+) \(x^2+1=0\)
<=>\(x^2=-1\)
Ta có: \(x^2\) >= 0, \(-1< 0\). Mà vế trái = vế phải
=> \(x^2=-1\) ( Vô nghiệm)
Vậy \(x=0\)
29 tháng 5 2017
a) \(x+5x^2=0\)
\(x\left(1+5x\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(1+5x=0\)
\(\Leftrightarrow x=0\) hoặc \(x=\dfrac{-1}{5}\)
b) \(x+1=\left(x+1\right)^2\)
\(\Leftrightarrow x+1-\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(x+1\right)\left[1-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(1-x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)-x=0\)
\(\Leftrightarrow x+1=0\) hoặc \(-x=0\)
\(\Leftrightarrow x=-1\) hoặc \(x=0\)
HH
25 tháng 6 2018
\(x^3+9x=0\)
<=> \(x\left(x^2+9\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\x^2+9=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x\in\varnothing\end{cases}}\)
<=> \(x=0\)
\(9x^2-4-2\left(3x-2\right)^2=0\)
<=> \(\left(9x^2-4\right)-2\left(3x-2\right)^2=0\)
<=> \(\left[\left(3x\right)^2-2^2\right]-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)
<=> \(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)
<=> \(\left(3x-2\right)\left(-3x+6\right)=0\)
<=> \(\left(3x-2\right)3\left(-x+2\right)=0\)
<=> \(3\left(3x-2\right)\left(2-x\right)=0\)
<=> \(\orbr{\begin{cases}3x-2=0\\2-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=2\\x=2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{2}{3}\\x=2\end{cases}}\)
\(\left(x^3-x^2\right)-4x+8x-4=0\)
<=> \(\left(x^3-x^2\right)+\left(4x-4\right)=0\)
<=> \(x^2\left(x-1\right)+4\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x^2+4\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x^2+4=0\end{cases}}\)
<=> \(x=1\)
\(\left(25x^2-10x\right):\left(-5x\right)-3\left(x-2\right)=4\)
<=> \(5x\left(5x-2\right)\left(-\frac{1}{5x}\right)-3\left(x-2\right)=4\)
<=> \(-\left(5x-2\right)-3\left(x-2\right)=4\)
<=> \(\left(5x-2\right)+3\left(x-2\right)=-4\)
<=> \(5x-2+3x-6=-4\)
<=> \(8x-8=-4\)
<=> \(8\left(x-1\right)=-4\)
<=> \(x-1=-\frac{1}{2}\)
<=> \(x=-\frac{3}{2}\)
TT
4
29 tháng 10 2016
a)5x(x-2)+3x-6=0
5x(x-2)+3(x-2)=0
(5x+3)(x-2)=0
=> 5x+3=0 hoặc x-2=0
5x=-3 x=0+2
x=-3/5 x=2
Vậy x=-3/5 hoặc x=2
b)x3-9x=0
x(x2-9)=0
=>x=0 hoặc x2-9=0
x2=9
=>x=3 hoặc x=-3
Vậy x=0 hoặc x=3 hoặc x=-3
29 tháng 10 2016
a) 5x(x - 2) + 3x - 6 = 5x(x - 2) + 3(x - 2) = (5x + 3)(x - 2) = 0 =>\(\orbr{\begin{cases}5x+3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-0,6\\x=2\end{cases}}}\)
b) x3 - 9x = x(x2 - 9) = x(x - 3)(x + 3) => x = 0 hoặc x - 3 = 0 hay x + 3 = 0 =>\(x\in\left\{-3;0;3\right\}\)
11 tháng 7 2018
\(x^3+x=0\)
\(\Rightarrow x.\left(x^2+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varnothing\end{cases}}\)
\(x^2-2x-3=0\)
\(\Rightarrow x.\left(x-2\right)=3\)
Vì \(x>x-2\)và \(x\inƯ\left(3\right)=\left\{3;-3\right\}\)
Các phần sau tương tự
KT
11 tháng 7 2018
\(x^3+x=0\)
\(\Leftrightarrow\)\(x\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x=0\)
\(x^2-2x-3=0\)
\(\Leftrightarrow\)\(\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
\(2x^2+5x-3=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+3=0\\2x-1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=\frac{1}{2}\end{cases}}\)
Vậy...
\(x+5x^2=0\)
\(\Leftrightarrow\)\(x\left(5x+1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\5x+1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-\frac{1}{5}\end{cases}}\)
Vậy...
2 x 2 + 5x – 3 = 0
2 x 2 + 6x – x – 3 = 0
2x(x + 3) − (x + 3) = 0
(x + 3) (2x − 1) = 0
x + 3 = 0 hoặc 2x − 1= 0
•x + 3 = 0 ⇒ x = −3
•2x – 1 = 0 ⇒ x = 1/2
Vậy x = −3 hoặc x = 1/2