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NH
11 tháng 9 2019
a, \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
b, \(x^3+x^2+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x^2=-1\left(voly\right)\end{cases}\Leftrightarrow}x=-1\)
c, \(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\2-x=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3\\x=2\end{cases}}\)
d, \(2x\left(3x-5\right)=10-6x\)
\(\Leftrightarrow6x^2-10x-10+6x=0\)
\(\Leftrightarrow\left(6x^2+6x\right)-\left(10x+10\right)=0\)
\(\Leftrightarrow6x\left(x+1\right)-10\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(6x-10\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\6x-10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\6x=10\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-1\\x=\frac{5}{3}\end{cases}}\)
14 tháng 8 2016
a) (x-2)3 - 6(x+1)2 - x3 + 12 = 0
<=> x3-6x2+12x-8-6(x2+2x+1)-x3+12=0
<=> x3-6x2+12x-8-6x2-12x-6-x3+12=0
<=> -12x2+4=0
<=> \(x=\frac{1}{\sqrt{3}},x=-\frac{1}{\sqrt{3}}\)
vậy pt có 2 nghiệm....
b) x3 - 6x2 + 12x - 8 = 0
<=> (x3-2x2)-(4x2-8x)+(4x+8)=0
<=> (x-2)(x2-4x+4)=(x-2)3=0
=> x=2 là nghiệm
c) 8x3 - 12x2 + 6x - 1 = 0
<=> (2x-1)3=0
<=> x=1/2
14 tháng 8 2016
a) \(\left(x-2\right)^3-6\left(x+1\right)^2-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-6\left(x^2+2x+1\right)-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-6x^2-12x-6-x^3+12=0\)
\(\Leftrightarrow-12x^2-2=0\)
\(\Leftrightarrow-2\left(6x^2+1\right)=0\)
\(\Leftrightarrow6x^2+1=0\) (vô nghiệm)
Vậy không có giá trị nào của x thỏa mãn pt
b) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy x=2
c) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)
Vậy \(=\frac{1}{2}\)
13 tháng 7 2015
1a/ x3+x2+x+1=0
x2(x+1).(x+1)=0
=> x2(x+1)=0 x =1
hoặc =>[
x+1=0 x=-1
b/(x+2)2=x+2
x2+2.x.2+22 =x+2
x+x+4x+4=x+2
6x+4=x+2
....
c/(x+1)(6x2+2x)+(x-1)(6x2+2x)=0
x2-12 + (6x2+2x)2=0
=> x2-1 = 0 x=1
hoặc => [
(6x2+2x)2=0 x= 0
26 tháng 8 2019
a) \(3x^3-6x^2=0\)
\(3x^2\left(x-2\right)=0\)
\(\orbr{\begin{cases}3x^2=0\\x-2=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
b) \(x\left(x-4\right)-12x+48=0\)
\(x^2-4x-12x+48=0\)
\(x^2-16x+48=0\)
\(\left(x-12\right)\left(x-4\right)=0\)
\(\orbr{\begin{cases}x-12=0\\x-4=0\end{cases}}\)
\(\orbr{\begin{cases}x=12\\x=4\end{cases}}\)
c) Viết thiếu nha :v
d) \(2x\left(x-5\right)-x\left(2x+3\right)=16\)
\(2x^2-10x-x^2-2x^2-3x=16\)
\(-13x=16\)
\(x=-\frac{16}{13}\)
e) \(\left(4x^2-1\right)-\left(x-1\right)^2=-3\)
\(4x^2-1-x^2+2x-1=-3\)
\(3x^2-2+2x=-3\)
\(3x^2-2+2x+3=0\)
\(3x^2+1+2x=0\)
Vì \(3x^2+1+2x>0\)nên:
\(x\in\varnothing\)
26 tháng 8 2019
A) 3x3 - 6x2 = 0
=> 3x2(x - 2) = 0
=> \(\orbr{\begin{cases}3x^2=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
b) x(x - 4) - 12x + 48 = 0
=> x(x - 4) - 12(x - 4) = 0
=> (x - 12)(x - 4) = 0
=> \(\orbr{\begin{cases}x-12=0\\x-4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=12\\x=4\end{cases}}\)
c) x(x - 4) - (x2 - 8) = x2 - 4x - x2 + 8 = 4x + 8
16 tháng 8 2018
Bài 1:
a) (3x-2).(4x+5)-6x.(2x-1) = 12x^2 +15x - 8x -10 - 12x^2 + 6x = 13x - 10
b) (2x-5)^2 - 4.(x+3).(x-3) = 4x^2 - 20x + 25 - 4x^2 + 12x -12x + 36 = -20x + 61
Bài 2:
a)(2x-1)^2-(x+3)^2 = 0
<=> (2x-1-x-3).(2x-1+x+3) =0
<=>(x-4).(3x+2) = 0
<=> x-4 = 0 hoặc 3x+2=0
*x-4=0 => x=4
*3x+2 = 0 => 3x=-2 => x=-2/3
b)x^2(x-3)+12-4x=0 <=> x^2(x-3) - 4(x-3) =0 <=> (x-3).(x-2)(x+2) <=> x-3=0 hoặc x-2=0 hoặc x+2 =0
*x-3=0 => x=3
*x-2=0 =>x=2
*x+2=0 =>x=-2
c) 6x^3 -24x =0 <=> 6x(x^2 -4)=0 <=> 6x(x-2)(x+2)=0 <=> x=0 hoặc x-2 =0 hoặc x+2=0 <=> x=0 hoặc x=2 hoặc x=-2
TD
2
4 tháng 10 2016
a. x(x - 3) + x2 + 5 = 0
\(\Leftrightarrow x^2-3x+x^2+5=0\)
\(\Leftrightarrow2x^2-3x+5=0\)
\(\Leftrightarrow2\left(x-\frac{3}{4}\right)^2+\frac{31}{8}>0\)
Vô nghiệm
b)x2-6x=0
\(\Rightarrow x\left(x-6\right)=0\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x-6=0\end{array}\right.\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=0\\x=6\end{array}\right.\)
c)2x3+5x2-12x=0
\(\Leftrightarrow x\left(2x^2+5x-12\right)=0\)
\(\Leftrightarrow x\left(2x^2-3x+8x-12\right)=0\)
\(\Leftrightarrow x\left[x\left(2x-3\right)+4\left(2x-3\right)\right]=0\)
\(\Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+4=0\\2x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-4\\x=\frac{3}{2}\end{array}\right.\)
4 tháng 10 2016
a) x^2 - 3x + x^2 +5 =0
=> -3x + 5 = 0
=> -3x = -5
=> x= 5/3
b)x^2 - 6x = 0
x( x - 6 ) = 0
=> x =0
hoặc x-6 =0 => x = 6
vậy x =0 hoặc x =6
=> ...........
A
1 tháng 10 2020
a, \(x^2-12x-2x+24=0\Leftrightarrow x^2-14x+24=0\Leftrightarrow\left(x-12\right)\left(x-2\right)=0\)
TH1 : x = 12 ; TH2 : x = 2
b, \(x^2-5x-24=0\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
TH1 : x = 8 ; TH2 : x = -3
c, \(4x^2-12x-7=0\Leftrightarrow\left(2x+1\right)\left(2x-7\right)=0\)
TH1 : x = -1/2 ; TH2 : x = 7/2
d, \(x^3+6x^2+12x+8=0\Leftrightarrow\left(x+2\right)^3=0\Leftrightarrow x=-2\)
Tương tự HĐT thôi :)
1 tháng 10 2020
a) x2 - 12x - 2x + 24 = 0
<=> x( x - 12 ) - 2( x - 12 ) = 0
<=> ( x - 12 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x-12=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=12\\x=2\end{cases}}\)
b) x2 - 5x - 24 = 0
<=> x2 + 3x - 8x - 24 = 0
<=> x( x + 3 ) - 8( x + 3 ) = 0
<=> ( x + 3 )( x - 8 ) = 0
<=> \(\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
c) 4x2 - 12x - 7 = 0
<=> 4x2 + 2x - 14x - 7 = 0
<=> 2x( 2x + 1 ) - 7( 2x + 1 ) = 0
<=> ( 2x + 1 )( 2x - 7 ) = 0
<=> \(\orbr{\begin{cases}2x+1=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
d) x3 + 6x2 + 12x + 8 = 0
<=> ( x + 2 )3 = 0
<=> x + 2 = 0
<=> x = -2
e) ( x + 2 )2 - x2 + 4 = 0
<=> x2 + 4x + 4 - x2 + 4 = 0
<=> 4x + 8 = 0
<=> 4x = -8
<=> x = -2
f) 2( x + 5 ) = x2 + 5x
<=> x2 + 5x - 2x - 10 = 0
<=> x( x + 5 ) - 2( x + 5 ) = 0
<=> ( x + 5 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
m) 16( 2x - 3 )2 - 25( x - 5 )2 = 0
<=> 42( 2x - 3 )2 - 52( x - 5 )2 = 0
<=> [ 4( 2x - 3 ) ]2 - [ 5( x - 5 ) ]2 = 0
<=> ( 8x - 12 )2 - ( 5x - 25 )2 = 0
<=> [ 8x - 12 - ( 5x - 25 ) ][ 8x - 12 + ( 5x - 25 ) ] = 0
<=> ( 8x - 12 - 5x + 25 )( 8x - 12 + 5x - 25 ) = 0
<=> ( 3x + 13 )( 13x - 37 ) = 0
<=> \(\orbr{\begin{cases}3x+13=0\\13x-37=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{3}\\x=\frac{37}{13}\end{cases}}\)
n) x2 - 6x + 4 = 0
<=> ( x2 - 6x + 9 ) - 5 = 0
<=> ( x - 3 )2 - ( √5 )2 = 0
<=> ( x - 3 - √5 )( x - 3 + √5 ) = 0
<=> \(\orbr{\begin{cases}x-3-\sqrt{5}=0\\x-3+\sqrt{5}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)
NC
1 tháng 10 2020
a) \(x^2-12x-2x+24=0\)
\(\Leftrightarrow x\left(x-12\right)-2\left(x-12\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=12\\x=2\end{cases}}\)
b) \(x^2-5x-24=0\)
\(\Leftrightarrow\left(x^2+3x\right)-\left(8x+24\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
c) \(4x^2-12x-7=0\)
\(\Leftrightarrow\left(4x^2-14x\right)+\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)
d) \(x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Rightarrow x=-2\)
NC
1 tháng 10 2020
e) \(\left(x+2\right)^2-x^2+4=0\)
\(\Leftrightarrow4x+8=0\)
\(\Rightarrow x=-2\)
f) \(2\left(x+5\right)=x^2+5x\)
\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
m) \(16\left(2x-3\right)^2-25\left(x-5\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}8x-12=5x-25\\8x-12=25-5x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=-13\\13x=37\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{13}{3}\\x=\frac{37}{13}\end{cases}}\)
n) \(x^2-6x+4=0\)
\(\Leftrightarrow\left(x-3\right)^2-5=0\)
\(\Leftrightarrow\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)
a) \(x^2-6x+5=\left(x^2-6x+3^2\right)-4=\left(x+3\right)^2-4=0\)
<=> (x + 3)2 = 4 <=> x + 3 = + 2
<=> x = -1 hoặc x = -5
b) <=> x3 + x2 + x = 3
<=> x = 1