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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) \(x^2-6x+4=0\)
\(\Leftrightarrow\left(x-3\right)^2-5=0\)
\(\Leftrightarrow\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)
VM
25 tháng 10 2019
\(2x^2-6x=0\)
\(\Rightarrow2x.\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0:2\\x=0+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}.\)
\(2x.\left(x+2\right)-3.\left(x+2\right)=0\)
\(\Rightarrow\left(x+2\right).\left(2x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\2x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0-2\\2x=3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3:2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-2;\frac{3}{2}\right\}.\)
\(x^3-16x=0\)
\(\Rightarrow x.\left(x^2-16\right)=0\)
\(\Rightarrow x.\left(x^2-4^2\right)=0\)
\(\Rightarrow x.\left(x-4\right).\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=0+4\\x=0-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{0;4;-4\right\}.\)
Chúc bạn học tốt!
22 tháng 1 2020
a) Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow\left(x^2-x\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
b) Ta có: \(-x^2+5x-6=0\)
\(\Leftrightarrow-\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow-\left(x^2-2x-3x+6\right)=0\)
\(\Leftrightarrow-\left[\left(x^2-2x\right)-\left(3x-6\right)\right]=0\)
\(\Leftrightarrow-\left[x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow-\left[\left(x-2\right)\left(x-3\right)\right]=0\)
\(\Leftrightarrow-\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy: x∈{2;3}
c) Ta có: \(4x^2-12x+5=0\)
\(\Leftrightarrow4x^2-10x-2x+5=0\)
⇔(4x2-10x)-(2x-5)=0
\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{5}{2}\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{2};\frac{5}{2}\right\}\)
d) Ta có: \(2x^2+5x+3=0\)
\(\Leftrightarrow2x^2+2x+3x+3=0\)
\(\Leftrightarrow\left(2x^2+2x\right)+\left(3x+3\right)=0\)
\(\Leftrightarrow2x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;\frac{-3}{2}\right\}\)
e) Ta có: \(x^3+2x^2-x-2=0\)
\(\Leftrightarrow\left(x^3+2x^2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow x^2\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{-2;1;-1\right\}\)
g) Ta có: \(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow9x^2-6x+1-20x^2-20x-5+12x^2-3-x^2+2x-1=0\)
\(\Leftrightarrow-24x-8=0\)
\(\Leftrightarrow-8\left(3x+1\right)=0\)
⇔3x+1=0
\(\Leftrightarrow3x=-1\)
\(\Leftrightarrow x=-\frac{1}{3}\)
Vậy: \(x=-\frac{1}{3}\)
NH
22 tháng 1 2020
h) \(2x^3-7x^2+7x-2=0\)
\(\Leftrightarrow2x^3-4x^2-3x^2+6x+x-2=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-3x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-2x-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x\left(x-1\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\frac{1}{2}\end{matrix}\right.\)
Vậy S = {2; 1; \(\frac{1}{2}\)}
i) \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+\frac{1}{2}\right)^2+\frac{23}{4}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\\\left(x+\frac{1}{2}\right)^2=\frac{-23}{4}\left(loai\right)\end{matrix}\right.\)
Vậy S = {1;-2}
S
1 tháng 7 2020
a) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Leftrightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
b) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Leftrightarrow-x-21=0\)
\(\Leftrightarrow x=-21\)
Vậy ...
c) \(5x\left(12x+7\right)-3x\left(2x-5\right)=-100\)
\(\Leftrightarrow60x^2+35x-6x^2+15x+100=0\)
\(\Leftrightarrow54x^2+50x+100=0\)
\(\Leftrightarrow54\left(x^2+\frac{25}{27}x+\frac{625}{2916}\right)+\frac{290975}{2916}=0\)
\(\Leftrightarrow54\left(x+\frac{25}{54}\right)^2+\frac{290975}{2916}=0\left(ktm\right)\)
Vậy phương trình vô nghiệm.
d) \(x\left(x-1\right)-x^2+2x=5\)
\(\Leftrightarrow x^2-x-x^2+2x-5=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
Vậy ...
e) \(2x^3\left(2x-3\right)-x^2\left(4x^2-6x+2\right)=0\)
\(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)
\(\Leftrightarrow-2x^2=0\)
\(\Leftrightarrow x=0\)
Vậy ...
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
\(\Leftrightarrow x\left(2x^2+5x-12\right)=0\)
\(\Leftrightarrow x\left(2x^2-3x+8x-12\right)=0\)
\(\Leftrightarrow x\left[x\left(2x-3\right)+4\left(2x-3\right)\right]=0\)
\(\Leftrightarrow x\left(x+4\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+4=0\\2x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-4\\x=\frac{3}{2}\end{array}\right.\)
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
=> ...........