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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 :)
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
<=> \(\orbr{\begin{cases}x-3-\sqrt{5}=0\\x-3+\sqrt{5}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)
\(2x^2-6x=0\)
\(\Rightarrow2x.\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0:2\\x=0+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
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!
\(1.6x\left(x-10\right)-2x+20=0\)
⇔\(6x\left(x-10\right)-2\left(x-10\right)=0\)
⇔ \(2\left(x-10\right)\left(3x-1\right)=0\)
⇔ x = 10 hoặc x = \(\dfrac{1}{3}\)
KL....
\(2.3x^2\left(x-3\right)+3\left(3-x\right)=0\)
⇔ \(3\left(x-3\right)\left(x^2-1\right)=0\)
⇔ \(x=+-1\) hoặc \(x=3\)
KL....
\(3.x^2-8x+16=2\left(x-4\right)\)
⇔ \(\left(x-4\right)^2-2\left(x-4\right)=0\)
⇔ \(\left(x-4\right)\left(x-6\right)=0\)
⇔ \(x=4\) hoặc \(x=6\)
KL.....
\(4.x^2-16+7x\left(x+4\right)=0\)
\(\text{⇔}4\left(x+4\right)\left(2x-1\right)=0\)
⇔ \(x=-4hoacx=\dfrac{1}{2}\)
KL.....
\(5.x^2-13x-14=0\)
⇔ \(x^2+x-14x-14=0\)
\(\text{⇔}\left(x+1\right)\left(x-14\right)=0\)
\(\text{⇔}x=14hoacx=-1\)
KL......
Còn lại tương tự ( dài quá ~ )
\(2x^3+5x^2-12x=0\)
\(\Rightarrow x\cdot\left(2x^2+5x-12\right)=0\)
\(\Rightarrow x\cdot\left(2x^2-3x+8x-12\right)=0\)
\(\Rightarrow x\cdot\left[x\cdot\left(2x-3\right)+4\cdot\left(2x-3\right)\right]=0\)
\(\Rightarrow x\cdot\left(2x-3\right)\cdot\left(x+4\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\2x-3=0\\x+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{3}{2}\\x=-4\end{cases}}\)
\(x^2-5x-24=0\)
\(\Rightarrow x^2+3x-8x-24=0\)
\(\Rightarrow x\cdot\left(x+3\right)-8\cdot\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\cdot\left(x-8\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+3=0\\x-8=0\end{cases}\Rightarrow\hept{\begin{cases}x=-3\\x=8\end{cases}}}\)
\(x^2-6x+8=0\)
\(\Rightarrow x^2-2x-4x+8=0\)
\(\Rightarrow x\cdot\left(x-2\right)-4\cdot\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\cdot\left(x-4\right)=0\)
\(\Rightarrow\hept{\begin{cases}x-2=0\\x-4=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\x=4\end{cases}}}\)
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) \(x^3+6x^2+12x+8=0\)
\(\Leftrightarrow\left(x+2\right)^3=0\)
\(\Rightarrow x=-2\)
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) \(x^2-6x+4=0\)
\(\Leftrightarrow\left(x-3\right)^2-5=0\)
\(\Leftrightarrow\left(x-3+\sqrt{5}\right)\left(x-3-\sqrt{5}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{5}\\x=3-\sqrt{5}\end{cases}}\)