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\(x^2-3x=0\)
\(\Leftrightarrow x\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
\(x^5-9x=0\)
\(\Leftrightarrow x\left(x^4-9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^4-9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt[4]{9}\end{cases}}\)
9x2 - 4 - ( 3x - 2 )( x + 5 ) = 0
<=> ( 3x - 2 )( 3x + 2 ) - ( 3x - 2 )( x + 5 ) = 0
<=> ( 3x - 2 )( 3x + 2 - x - 5 ) = 0
<=> ( 3x - 2 )( 2x - 3 ) = 0
<=> \(\orbr{\begin{cases}3x-2=0\\2x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{3}{2}\end{cases}}\)
x3 + 64 + ( x + 4 )( 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 4x + 16 ) + ( x + 4 )( 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 4x + 16 + 2x - 3 ) = 0
<=> ( x + 4 )( x2 - 2x + 13 ) = 0
<=> \(\orbr{\begin{cases}x+4=0\\x^2-2x+13=0\end{cases}}\Leftrightarrow x=-4\)( vì x2 - 2x + 13 = ( x2 - 2x + 1 ) + 12 = ( x - 1 )2 + 12 ≥ 12 > 0 ∀ x )
( x - 3 )( x2 + 4x + 9 ) + 2( x2 - 9 ) - 10( x - 3 ) = 0
<=> ( x - 3 )( x2 + 4x + 9 ) + 2( x - 3 )( x + 3 ) - 10( x - 3 ) = 0
<=> ( x - 3 )( x2 + 4x + 9 + 2x + 6 - 10 ) = 0
<=> ( x - 3 )( x2 + 6x + 5 ) = 0
<=> ( x - 3 )( x + 1 )( x + 5 ) = 0
<=> x = 3 hoặc x = -1 hoặc x = -5
<=> ( x - 3 )(
M = ( x + 4 )( x - 4 ) - 2x( 3 + x ) + ( x + 3 )2
= x2 - 16 - 6x - 2x2 + x2 + 6x + 9
= -7 ( đpcm )
N = ( x2 + 4 )( x + 2 )( x - 2 ) - ( x2 + 3 )( x2 - 3 )
= ( x2 + 4 )( x2 - 4 ) - ( x4 - 9 )
= x4 - 16 - x4 + 9
= -7 ( đpcm )
P = ( 3x - 2 )( 9x2 + 6x + 4 ) - 3( 9x3 - 2 )
= 27x3 - 8 - 27x3 + 6
= -2 ( đpcm )
Q = ( 3x + 2 )2 + ( 6x + 10 )( 2 - 3x ) + ( 2 - 3x )2
= 9x2 + 12x + 4 + 12x - 18x2 + 20 - 30x + 4 - 12x + 9x2
= -18x + 28 ( có phụ thuộc vào biến )
a/=> 9x2 - 6x + 1 - (9x2 + 12x + 4)=0 => 9x2 - 6x + 1 - 9x2 - 12x - 4 =0 => -18x - 3 =0 => -18x = 3 => x = -1/6 b/=>4x2 + 4x + 1 - (x2 - 2x + 1)=0 => 4x2 + 4x + 1 - x2 + 2x - 1 =0 => 3x2 + 6x =0 => 3x(x+2)=0 => trường hợp 1: 3x=0=>x=0 ; trường hợp 2: x+2=0=>x=-2 c/=> x2 - 2*2*x + 22=0 => (x - 2)2 =0 => x-2=0 => x=2 d/=> x2 - 2*5*x + 52 =0 => (x - 5)2 =0 => x-5=0 => x=5 e/=> 9x2 + 6x - 3 =0 => 9x2 - 3x + 9x - 3 =0 => 3x(3x - 1) + 3(3x - 1) =0 => (3x + 3)(3x - 1) =0 => trường hợp1: 3x+3=0 =>3x=-3=>x=-1 ; trường hợp2: 3x-1=0=>x=1/3
b) \(3x\left(x+5\right)-2x-10=0\)
\(\Leftrightarrow3x\left(x+5\right)-2\left(x+5\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-5\end{cases}}\)
c) \(x^3-9x=0\)
\(\Leftrightarrow x\left(x^2-9\right)=0\)
\(\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)
TH1: \(x=0\)
TH2: \(x-3=0\Rightarrow x=3\)
\(x+3=0\Rightarrow x=-3\)
Vậy:..
d) \(\left(5+2x\right)\left(2x-7\right)=4x^2-25\)
\(\Leftrightarrow\left(5+2x\right)\left(2x-7\right)=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(2x+5\right)\left(2x-7-2x+5\right)=0\)
\(\Leftrightarrow-2\left(2x+5\right)=0\)
\(\Leftrightarrow2x+5=0\)
\(\Leftrightarrow x=-\frac{5}{2}\)
e) \(x^2-11x+30=0\)
\(\Leftrightarrow x^2-5x-6x+30=0\)
\(\Leftrightarrow x\left(x-5\right)-6\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=5\end{cases}}\)
