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:
\(x\in\varnothing\)
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
\(a,3x^3-6x^2=0\Rightarrow3x^2\left(x-2\right)=0.\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
\(b,x\left(x-4\right)-12x+48=0\)
\(\Rightarrow x\left(x-4\right)-12\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-4=0\\x-12=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=12\end{cases}}}\)
\(c,x\left(x-4\right)-\left(x^2-8\right)=0\)
\(\Rightarrow c,x^2-4x-x^2+8=0\)
\(\Rightarrow-4x+8=0\)
\(\Rightarrow-4\left(x-2\right)=0\)
\(\Rightarrow x=2\)
\(d,2x\left(x-5\right)-x\left(2x+3\right)=16\)
\(\Rightarrow2x^2-10x-2x^2-3x=16\)
\(\Rightarrow-13x=16\Leftrightarrow x=-\frac{16}{13}\)
\(e,\left(4x^2-1\right)-\left(x-1\right)^2=-3\)
\(\Rightarrow4x^2-1-x^2+2x-1=-3\)
\(\Rightarrow3x^2+2x+1=0\)
\(\Rightarrow x^2+\frac{2}{3}x+\frac{1}{3}=0\)
\(\Rightarrow x^2+2.x.\frac{1}{3}+\frac{1}{9}+\frac{2}{9}=0\)
\(\Rightarrow\left(x+\frac{1}{3}\right)^2+\frac{2}{9}=0\)( vô lý )
Vậy phương trình vô nghiệm
a) 3x3 - 6x2 = 0
<=> 3x2( x - 2 ) = 0
<=> \(\orbr{\begin{cases}3x^2=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
b) x( x - 4 ) - 12x + 48 = 0
<=> x(x - 4 ) - ( 12x - 48 ) = 0
<=> x(x - 4 ) - 12( x - 4 ) = 0
<=> ( x - 12 )( x - 4 ) = 0
<=> \(\orbr{\begin{cases}x-12=0\\x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=12\\x=4\end{cases}}\)
c) x( x - 4 ) - ( x2 - 8 ) = 0
<=> x2 - 4x - x2 + 8 = 0
<=> -4x + 8 = 0
<=> -4x = -8
<=> x = 2
d) 2x( x - 5 ) - x( 2x + 3 ) = 16
<=> 2x2 - 10x - 2x2 - 3x = 16
<=> -13x = 16
<=> x = -16/13
e) ( 4x2 - 1 ) - ( x - 1 )2 = -3
<=> 4x2 - 1 - x2 + 2x - 1 + 3 = 0
<=> 3x2 + 2x + 3 = 0
<=> 3( x2 + 2/3x + 1/9 ) + 8/3 = 0
<=> 3( x + 1/3 )2 + 8/3 >= 8/3 > 0
<=> Vô nghiệm
