Tìm x biết
a. \(x^4-16x^2=0\)
b. \(\left(x-5\right)^3-x+5=0\)
c. \(5.\left(x-2\right)=x^2-4\)
d. \(x-3=\left(3-x\right)^2\)
e. \(x^2.\left(x-5\right)+5-x=0\)
g.\(3x^4-9x^3=-9x^2+27x\)
h. \(x^2.\left(x+8\right)+x^2=-8x\)
i.\(\left(x+3\right).\left(x^2-3x+5\right)=x^2+3x\)
k.\(2.\left(x+3\right)-x^2-3x=0\)
l. \(8x^3-50x=0\)
\(a.x^4-16x^2=0\Leftrightarrow\left(x^2+4x\right)\left(x^2-4x\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+4=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\\x=4\end{matrix}\right.\)
\(b.\left(x-5\right)^3-x+5=0\)
\(\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
a) x4 - 16x2 = 0
<=> x2 ( x2 - 16 ) = 0
<=> \(\left[{}\begin{matrix}x^2=0\\x^2-16=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=0\\x=-4\\x=4\end{matrix}\right.\)
Vậy...
b) ( x - 5)3 - x + 5 = 0
<=> ( x - 5)3 - (x - 5) = 0
<=> (x - 5) [ (x - 5)2 - 1] =0
<=> \(\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
Vậy...
c) 5(x - 2) = x2 - 4
<=> 5(x - 2) - (x2 - 4) = 0
<=> (x - 2)( 5 - x - 2) = 0
<=> (x - 2)( 3 - x ) = 0
<=> \(\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Vậy...
d) x - 3 = (3 - x)2
<=> x - 3 - (x - 3)2 = 0
<=> (x - 3)(1 - x + 3) = 0
<=> (x - 3)( 4 - x ) = 0
<=> \(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy...
e) x2 (x - 5) + 5 - x = 0
<=> x2 (x - 5) - (x - 5) = 0
<=> (x2 - 1)( x - 5) = 0
<=> \(\left[{}\begin{matrix}\left(x-1\right)\left(x+1\right)=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=5\end{matrix}\right.\)
,
g) 3x4 - 9x3 = -9x2 + 27x
<=> 3x4 - 9x3 + 9x2 - 27x= 0
<=> 3x3 ( x - 3) + 9x (x - 3) = 0
<=> (3x3 + 9x)(x - 3) = 0
<=> 3x(x2 + 3)(x-3) = 0
<=> \(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
Vậy...
h) \(x^2\left(x+8\right)+x^2+8x=0\)
\(\Leftrightarrow x^2\left(x+8\right)+x\left(x+8\right)=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x+8\right)=0\)
\(\Leftrightarrow x\left(1+x\right)\left(x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-8\end{matrix}\right.\)
Vậy...
i) \(\left(x+3\right)\left(x^2-3x+5\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+5-x\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-4x+4+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\\left(x-2\right)^2=-1\left(voli\right)\end{matrix}\right.\)
Vậy x = -3
k) 2(x + 3) - x(x + 3) = 0
<=> (2 - x)(x + 3) = 0
<=> \(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
Vậy...
l) 2x (4x2 - 25) = 0
<=> \(\left[{}\begin{matrix}x=0\\x=2,5\end{matrix}\right.\)
