Bài 14 : Phân tích đa thức sau thành nhân tử :
f, ( x\(^2\) - 3 )\(^2\) + 4.( x\(^2\) - 3 )y + 4y\(^2\)
g, 49.( y - 4 )\(^2\) - 9.( x + 2 )\(^2\)
z, 39xyz\(^2\) - 26x\(^2\)yz + 13xy\(^2\)z
h, 5x\(^2\).( x + 2y ) - 15x.( 2y + x )
Bài 10 : Tìm x, biết :
a, 5 ( x + 3 ) - 2x ( 3 + x ) = 0
b, 2x\(^2\) - 6x = 0
c, ( x - 1 )\(^2\) = x - 1
d, 4x( x - 2020 ) - x + 2020 = 0
e, 8( 2x - 1 ) + x( 1 - 2x ) = 0
f, 6(x + 3)\(^2\) = 3 + x
Bài 11 : Tìm x , biết :
a, x\(^2\) - x = 0
b, 3x - x\(^2\) = 0
c, x\(^3\) + x = 0
d, 4x + 4x\(^3\) = 0
Bài 14:
f: \(\left(x^2-3\right)^2+4\cdot\left(x^2-3\right)\cdot y+4y^2\)
\(=\left(x^2-3\right)^2+2\cdot\left(x^2-3\right)\cdot2y+\left(2y\right)^2\)
\(=\left(x^2-3+2y\right)^2\)
g: \(49\left(y-4\right)^2-9\left(x+2\right)^2\)
\(=\left(7y-28\right)^2-\left(3x+6\right)^2\)
\(=\left(7y-28-3x-6\right)\left(7y-28+3x+6\right)\)
\(=\left(7y-3x-34\right)\left(7y+3x-22\right)\)
z: \(39xyz^2-26x^2yz+13xy^2z\)
\(=13xyz\cdot3z-13xyz\cdot2x+13xyz\cdot y\)
\(=13xyz\left(3z-2x+y\right)\)
h: \(5x^2\left(x+2y\right)-15x\left(x+2y\right)\)
\(=\left(x+2y\right)\left(5x^2-15x\right)\)
\(=\left(x+2y\right)\cdot5x\cdot\left(x-3\right)\)
Bài 10:
a: \(5\left(x+3\right)-2x\left(x+3\right)=0\)
=>(x+3)(5-2x)=0
=>\(\left[{}\begin{matrix}x+3=0\\5-2x=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-3\\2x=5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-3\\x=\dfrac{5}{2}\end{matrix}\right.\)
b: \(2x^2-6x=0\)
=>2x(x-3)=0
=>x(x-3)=0
=>\(\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
c: \(\left(x-1\right)^2=x-1\)
=>\(\left(x-1\right)^2-\left(x-1\right)=0\)
=>\(\left(x-1\right)\left(x-1-1\right)=0\)
=>(x-1)(x-2)=0
=>\(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
d: \(4x\left(x-2020\right)-x+2020=0\)
=>\(4x\left(x-2020\right)-\left(x-2020\right)=0\)
=>(x-2020)(4x-1)=0
=>\(\left[{}\begin{matrix}x-2020=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{4}\end{matrix}\right.\)
e: \(8\left(2x-1\right)+x\left(1-2x\right)=0\)
=>\(x\left(1-2x\right)-8\left(1-2x\right)=0\)
=>(1-2x)(x-8)=0
=>\(\left[{}\begin{matrix}1-2x=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=8\end{matrix}\right.\)
f: \(6\left(x+3\right)^2=x+3\)
=>\(6\left(x+3\right)^2-\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(6x+18-1\right)=0\)
=>(x+3)(6x+17)=0
=>\(\left[{}\begin{matrix}x+3=0\\6x+17=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\6x=-17\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=-3\\x=-\dfrac{17}{6}\end{matrix}\right.\)
Bài 11:
a: \(x^2-x=0\)
=>\(x\left(x-1\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
b: \(3x-x^2=0\)
=>x(3-x)=0
=>\(\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
c: \(x^3+x=0\)
=>\(x\left(x^2+1\right)=0\)
mà \(x^2+1>=1>0\forall x\)
nên x=0
d: \(4x^3+4x=0\)
=>\(4x\left(x^2+1\right)=0\)
mà \(4\left(x^2+1\right)>=4>0\forall x\)
nên x=0