Phân tích đa thức thành nhân tử:
(4x^2-4x+1)-(x-1)
Mn giúp mình với nhé.
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Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
Ta có: 4x2 - y2 + 4x + 4y - 3
= (4x2 - 4x + 1) - (y2 - 4y + 4)
= (2x - 1)2 - (y - 2)2
= (2x - 1 -y + 2)(2x - 1 + y - 2)
= (2x - y + 1)(2x + y - 3)
\(4x^2-y^2+4x+4y-3\)
\(=\left(4x^2+4x+1\right)-\left(y^2-4y+4\right)\)
\(=\left(2x+1\right)^2-\left(y-2\right)^2\)
\(=\left(2x+1+y-2\right)\left(2x+1-y+2\right)\)
\(=\left(2x+y-1\right)\left(2x-y+3\right)\)
\(\left(1+x^2\right)^2-4x\left(1-x^2\right)\)
\(\Leftrightarrow\left(1+x^2\right)^2+4x\left(1+x^2\right)\)
\(\Leftrightarrow\left(1+x^2\right)\times\left[\left(1+x^2\right)+4\right]\)
( 1+x2 )2 -4x( 1- x2 )
=x4+2x2+1-4x+4x3
=x3+2x2-x+2x3+4x2-2x-x2-2x+1
=x(x2+2x-1)+2x(x2+2x-1)-(x2+2x-1)
=(x2+2x-1)(x2+2x-1)
=(x2+2x-1)2
\(\left(1+x\right)^2-4x\left(1-x^2\right)\)
\(=\left(1+x\right)^2-4x\left(1-x\right)\left(1+x\right)\)
\(=\left(1+x\right)\left(1+x-4\left(1-x\right)\right)\)
\(=\left(1+x\right)\left(1+x-4+4x\right)\)
\(=\left(1+x\right)\left(5x-3\right)\)
\(\left(x^2+2x\right)^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+4x^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+2x^2-4x-3=0\Leftrightarrow\left(x-1\right)\left(x+1\right)^2\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)
Ta có: \(\left(x^2+2x\right)^2-2x^2-4x-3=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\\x=1\end{matrix}\right.\)
a) (2x - 1)2 - (x + 3)2
= (2x - 1 - x - 3).(2x - 1 + x + 3)
= (x - 4).(3x + 2)
b) x2.(x - 3) + 12 - 4x
= x2.(x - 3) - 4x + 12
= x2.(x - 3) - 4.(x - 3)
= (x - 3).(x2 - 4)
= (x - 3).(x - 2).(x + 2)
Áp dụng HĐT:
a2 - b2 = (a - b)(a + b)
\(\left(2x-1\right)^2-\left(x+3\right)^2\)
\(=\left(2x-1-x-3\right)\left(2x-1+x+3\right)\)
\(=\left(x-4\right)\left(3x+2\right)\)
\(-4x^2+12xy-9y^2+25=25-\left(2x-3y\right)^2=\left(5-2x+3y\right)\left(5+2x-3y\right)\)
\(\left(4x^2-4x+1\right)-\left(x-1\right)^2\)
\(=\left(2x-1\right)^2-\left(x-1\right)^2\)
\(=\left(2x-1-x+1\right)\left(2x-1+x-1\right)\)
\(=x\left(3x-2\right)\)