Phân tích đa thức thành nhân tử
1-4x2
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a/ \(4x^2-9\)
\(=\left(2x-3\right)\left(2x+3\right)\)
b/ \(3x\left(3x-2\right)+1\)
\(=9x^2-6x+1\)
\(=\left(3x-1\right)^2\)
\(a,=\left(2x-3\right)\left(2x+3\right)\)
\(b,=9x^2-6x+1=\left(3x-1\right)^2\)
-Đặt \(t=\left(x^2-x+1\right)\)
\(\left(x^2-x+1\right)^2-5x\left(x^2-x+1\right)+4x^2\)
\(=t^2-5xt+4x^2\)
\(=t^2-4xt-xt+4x^2\)
\(=t\left(t-4x\right)-x\left(t-4x\right)\)
\(=\left(t-4x\right)\left(t-x\right)\)
\(=\left(x^2-x+1-4x\right)\left(x^2-x+1-x\right)\)
\(=\left(x^2-5x+1\right)\left(x^2-2x +1\right)\)
\(=\left(x^2-5x+1\right)\left(x-1\right)^2\)
a) $4x^2+4x+1$
$=(2x)^2+2\cdot2x\cdot1+1^2$
$=(2x+1)^2$
b) $x^2+6x-y^2+9$
$=(x^2+6x+9)-y^2$
$=(x^2+2\cdot x\cdot3+3^2)-y^2$
$=(x+3)^2-y^2$
$=(x+3-y)(x+3+y)$
$\text{#}Toru$
a: \(4x^2+4x+1\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)
\(=\left(2x+1\right)^2\)
b: \(x^2+6x-y^2+9\)
\(=\left(x^2+6x+9\right)-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3+y\right)\left(x+3-y\right)\)
\(4x^2-7x-2\\ =4x^2-8x+x-2\\ =4x\left(x-2\right)+\left(x-2\right)\\ =\left(x-2\right)\left(4x+1\right)\)
a: \(=x\left(x^2+4x+4-z^2\right)\)
\(=x\left(x+2-z\right)\left(x+2+z\right)\)
c: \(x^4+x^3-4x^2+x+1\)
\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)
\(=\left(x-1\right)\left(x^3+2x^2-2x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\right]\)
\(=\left(x-1\right)^2\cdot\left(x^2+3x+1\right)\)
(1-2x)(1+2x)