x4-3x3+3x2-x
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\(3x^3+3x^2-3x-9=3\left(x^3+x^2-x-3\right)\)
Check lại đề hộ mình nhé:vv
P(x) = 3x2 – 5 + x4 – 3x3 – x6 – 2x2 – x3
= – x6 + x4 + (– 3x3 – x3) + (3x2 – 2x2) – 5
= – x6 + x4 – 4x3 + x2 – 5.
= – 5+ x2 – 4x3 + x4 – x6
Và Q(x) = x3 + 2x5 – x4 + x2 – 2x3 + x –1
= 2x5 – x4 + (x3 – 2x3) + x2 + x –1
= 2x5 – x4 – x3 + x2 + x –1.
= –1+ x + x2 – x3 – x4 + 2x5
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
a: \(x^2-y^2-x-y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
f: \(x^3-5x^2-5x+1\)
\(=\left(x+1\right)\left(x^2-x+1\right)-5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-6x+1\right)\)
a) \(=\left(x+6y\right)\left(x-6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x-6y-1\right)\)
b) \(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c) \(=2\left(x-y\right)^2-18\)
\(=2\left[\left(x-y\right)^2-3^2\right]\)
\(=2\left(x-y+3\right)\left(x-y-3\right)\)
a: \(x^2-36y^2-x+6y\)
\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x+6y-1\right)\)
b: \(x^3-8x^2+16x\)
\(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c: \(2x^2-4xy+2y^2-18\)
\(=2\left(x^2-2xy+y^2-9\right)\)
\(=2\left(x-y-3\right)\left(x-y+3\right)\)
d: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
a) \(...=P\left(x\right)=2x^4-x^4+3x^3+4x^2-3x^2+3x-x+3\)
\(P\left(x\right)=x^4+3x^3+x^2+2x+3\)
\(...=Q\left(x\right)=x^4+x^3+3x^2-x^2+4x+4-2\)
\(Q\left(x\right)=x^4+x^3+2x^2+4x+2\)
b) \(P\left(x\right)+Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)+\left(x^4+x^3+2x^2+4x+2\right)\)
\(\Rightarrow P\left(x\right)+Q\left(x\right)=2x^4+4x^3+3x^2+6x+5\)
\(P\left(x\right)-Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)-\left(x^4+x^3+2x^2+4x+2\right)\)
\(\)\(\Rightarrow P\left(x\right)-Q\left(x\right)=x^4+3x^3+x^2+2x+3-x^4-x^3-2x^2-4x-2\)
\(\Rightarrow P\left(x\right)-Q\left(x\right)=2x^3-x^2-2x+1\)
a) x² - 9
= x² - 3²
= (x - 3)(x + 3)
b) 4x² - 1
= (2x)² - 1²
= (2x - 1)(2x + 1)
c) x⁴ - 16
= (x²)² - 4²
= (x² - 4)(x² + 4)
= (x² - 2²)(x² + 4)
= (x - 2)(x + 2)(x + 4)
d) x² - 4x + 4
= x² - 2.x.2 + 2²
= (x - 2)²
e) x³ - 8
= x³ - 2³
= (x - 2)(x² + 2x + 4)
f) x³ + 3x² + 3x + 1
= x³ + 3.x².1 + 3.x.1² + 1³
= (x + 1)³
Trả lời:
x4 - 3x3 + 3x2 - x
= x ( x3 - 3x2 + 3x - 1 )
= x ( x - 1 )3
Ta có :
\(x^4-3x^3+3x^2-x\)
\(=x\left(x^3-3x^2+3x-1\right)\)
\(=x\left(x-1\right)^3\)
Vậy ..........