A = \(4x^2-3x+7x^2+2x-5\)

\(11x^2-3x+2x-5\)

\(11x^2-x-5\)

B = \(3x+7y-6x-8+y-2\)

\(3x+7y-6x-10+y\)

\(- 3x+7y-10+y\)

\(3x+8y-10\)

C =  chịu

D= \(6x^4-3x^2+x^2-4x+3.4-x+2\)

\(6x^4-3x^2+x^2-4x;12-x+2\\ \)

\(6x^4-3x^2+x^2-4x+14-x\)

\(6x^4-2x^2-4x+14-x\)

\(6x^4-2x^2-5x+14\)

(5x³ – 4x²) : 2x² + (3x⁴ + 6x) : 3x – x(x² – 1)

= 5x³ : 2x² + (-4x²): 2x² + 3x⁴ : 3x + 6x : 3x – [x. x² + x . (-1)]

= (5:2) . (x³ : x²) + [(-4) : 2] . (x² : x²) + (3 : 3) . (x⁴ : x) + (6 : 3). (x:x) – ( x³ – x)

= \(\dfrac{5}{2}\)x – 2 + x³ + 2 – x³ + x

= (x³ – x³) + (\(\dfrac{5}{2}\)x + x) + (-2 + 2)

= 0 + \(\dfrac{7}{2}\)x + 0

= \(\dfrac{7}{2}\)x

Bạn nên tách lẻ từng bài ra để được hỗ trợ tốt hơn, không nên đăng 1 loạt bài như thế này nhé.

2:

a: P(x)=3x^2-4x-1

Q(x)=-3x^2-4x-2

b:F(x)=-3x^2-4x-2-3x^2+4x+1=-6x^2-1

Q(x)=3x^2-4x-1+3x^2+4x+2=6x^2+1

c: F(-2)=-6*4-1=-25

Q(3)=-27-12-2=-41

Cho `f(x)=0`

`=>(x^2-2)(3x^4+6)=0`

   Mà `3x^4+6 > 0 AA x`

`=>x^2=2`

`=>x^2=2`

`=>x=+-\sqrt{2}`

Vậy nghiệm của đa thức `f(x)` là `x=\sqrt{2}` hoặc `x=-\sqrt{2}`

cho f(X) = 0

\(=>\left(2x-2\right)\left(3x.4+6\right)=0\)

\(=>\left[{}\begin{matrix}2x-2=0\\12x+6=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=2\\12x=-6\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Lời giải:
a.

$x^8+x^4+1=(x^4)^2+2x^4+1-x^4$
$=(x^4+1)^2-(x^2)^2=(x^4+1-x^2)(x^4+1+x^2)$

$=(x^4+1-x^2)[(x^2+1)^2-x^2]$

$=(x^4-x^2+1)(x^2+1-x)(x^2+1+x)$

b. 

$x^{12}-3x^6-1=(x^6-\frac{3}{2})^2-\frac{13}{4}$

$=(x^6-\frac{3}{2}-\frac{\sqrt{13}}{2})(x^6-\frac{3}{2}+\frac{\sqrt{13}}{2})$

c.

$3x^4+10x^2-25=(3x^4+15x^2)-(5x^2+25)$

$=3x^2(x^2+5)-5(x^2+5)=(x^2+5)(3x^2-5)$

$=(x^2+5)(\sqrt{3}x-\sqrt{5})(\sqrt{3}x+\sqrt{5})$

c.

$x^2-5y^2-y^4+2xy-9$

$=(x^2+2xy+y^2)-(y^4+6y^2+9)$
$=(x+y)^2-(y^2+3)^2$
$=(x+y+y^2+3)(x+y-y^2-3)$

\(a,x^8+x^4+1\\ =\left(x^8+2x^4+1\right)-x^4\\ =\left(x^4+1\right)^2-x^4\\ =\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\\ b,x^{12}-3x^6-1\\ =\left(x^{12}-2x^6+1\right)-x^6-2\\ =\left(x^6-1\right)^2-x^6-2\\ =\left(x^6-x^3-1\right)\left(x^6+x^3-1\right)-2???\\ c,3x^4+10x^2-25\\ =4x^4-\left(x^4-10x^2+25\right)\\ =4x^4-\left(x^2-5\right)^2\\ =\left(2x^2-x^2+5\right)\left(2x^2+x^2-5\right)\\ =\left(x^2+5\right)\left(3x^2-5\right)\\ d,x^2-5y^2-y^4+2xy-9\\ =\left(x^2+2xy+y^2\right)-\left(y^4+6y^2+9\right)\\ =\left(x+y\right)^2-\left(y^2+3\right)^2\\ =\left(x+y+y^2+3\right)\left(x+y-y^2-3\right)\)

a) x⁸+x⁴+1 = (x⁴+1)²-x⁴ = (x⁴-x²+1)(x⁴+x²+1)

b) x¹²-3x⁶-1 = (x⁶-1)²-x⁶ = (x⁶-x³-1)(x⁶+x³-1)

c) 3x⁴+10x²-25 = 4x⁴-(x⁴-10x²+25) = 4x⁴- (x²-5)² = (x²+5)(3x²-5)

d) x²-5y²-y⁴+2xy-9 = (x+y)²-(y²+3)² = (x+y-y²-3)(x+y+y²+3)

a, \(A=2x^3-9x^5+3x^5-3x^2+7x^2-12=-6x^5+2x^3+4x^2-12\)

b, \(B=2x^4+x^2+2x-2x^3-2x^2+x^2-2x+1=2x^4-2x^3+1\)

c, \(C=2x^2+x-x^3-2x^2+x^3-x+3=3\)

a)      A = 3x4 + 5x2y2 + 2y4 + 2y2 = 3x2(x2 + y2) + 2y2(x2 + y2) +2y2

= 3x2.2 + 2y2.2 + 2y2 = 6x2 + 6y2 = 6(x2 + y2) = 6.2 = 12

b) Ta thấy x4 ≥ 0; x2 ≥ 0. => 3x4  +  x2 + 2018 > 0 với mọi x

Vậy đa thức A(x) không có nghiệm.

c) Tìm được P(x) = -2x + 3