Phân tích đa thức thành nhân tử a) x^4+x^2+2x+6 b) x^8+3x^4+4
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\(b,=x^4-2x^3-x^3+2x^2+3x^2-6x-3x+6\\ =\left(x-2\right)\left(x^3-x^2+3x-3\right)\\ =\left(x-2\right)\left(x-1\right)\left(x^2+3\right)\\ c,=x^4-2x^3+4x^3-8x^2+4x^2-8x+3x-6\\ =\left(x-2\right)\left(x^3+4x^2+4x+3\right)\\ =\left(x-2\right)\left(x^3+3x^2+x^2+3x+x+3\right)\\ =\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a, 3x(x^2-4)=0
+3x=0=>x=0
+x^2-4=0
=>x^2=4
=>x=+-2
c,x(x+2)-3x-6=0
x(x+2)-3(x+2)=0
(x+2)(x-3)=0
TH1 :x+2=0
x=-2
TH2 : x-3=0
x=3
câu b bạn chờ mình chúc nha
nhớ k cho mình
a: \(x^4-2x^3+x^2-2x\)
\(=\left(x^4-2x^3\right)+\left(x^2-2x\right)\)
\(=x^3\left(x-2\right)+x\left(x-2\right)\)
\(=x\left(x-2\right)\left(x^2+1\right)\)
b: \(x^4+x^3-8x-8\)
\(=\left(x^4+x^3\right)-\left(8x+8\right)\)
\(=x^3\left(x+1\right)-8\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3-8\right)\)
\(=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(2x^3-35x+75=2x^2\left(x+5\right)-10x\left(x+5\right)+15\left(x+5\right)=\left(x-5\right)\left(2x^2-10+15\right) \)
câu b tớ thêm chút
a) x8+3x4+4
=x8-x4+4x4+4
=(x4-1)(x4+1)+4.(x4+1)
=(x4+1)(x4-1+4)
=(x4+1)(x4+3)
b) x6-x4-2x3+2x2
=x4.(x2-1)-2x2.(x-1)
=x4.(x-1)(x+1)-2x2(x-1)
=x2.(x-1)[x2(x+1)-2]
=x2.(x-1)(x3+x2-2)
=x2.(x-1)(x3-1+x2-1)
=x2.(x-1)[(x-1)(x2+x+1)+(x-1)(x+1)]
=x2.(x-1)(x-1)(x2+x+1+x+1)
=x2.(x-1)2.(x2+2x+2)
a: \(=x^4-5x^3+4x^3-20x^2+7x^2-35x+4x-20\)
\(=\left(x-5\right)\left(x^3+4x^2+7x+4\right)\)
\(=\left(x-5\right)\left(x^3+x^2+3x^2+3x+4x+4\right)\)
\(=\left(x-5\right)\left(x+1\right)\left(x^2+3x+4\right)\)
b: Đề sai rồi bạn
a: \(x^4+x^2+2x+6\)
\(=x^4-2x^3+3x^2+2x^3-4x^2+6x+2x^2-4x+6\)
\(=\left(x^2-2x+3\right)\left(x^2+2x+2\right)\)