1.\(PTĐT\) thành nhân tử
a) \(x^4+2x^3-16x^2-2x+15\)
b) \(2x^4-x^3-9x^2+13x-5\)
c) \(x^4+6x^3+11x^2+6x+1\)
2. CMR; ∀ n ∈ Z thì
a) \(n^4+2n^3-n^2-2n\) ⋮ 24
b) \(n^4-4n^3-4n^2+16n\) ⋮ 384
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10 tháng 11 2021
1.
\(a,=x^4-3x^3+5x^3-15x^2-x^2+3x-5x+15\\ =\left(x-3\right)\left(x^3+5x^2-x-5\right)\\ =\left(x-3\right)\left(x+5\right)\left(x^2-1\right)\\ =\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+5\right)\\ b,=2x^4-2x^3+x^3-x^2-8x^2+8x+5x-5\\ =\left(x-1\right)\left(2x^3+x^2-8x+5\right)\\ =\left(x-1\right)\left(2x^3+5x^2-4x^2-10x+2x+5\right)\\ =\left(x-1\right)\left(2x+5\right)\left(x^2-2x+1\right)\\ =\left(x-1\right)^3\left(2x+5\right)\)
2.
\(a,=n^3\left(n+2\right)-n\left(n+2\right)=n\left(n^2-1\right)\left(n+2\right)\\ =\left(n-1\right)n\left(n+1\right)\left(n+2\right)\)
Đây là tích 4 số nguyên liên tiếp nên chia hết cho \(1\cdot2\cdot3\cdot4=24\)
Suy ra đpcm
Bổ sung điều kiện câu b: n chẵn và n>4
\(b,=n\left(n^3-4n^2-4n+16\right)=n\left[n^2\left(n-4\right)-4\left(n-4\right)\right]\\ =\left(n-4\right)\left(n-2\right)n\left(n+2\right)\)
Với n chẵn và \(n>4\) thì đây là tích 4 số nguyên chẵn liên tiếp nên chia hết cho \(2\cdot4\cdot6\cdot8=384\)
20 tháng 10 2021
a) \(x^5+4x+5=\left(x^5+x^4\right)-\left(x^4+x^3\right)+\left(x^3+x^2\right)-\left(x^2+x\right)+\left(5x+5\right)=x^4\left(x+1\right)-x^3\left(x+1\right)+x^2\left(x+1\right)-x\left(x+1\right)+5\left(x+1\right)=\left(x^4-x^3+x^2-x+5\right)\left(x+1\right)\)
b) \(x^4+6x^3+11x^2+6x+1=\left(x^4+3x^3+x^2\right)+\left(3x^3+9x^2+3x\right)+\left(x^2+3x+1\right)=x^2\left(x^2+3x+1\right)+3x\left(x^2+3x+1\right)+\left(x^2+3x+1\right)=\left(x^2+3x+1\right)^2\)
c) \(64x^4+1=\left[\left(8x^2\right)^2+16x^2+1\right]-16x^2=\left(8x^2+1\right)^2-\left(4x\right)^2=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)d) \(81x^4+4=\left[\left(9x^2\right)^2+36x^2+2^2\right]-36x^2=\left(9x^2+2\right)^2-\left(6x\right)^2=\left(9x^2-6x+2\right)\left(9x^2+6x+2\right)\)
1 tháng 12 2015
câu d nè bạn
\(x^3+9x^2+23x+15=x^3+5x^2+4x^2+20x+3x+15\)
=\(x^2\left(x+5\right)+4x\left(x+5\right)+3\left(x+5\right)\)
=\(\left(x^2+4x+3\right)\left(x+5\right)=\left(x+1\right)\left(x+3\right)\left(x+5\right)\)
câu c nè
\(x^3-6x^2-x+30=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)=\left(x^2-x-6\right)\left(x-5\right)\)
=\(\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
tick rui minh làm tiếp cho
22 tháng 10 2023
2:
a: \(x^2-12x+20\)
\(=x^2-2x-10x+20\)
=x(x-2)-10(x-2)
=(x-2)(x-10)
b: \(2x^2-x-15\)
=2x^2-6x+5x-15
=2x(x-3)+5(x-3)
=(x-3)(2x+5)
c: \(x^3-x^2+x-1\)
=x^2(x-1)+(x-1)
=(x-1)(x^2+1)
d: \(2x^3-5x-6\)
\(=2x^3-4x^2+4x^2-8x+3x-6\)
\(=2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+4x+3\right)\)
e: \(4y^4+1\)
\(=4y^4+4y^2+1-4y^2\)
\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)
\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)
f; \(x^7+x^5+x^3\)
\(=x^3\left(x^4+x^2+1\right)\)
\(=x^3\left(x^4+2x^2+1-x^2\right)\)
\(=x^3\left[\left(x^2+1\right)^2-x^2\right]\)
\(=x^3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x+2\right)\left(x-1\right)\)
h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)
\(=\left(x^2+2x+1-1\right)^2-2\left(x+1\right)^2-1\)
\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-2\left(x+1\right)^2+1-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-4\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)
\(=\left(x+1\right)^2\left(x+1+2\right)\left(x+1-2\right)\)
\(=\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)\)
i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-\left(x+2y\right)-3\left(x+2y\right)+3\)
\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)
\(=\left(x+2y-1\right)\left(x+2y-3\right)\)
j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)
\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-3\)
\(=\left(x^2-3x+3\right)\left(x^2-3x-1\right)\)
25 tháng 2 2023
a: \(=-2x^2\cdot3x+2x^2\cdot4X^3-2x^2\cdot7+2x^2\cdot x^2\)
\(=8x^5+2x^4-6x^3-14x^2\)
b: \(=2x^3-3x^2-5x+6x^2-9x-15\)
\(=2x^3+3x^2-14x-15\)
c: \(=\dfrac{-6x^5}{3x^3}+\dfrac{7x^4}{3x^3}-\dfrac{6x^3}{3x^3}=-2x^2+\dfrac{7}{3}x-2\)
d: \(=\dfrac{\left(3x-2\right)\left(3x+2\right)}{3x+2}=3x-2\)
e: \(=\dfrac{2x^4-8x^3-6x^2-5x^3+20x^2+15x+x^2-4x-3}{x^2-4x-3}\)
=2x^2-5x+1
MC
21 tháng 10 2018
mk chỉ phân tích thôi bạn tự chia nha!
a, \(16x^4-81=(4x^2)^2-9^2=(4x^2-9)(4x^2+9)\)
\(=[(2x)^2-3^2](4x^2+9)\)
\(=(2x+3)(2x-3)(4x^2+9)\)
b, \(x^3-3x^2+3x-1=(x-1)^3\)
\(x^2-2x+1=(x-1)^2\)
c, \(18x^5+9x^4+3x^3+6x^2+3x+1=(18x^5+9x^4+3x^3)+(6x^2+3x+1)\)
\(=(6x^2+3x+1)(3x^3+1)\)
câu c bạn đánh sai 1 dấu phép toán kìa!!!!
4 tháng 10 2021
1: \(\dfrac{2x^3+11x^2+18x-3}{2x+3}\)
\(=\dfrac{2x^3+3x^2+8x^2+12x+6x+9-12}{2x+3}\)
\(=x^2+4x+3-\dfrac{12}{2x+3}\)
Bài 1:
c: \(=\left(x^2+3x+1\right)^2\)