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S
14 tháng 8 2019
\(x^4y^4+64=x^4y^4+16x^2y^2+64-16x^2y^2=\left(x^2y^2+8\right)^2-16x^2y^2=\left(x^2y^2-4xy+8\right)\left(x^2y^2+4xy+8\right)\)
\(x^8+x+1=x^8-x^2+\left(x^2+x+1\right)=x^2\left(x^6-1\right)+\left(x^2+x+1\right)=x^2\left(x^3-1\right)\left(x^3+1\right)+\left(x^2+x+1\right)=x^2\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)+x^2+x+1=\left(x^2+x+1\right)\text{[}x^2\left(x+1\right)\left(x-1\right)\left(x^2-x+1\right)+1\text{]}\)
\(g,tach:x^2+x+1\)
\(x^4+4y^4=x^4+4x^2y^2+4y^4-4x^2y^2=\left(x^2+2y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\) \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)
L
15 tháng 11 2019
1)
=a^4+2a^2+1-a^2
=(a^2+1)^2-a^2
=(a^2-a+1)(a^2+a+1)
2)
=a^4+4b^4-4a^2b^2
=(a^2+2b^2)^2-4a^2b^2
=(a^2-2ab+2b^2)(a^2+2ab+2b^2)
3)
=(8x^2+1)^2-16x^2
=(8x^2-4x+1)(8x^2+4x+1).
4)
=x^5+x^4+x^3-x^3+1
=x^2(x^2+x+1)-(x-1)(x^2+x+1)
=(x^2-x+1)(x^2+x+1)
5).
=x^7-x+x^2+x+1
=x(x^6-1)+x^2+x+1
=x(x^3-1)(x^3+1)+x^2+x+1
=x(x-1)(x^2+x+1)(x^3+1)+x^2+x+1
=(x^2+x+1)[(x^2-x)(x^3+1)+1]
6)
=x^8-x^2+x^2+x+1
=x^2(x-1)(x^2+x+1)(x^3+1)+x^2+x+1
Xong nhóm x^2+x+1 vào.
7)
=x^4-(2x-1)^2
=(x^2-2x+1)(x^2+2x-1)
8)
=(a^8+b^8)^2-a^8b^8
=(a^8-a^4b^4+b^8)(a^8+a^4b^4+b^8).
NH
13 tháng 11 2019
1, a4 + a2 + 1
= a4 + 2a2 + 1 - a2
= (a2)2 + 2a2 + 1 - a2
= (a2 + 1)2 - a2
= (a2 + 1 - a)(a2 + 1 + a)
2, a4 + 4b4
= (a2)2 + 2. a2 . b2 + (2b)2 - a2 . b2
= (a2 + 2b)2 - (ab)2
= (a2 + 2b - ab)(a2 + 2b + ab)
3, 64x4 + 1
= (8x2)2 + 16x2 + 1 - 16x2
= (8x2 + 1)2 - (4x)2
= (8x2 + 1 - 4x)(8x2 + 1 + 4x)
4, x5 + x4 + 1
= x5 + x4 + x3 - x3 - x2 - x + x + x2 + 1
= (x5 + x4 + x3) - (x3 + x2 + x) + (x + x2 + 1)
= x3(x2 + x + 1) - x(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x3 - x + 1)
5, x7 + x2 + 1
= x7 – x + x2 + x + 1
= x(x6 – 1) + (x2 + x + 1)
= x(x3 – 1)(x3 + 1) + (x2 + x + 1)
= x(x3 + 1)(x – 1) (x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)[ x(x3 + 1)(x – 1) + 1]
= (x2 + x + 1)(x5 – x4 + x3 – x2 + x – 1)
6, x8 + x + 1
= x8 + x7 + x6 - x7 - x6 - x5 + x5 + x4 + x3 - x4 - x3 - x2 + x2 + x + 1
= (x8 + x7 + x6) - (x7 + x6 + x5) + (x5 + x4 + x3 ) - (x4 + x3 + x2) + (x2 + x + 1)
= x6(x2 + x + 1) - x5(x2 + x + 1) + x3(x2 + x + 1) - x2(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x6 - x5 + x3 - x2 + 1)
7, x4 - 4x2 + 4x - 1
= x4 - (4x2 - 4x + 1)
= (x2)2 - (2x - 1)2
= (x2 - 2x + 1)(x2 + 2x - 1)
= (x - 1)2 (x2 + 2x - 1)
8, a16 + a8b8 + b16
= (a16 + 2a8b8 + b16) - a8b8
= (a8 + b8)2 - (a4b4)2
= (a8 + b8 - a4b4)(a8 + b8 + a4b4)
= (a8 + b8 - a4b4)[(a8 + b8 + 2a4b4) - a4b4]
= (a8 + b8 - a4b4)[(a4 + b4)2 - (a2b2)2]
= (a8 + b8 - a4b4)(a4 + b4 - a2b2)(a4 + b4 + a2b2)
= (a8 + b8 - a4b4)(a4 + b4 - a2b2)[(a4 + b4 + 2a2b2) - a2b2]
= (a8 + b8 - a4b4)(a4 + b4 - a2b2)[(a2 + b2) - (ab)2]
= (a8 + b8 - a4b4)(a4 + b4 - a2b2)(a2 + b2 - ab)(a2 + b2 + ab)
AH
Akai Haruma
Giáo viên
14 tháng 8 2019
Bài 1, dạng 1:
a) Biểu thức không phân tích được thành nhân tử.
b)
\(x^4y^4+64=(x^2y^2)^2+8^2=(x^2y^2)^2+8^2+2.x^2y^2.8-16x^2y^2\)
\(=(x^2y^2+8)^2-(4xy)^2=(x^2y^2+8-4xy)(x^2y^2+8+4xy)\)
c)
\(x^4y^4+4=(x^2y^2)^2+2^2=(x^2y^2)^2+2^2+2.x^2y^2.2-4x^2y^2\)
\(=(x^2y^2+2)^2-(2xy)^2=(x^2y^2+2-2xy)(x^2y^2+2+2xy)\)
f)
\(x^8+x+1=x^8-x^2+x^2+x+1\)
\(=x^2(x^6-1)+(x^2+x+1)=x^2(x^3-1)(x^3+1)+(x^2+x+1)\)
\(=x^2(x-1)(x^2+x+1)(x^3+1)+(x^2+x+1)\)
\(=(x^2+x+1)[x^2(x-1)(x^3+1)+1]=(x^2+x+1)(x^6-x^5+x^3-x^2+1)\)
g)
\(x^8+x^7+1=x^8-x^2+x^7-x+x^2+x+1\)
\(=x^2(x^6-1)+x(x^6-1)+x^2+x+1\)
\(=(x^6-1)(x^2+x)+x^2+x+1\)
\(=(x^3-1)(x^3+1)(x^2+x)+x^2+x+1\)
\(=(x-1)(x^2+x+1)(x^3+1)(x^2+x)+(x^2+x+1)\)
\(=(x^2+x+1)[(x-1)(x^3+1)(x^2+x)+1]=(x^2+x+1)(x^6-x^4+x^3-x+1)\)
h)
Biểu thức không phân tích được thành nhân tử.
k)
\(x^4+4y^4=(x^2)^2+(2y^2)^2+2x^2.2y^2-4x^2y^2\)
\(=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)\)
l)
\(4x^4+1=(2x^2)^2+1^2+2.2x^2.1-4x^2\)
\(=(2x^2+1)^2-(2x)^2=(2x^2+1-2x)(2x^2+1+2x)\)
AH
Akai Haruma
Giáo viên
14 tháng 8 2019
Bài 2 dạng 4
a)
\(a^2-b^2-2x(a-b)=(a^2-b^2)-2x(a-b)=(a-b)(a+b)-2x(a-b)\)
\(=(a-b)(a+b-2x)\)
b)
\(a^2-b^2-2x(a+b)=(a^2-b^2)-2x(a+b)\)
\(=(a-b)(a+b)-2x(a+b)=(a+b)(a-b-2x)\)
4 tháng 8 2020
Bài 5:
a) Ta có: \(x^4+4\)
\(=x^4+4\cdot x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b) Ta có: \(x^4+64\)
\(=x^4+16x^2+64-16x^2\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
c) Ta có: \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+1\)
\(=x^6\left(x^2+x+1\right)-\left(x^6-1\right)\)
\(=x^6\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x^2+x+1\right)\left[x^6-\left(x-1\right)\left(x^3+1\right)\right]\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x-x^3-1\right)\)
d) Ta có: \(x^8+x^4+1\)
\(=x^8+x^4+x^6-x^6+1\)
\(=x^4\left(x^4+x^2+1\right)-\left(x^6-1\right)\)
\(=x^4\left(x^4+x^2+1\right)-\left(x^2-1\right)\left(x^4+x^2+1\right)\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^4-x^2+1\right)\)
g) Ta có: \(x^4+2x^2-24\)
\(=x^4+6x^2-4x^2-24\)
\(=x^2\left(x^2+6\right)-4\left(x^2+6\right)\)
\(=\left(x^2+6\right)\left(x^2-4\right)\)
\(=\left(x^2+6\right)\left(x-2\right)\left(x+2\right)\)
i) Ta có: \(a^4+4b^4\)
\(=a^4+4a^2b^2+4b^4-4a^2b^2\)
\(=\left(a^2+2b^2\right)^2-\left(2ab\right)^2\)
\(=\left(a^2-2ab+2b^2\right)\left(a^2+2ab+2b^2\right)\)
15 tháng 9 2018
\(A=x^2-4y^4=\left(x-2y^2\right)\left(x+2y^2\right)\)
\(B=8x^3+1=\left(2x+1\right)\left(4x^2-2x+1\right)\)
\(C=54x^3-16y^3=2\left(27x^3-8y^3\right)=2\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)\)
\(D=x^2-6x+8=\left(x^2-6x+9\right)-1=\left(x-3\right)^2-1=\left(x-3-1\right)\left(x-3+1\right)=\left(x-4\right)\left(x-2\right)\)
\(E=2x^2-5x+2=\left(2x^2-4x\right)-\left(x-2\right)=2x\left(x-2\right)-\left(x-2\right)=\left(x-2\right)\left(2x-1\right)\)
\(G=x^4+2x^2-3=\left(x^4+3x^2\right)-\left(x^2+3\right)=x^2\left(x^2+3\right)-\left(x^2+3\right)=\left(x^2+3\right)\left(x^2-1\right)=\left(x^2+3\right)\left(x-1\right)\left(x+1\right)\)
