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bÀI LÀM
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
\(x^3+x^2+4=x^3+2x^2-x^2-2x+2x+4\)
\(=x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)\)
\(=\left(x^2-x+2\right)\left(x+2\right)\)
\(x^8+64=x^8+16x^4+64-16x^4\)
\(=\left(x^4+8\right)^2-\left(4x^2\right)^2\)
\(=\left(x^4-4x^2+8\right)\left(x^4+4x^2+8\right)\)
\(4a^4+b^4=4a^4+4a^2b^2+b^4-4a^2b^2\)
\(=\left(2a^2+b^2\right)^2-\left(2ab\right)^2\)
\(=\left(2a^2+b^2-2ab\right)\left(2a^2+b^2+2ab\right)\)
\(x^3-2x-4=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x^2+2x+2\right)\left(x-2\right)\)
Chúc bạn học tốt.
b) x8 +7x4+16
= x8+8x4-x4+16
= (x8+8x4+16) - x4
=(x4+4)2-x4
= (x4+4+x2)(x4+4-x2)
c) x5+x-1
= x5 - x4+x3+x4-x3+x2-x2+x-1
= x3(x2-x+1) + x2(x2-x+1) - (x2-x+1)
= (x2-x+1)(x3+x2 -1)
d)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+x2-x+1)
= x (x-1)(x2+x+1)
e) x5+x4+1
= x5+x4+x3 - x3+1
= x3(x2+x+1) - (x-1)(x2+x+1)
= (x2+x+1)(x3-x+1)
f) x8+x+1
= x8-x2+x2+x+1
= x2(x6-1)+(x2+x+1)
= x2(x3-1)(x3+1) +(x2+x+1)
= (x5+x2)(x-1)(x2+x+1) +(x2+x+1)
= (x2+x+1)(x6-x5+x3-x2+1)
A= x4 + 64
A= (x2)2 + 2.x2.8 +82 - (2.x2 .8)
A=(x2+8)2 -16x2
A =(x2+8+4x).(x2+8-4x)
-
G=(x2+y2+z2)2 (có sẵn hdt rồi mak_)
\(A=\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)
Đặt \(x^2+x=t\), ta có:
\(A=t^2-14t+24\)
\(=t^2-2t-12t+24\)
\(=t\left(t-2\right)-12\left(t-2\right)\)
\(=\left(t-2\right)\left(t-12\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-12\right)\)
\(B=\left(x^2+x\right)^2+4x^2+4x-12\)
\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
Đặt \(x^2+x=t\), ta có:
\(B=t^2+4t-12\)
\(=t^2+6t-2t-12\)
\(=t\left(t+6\right)-2\left(t+6\right)\)
\(=\left(t+6\right)\left(t-2\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(x^2+5x+4=t\), ta có:
\(C=t\left(t+2\right)+1\)
\(=t^2+2t+1\)
\(=\left(t+1\right)^2\)
\(=\left(x^2+5x+4+1\right)^2\)
\(=\left(x^2+5x+5\right)^2\)
\(D=\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+7=t\), ta có:
\(D=t\left(t+8\right)+15\)
\(=t^2+8t+15\)
\(=t^2+3t+5t+15\)
\(=t\left(t+3\right)+5\left(t+3\right)\)
\(=\left(t+3\right)\left(t+5\right)\)
\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
\(F=\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(x^2+x+1=t\), ta có:
\(F=t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2+4t-3t-12\)
\(=t\left(t+4\right)-3\left(t+4\right)\)
\(=\left(t+4\right)\left(t-3\right)\)
\(=\left(x^2+x+1+4\right)\left(x^2+x+1-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(E=x^4+2x^3+5x^2+4x-12\)
\(=x^4-x^3+3x^3-3x^2+8x^2-8x+12x-12\)
\(=x^3\left(x-1\right)+3x^2\left(x-1\right)+8x\left(x-1\right)+12\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+3x^2+8x+12\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+x^2+2x+6x+12\right)\)
\(=\left(x-1\right)\left[x^2\left(x+2\right)+x\left(x+2\right)+6\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
b, \(x^2-6x-2=x^2-6x+9-11=\left(x-3\right)^2-\sqrt{11}^2\)
\(=\left(x-3-\sqrt{11}\right)\left(x-3+\sqrt{11}\right)\)
c,\(9x^2+6x-1=\left(3x\right)^2+2.3x+1-2=\left(3x+1\right)^2-\sqrt{2}^2\)
\(=\left(3x+1-\sqrt{2}\right)\left(3x+1+\sqrt{2}\right)\)
d,\(x^8+64=\left(x^4\right)^2+8^2+16x^4-16x^4\)
\(=\left(x^4+8\right)^2-\left(4x^2\right)^2=\left(x^4+4x^2+8\right)\left(x^4-4x^2+8\right)\)
e,\(81x^4+4=\left(9x^2\right)^2+2^2+36x^2-36x^2=\left(9x^2+2\right)^2-\left(6x\right)^2\)
\(=\left(9x^2+2-6x\right)\left(9x^2+6x+2\right)\)
g,\(x^8+x^7+1\)
\(=\left(x^8+x^7+x^6\right)+\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)-\left(x^6+x^5+x^4\right)-\left(x^3+x^2+x\right)\)
\(=x^6\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)-x\left(x^2+x+1\right)\)\(\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
b/ 4x4 + 4x3 + 5x2 + 2x + 1
= (4x4 + 4x3 + x2) + 2(2x2 + x) + 1
= (2x2 + x)2 + 2(2x2 + x) + 1
= (2x2 + x + 1)2
c/ x8 + x + 1 = (x2 + x + 1)(x6 - x5 + x3 - x2 + 1)
e/ x4 - 8x + 63 = (x2 - 4x + 7)(x2 + 4x + 9)
\(a,...3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)\(=3\left(x^4+x^2+1\right)-\left(\left(x^4+x^2+1\right)+2\left(x^3+x^2+x\right)\right)\)
\(2\left(x^4+x^2+1\right)-2\left(x^3+x^2+x\right)=2\left(x^4-x^3-x+1\right)\) \(2\left(x^3\left(x-1\right)-\left(x-1\right)\right)=2\left(x-1\right)\left(x^3-1\right)\)
\(2\left(x-1\right)^2\left(x^2+x+1\right)\)