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\(a^3+a^2c-abc+b^2c+b^3\)
\(=\left(a^3+b^3\right)+\left(a^2c+b^2c-abc\right)\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+\)\(c\left(a^2+b^2-ab\right)\)
\(=\left(a^2+b^2-ab\right)\left(a+b+c\right)\)
x^10 + x^5 + 1
= x^10 + x^9 - x^9 + x^8 - x^8 + x^7 - x^7 + x^6 - x^6 + x^5 + x^5 - x^5 + x^4 - x^4 + x^3 - x^3 + x^2 - x^2 + x - x + 1
= (x^10 + x^9 + x^8) - (x^9 + x^8 + x^7) + (x^7 + x^6 + x^5) - (x^6 + x^5 + x^4) + (x^5 + x^4 + x^3) - (x^3 + x^2 + x) + (x^2 + x + 1)
= x^8 (x^2 + x + 1) - x^7 (x^2 + x + 1) + x^5 (x^2 + x + 1) - x^4 (x^2 + x + 1) + x^3 (x^2 + x + 1) - x (x^2 + x + 1) + (x^2 + x + 1)
= (x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)
\(x^4+x^2-2=x^4-x^2+2x^2-2 \)
\(=x^2\left(x^2-1\right)+2\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2+2\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x^2+2\right)\)
\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-3\)(1)
Đặt \(x^2+5x=t\)
\(\Rightarrow\left(1\right)=t\left(t+2\right)-3=t^2+2t-3\)
\(=t^2+3t-t-3=t\left(t+3\right)-\left(t+3\right)\)
\(=\left(t-1\right)\left(t+3\right)\)(2)
Mà \(x^2+5x=t\)nên \(\left(2\right)=\left(x^2+5x-1\right)\left(x^2+5x+3\right)\)
hay \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-3\)\(=\left(x^2+5x-1\right)\left(x^2+5x+3\right)\)
a) \(=x^4-x^3-2x^3+2x^2+2x^2-2x-x+1\)
\(=x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x^3-2x^2+2x-1\right)\left(x-1\right)\)
\(=\left(x^3-x^2-x^2+x+x-1\right)\left(x-1\right)\)
\(=\left(x^2-x+1\right)\left(x-1\right)^2\)
c)
\(=6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6\)
\(=6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(6x^3+17x^2-4x-3\right)\left(x-2\right)\)
\(=\left(6x^3+18x^2-x^2-3x-x-3\right)\left(x-2\right)\)
\(=\left(6x^2-x-1\right)\left(x+3\right)\left(x-2\right)\)
\(=\left(2x-1\right)\left(3x+1\right)\left(x+3\right)\left(x-2\right)\)
b)
\(=x^4+1011x^2+1011+\left(1010x^2-2020x+1010\right)\)
\(=x^4+1011x^2+1011+1010\left(x^2-2x+1\right)\)
\(=x^4+1011x^2+1011+1010\left(x-1\right)^2\)
CÓ: \(x^4+1010\left(x-1\right)^2+1011x^2\ge0\forall x\)
=> \(x^4+1010\left(x-1\right)^2+1011x^2+1011\ge1011>0\forall x\)
=> ĐA THỨC b > 0 => Ko ph được thành nhân tử.
a) a3+a2c-abc+b2c+b3 =(a3+b3)+(a2c-abc+b2c)=(a+b)(a2-ab+b2)+c(a2-ab+b2)=(a2-ab+b2)(a+b-c)
b) x3-7x-6 = x3+x2-x2-x-6x-6=x2(x+1)-x(x+1)-6(x+1)=(x+1)(x2-x-6)=(x+1)(x-3)(x+2)
c) x3-x2-14x+24=x3-2x2+x2-2x-12x+24=x2(x-2)+x(x-2)-12(x-2)=(x-2)(x2+x-12)=(x-2)(x+4)(x-3)
Phần b nhé
3x mũ 4 mak bạn