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\(x^2-y^2+4x+4\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
\(4x^2-y^2+8\left(y-2\right)\)
\(=4x^2-\left(y^2-8y+16\right)\)
\(=4x^2-\left(y-4\right)^2\)
\(=\left(2x+y-4\right)\left(2x-y+4\right)\)
a/ 9a^3 - 13a + 6 = 9a^3 - 6a^2 + 6a^2 - 4a - 9a + 6 = (9a^3 - 6a^2) + (6a^2 - 4a) - (9a - 6) = 3a^2(3a - 2) + 2a(3a - 2) - 3(3a - 2) = (3a^2 + 2a - 3)(3a - 2) Mình gửi luôn cho nóng^^Được câu nào hay câu đó. Yên tâm mình sẽ cố nghĩ &gửi nốt :)))
b/x^4 - 4x^3 + 8x + 3 = x^4 - 3x^3 - x^3 + 3x^2 - 3x^2 + 9x - x + 3 = (x^4 - 3x^3) - (x^3 - 3x^2) - (3x^2 - 9x) - (x - 3) = x^3(x - 3) - x^2(x - 3) - 3x(x - 3) - (x - 3) = (x^3 - x^2 - 3x - 1)(x - 3) Mình đang cố nghĩ nốt con c đây, có vẻ khó^^
\(A=x^3+4x^2-8x-8=\left(x^3-8\right)+4x\left(x-2\right)=\left(x^3-2^3\right)+4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)+4x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4+4x\right)=\left(x-2\right)\left(x^2+6x+4\right)\)
\(B=a^2+b^2-a^2b^2+ab-a-b=\left(ab-a\right)-\left(a^2b^2-a^2\right)+\left(b^2-b\right)\)
\(=a\left(b-1\right)-a^2\left(b^2-1\right)+b\left(b-1\right)=a\left(b-1\right)-a^2\left(b-1\right)\left(b+1\right)+b\left(b-1\right)\)
\(=\left(b-1\right)\left(a-a^2b-a^2+b\right)\)
\(C=x^4-x^3-x+1=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
Đoàn Thị Huyền Đoan: Hình như câu A bạn chép xuống bị sai đề rồi!
1) \(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Phân tích đa thức thành nhân tử
a) (1-2x)(1+2x)-x(x+2)(x-2)
\(=1-4x^2-x\left(x^2-4\right)\)
\(=1-4x^2-x^3+4x\)
\(=\left(1-x^3\right)+\left(4x-4x^2\right)\)
\(=\left(1-x\right)\left(1+x+x^2\right)+4x\left(1-x\right)\)
\(=\left(1-x\right)\left(1+x+x^2+4x\right)\)
\(=\left(1-x\right)\left(x^2+5x+1\right)\)
\(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a\left(a^3+6a^2b+12ab^2+8b^3\right)-b\left(8a^3+12a^2b+6ab^2+b^3\right)\)
\(=a^4+6a^3b+12a^2b^2+8b^3a-8a^3b-12a^2b^2+6ab^3-b^4\)
\(=a^4+6a^3b+8b^3a-8a^3b-6ab^3-b^4\)
\(=\left(a^4-b^4\right)+\left(6a^3b-6ab^3\right)+\left(8b^3a-8a^3b\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a^2-b^2\right)+8ab\left(b^2-a^2\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3\right)+6ab\left(a-b\right)\left(a+b\right)-8ab\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a^3+a^2b+ab^2+b^3+6a^2b+6ab^2-8a^2b-8ab^2\right)\)
\(=\left(a-b\right)\left(a^3-a^2b-ab^2+b^3\right)\)
\(=\left(a-b\right)\left[a^2\left(a-b\right)-b^2\left(a-b\right)\right]\)
\(=\left(a-b\right)^3\left(a+b\right)\)
Bài 2:
a) =a2b - a2c + b2c - ab2 + ac2 - bc2
=(a2b - bc2) - (a2c - ac2) + (b2c - ab2)
=b(a-c)(a+c) - ac(a-c) - b2(a-c)
=(a - c)(ab -bc - ac - b2)
b)=(1 - 2a + a2) - (b2 - 2bc + c2)
=(1 - a)2 - (b - c)2
=(c - b - a + 1)(b - c - a + 1)
b: \(=\left(x^2+4x-3\right)^2-2x\left(x^2+4x-3\right)-3x\left(x^2+4x-3\right)+6x^2\)
\(=\left(x^2+4x-3\right)\left(x^2+4x-3-2x\right)-3x\left(x^2+4x-3-2x\right)\)
\(=\left(x^2+2x-3\right)\left(x^2+4x-3-3x\right)\)
\(=\left(x^2+x-3\right)\left(x+3\right)\left(x-1\right)\)
c: \(=a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+\left(c-a\right)^3\)
\(=a^3-3a^2b+3ab^2-3b^2c+3bc^2-c^3+c^3-3a^2c+3ac^2-a^3\)
\(=-3a^2b+3ab^2-3b^2c+3bc^2-3a^2c+3ac^2\)
\(=-3\left(a^2b-ab^2+b^2c-bc^2+a^2c-ac^2\right)\)
1)a2(b-c)+b2(c-a)+c2(a-b)
=a2b-a2c+b2c-b2a+c2a-c2b
=(a2b-c2b)+(b2c-b2a)+(c2a-a2c)
=b.(a2-c2)-b2.(a-c)-ac.(a-c)
=b.(a-c)(a+c)-b2(a-c)-ac(a-c)
=(a-c)(ab+bc-b2-ac)
=(a-c)[(ab-ac)+(bc-b2)]
=(a-c)[a.(b-c)-b.(b-c)]
=(a-c)(b-c)(a-b)
a) Ta có:
VP=(a+b)3−3ab(a+b)
=a3+3a2b+3ab2+b3−3a2b−3ab2
=a3+b3=VT (đpcm)
b) Ta có:
VP=(a−b)3+3ab(a−b)
=a3−3a2b+3ab2−b3+3a2b−3ab2
=a3−b3=VT (đpcm)
Áp dụng:
Với ab=12 và a+b=−7 ta có:
a3+b3=(a+b)3−3ab(a+b)
=(−7)3−3.12.(−7)=−91