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b) a3 + b3 + c3 - 3abc
= ( a + b)3 - 3ab - 3ba + c - 3abc
= (a3 + 3a2b + 3ab2 + b3) + c3 - (3a2b + 3ab2 + 3ab)
= (a + b)3 + c2 - 3ab(a + b + c)
= (a + b + c) [ (a + b)2 - ( a + b )c + c^2 ] - 3ab(a + b + c)
= ( a + b + c ) ( a2 + b2 + 2ab - ac - bc + c2 -3ab )
= ( a + b + c ) ( a2 + b2 + c2 - ab - ac - bc
a) 12x3 + 4x2 + 9x + 3 = 4x2(3x + 1) + 3(3x + 1) = (4x2 + 3)(3x + 1)
b) x3 + 2x2 - x - 2 = x2(x + 2) - (x + 2) = (x2 - 1(x + 2) = (x - 1)(x + 1)(x + 2)
c) a3 + (a - b)3 = (a + a - b)[a2 - a(a - b) + (a - b)2] = (2a - b)(a2 - a2 + ab + a2 - 2ab + b2)
= (2a - b)(a2 - ab + b2)
a) 12x3 + 4x2 + 9x + 3
= 4x2(3x + 1) + 3(3x + 1)
= (4x2 + 3)(3x + 1)
b) x3 + 2x2 - x - 2
= x2(x + 2) - (x + 2)
= (x2 - 1)(x + 2)
c) a3 + (a - b)3
= a3 - a2(a - b) + a(a - b)2 + (a - b)a2 - (a - b)2a + (a - b)3
= a[(a2 - a(a - b) + (a - b)2] + (a - b)[a2 - a(a - b) + (a - b)2]
= (a + a - b)[(a2 - a(a - b) + (a - b)2]
a) 4x2 +4x+3=4x2 +4x+4-1=(2x-2)2 - 1=(2x-2-1)(2x-2+1)=(2x-3)(2x-1)
a) \(4x^2+4x+3\)
\(=\left(4x^2+4x+4\right)-1\)
\(=\left(2x+2\right)^2-1^2\)
\(=\left(2x+2+1\right)\left(2x+2-1\right)\)
\(\left(2x+3\right)\left(2x+1\right)\)
c) \(x^2-a^2+2ab-b^2\)
\(=x^2-\left(a-b\right)^2\)
\(=\left(x+a-b\right)\left(x-a+b\right)\)
\(c)\)
\(a^3+b^3+c^3-3abc\)
\(=a^3+3ab\left(a+b\right)+b^3+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ab-ac+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(d)\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=[\left(a+b\right)c]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)[a\left(b+c\right)+c\left(b+c\right)]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
a ) \(x^3-7x-6=x^3-x-6x-6=x^3-x-6\left(x+1\right)\)
\(=x\left(x^2-1\right)-6\left(x+1\right)=\left(x+1\right)\left[x\left(x-1\right)-6\right]\)
\(=\left(x+1\right)\left[\left(x^2-x-6\right)\right]=\left(x+1\right)\left[\left(x^2+2x-3x-6\right)\right]\)
\(=\left(x+1\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]=\left(x+1\right)\left(x+2\right)\left(x+3\right)\)
b )
\(x^3-19x-30=\left(x^3-9x\right)-\left(10x+30\right)=x\left(x^2-9\right)-10\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-3x-10\right)=\left(x+2\right)\left(x+3\right)\left(x-5\right)\)
c )
\(a^3-6a^2+11a-6=\left(a-3\right)\left(a-2\right)\left(a-1\right).\)
Câu 1 :
a) \(x^3-5x^2-14x\)
\(=x^3-7x^2+2x^2-14x\)
\(=x^2\left(x-7\right)+2x\left(x-7\right)\)
\(=\left(x-7\right)\left(x^2+2x\right)\)
\(=x\left(x-7\right)\left(x+2\right)\)
b) \(a^4+a^2+1\)
\(=\left(a^2\right)^2+2a^2+1-a^2\)
\(=\left(a^2+1\right)-a^2\)
\(=\left(a^2-a+1\right)\left(a^2+a+1\right)\)
c) \(x^4+64\)
\(=\left(x^2\right)^2+2\cdot x^2\cdot8+8^2-2\cdot x^2\cdot8\)
\(=\left(x^2+8\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+8\right)\left(x^2+4x+8\right)\)
Câu 2 :
a) \(\left(a-b\right)^2=a^2-2ab+b^2\)
Ta có : \(\left(a+b\right)^2=a^2+2ab+b^2\)
\(\Rightarrow a^2+b^2=\left(a+b\right)^2-2ab=7^2-2\cdot14=25\)
\(\Rightarrow\left(a-b\right)^2=25-2\cdot12=1\)
b) tương tự