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a,Ta có:
x³ + y³ + z³ - 3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz)
b, Từ:
x + y + z = 0
=> x + y = -z
<=> (x + y)^3 = (-z)^3
<=> x^3 + 3x^2y + 3xy^2 + y^3 = -z^3
<=> x^3 + y^3 + z^3 = -3x^2y - 3xy^2
<=> x^3 + y^3 + z^3 = -3xy(x+y)
<=> x^3 + y^3 + z^3 = -3xy(-z)
<=> x^3 + y^3 + z^3 = 3xyz
2 \(x^7+x^5+1=x^7+x^6+x^5-x^6+1=x^5\left(x^2+x+1\right)-\left(x^6-1\right)=x^5\left(x^2+x+1\right)-\left(x^3-1\right)\left(x^3+1\right)\)
\(=x^5\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)=\left(x^2+x+1\right)\left(x^5-\left(x-1\right)\left(x^3+1\right)\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
1 \(x^3-5x^2+3x+9=x^3+x^2-6x^2-6x+9x+9=x^2\left(x+1\right)-6x\left(x+1\right)+9\left(x+1\right)\)
\(=\left(x^2-6x+9\right)\left(x+1\right)=\left(x-3\right)^2\left(x+1\right)\)
Câu a trước đi ạ ^^
a) 7x - 6x2 - 2
= - 6x2 + 7x - 2
= (- 6x2 + 3x) + (4x - 2)
= 3x (- 2x + 1) + 2 (2x-1)
= - 3x ( 2x -1) + 2 (2x - 1)
= ( 2x -1 ) ( - 3x +2 )
\(x^4+x^3+2x^2+x+1\)
\(=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^2+1+x\right)\)
x^4+x^3+2x^2+x+1
=(x^4+2x^2+1)+(x^3+x)
=(x^2+1)^2+x(x^2+1)
=(x^2+1)(x^2+x+1)
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
\(\text{a) }3x-7x+2\\ \\=2-4x\\ \\=2\left(1-2x\right)\)
\(\text{b) }x^4+4\\ \\=x^4+4+4x^2-4x^2\\ \\=\left(x^4+4x^2+4\right)-4x^2\\ \\=\left(x^2+2\right)^2-\left(2x\right)^2\\ \\=\left(x^2+2+2x\right)\left(x^2+2-2x\right)\\ \)
\(\text{c) }4x-8y=4\left(x-2y\right)\)
\(\text{d) }x^3-x^2y-x+y\\ \\=\left(x^3-x^2y\right)-\left(x-y\right)\\ \\=x^2\left(x-y\right)-\left(x-y\right)\\ =\left(x-y\right)\left(x^2-1\right)\\ \\=\left(x-y\right)\left(x-1\right)\left(x+1\right)\)