Đáp án:
Giải thích các bước giải:
BT7:
a) `a^2 - ab + 2a - 2b`
`= (a^2 - ab) + (2a - 2b)`
`= a(a - b) + 2(a - b)`
`= (a - b)(a + 2)`
b) `x^4 - 4x^3 + 4x^2 - 16x`
`= (x^4 - 4x^3) + (4x^2 - 16x)`
`= x^3(x - 4) + 4x(x - 4)`
`= (x - 4)(x^3 + 4x)`
`= x(x - 4)(x + 4)`
BT8:
a) `4x^2 - 4xy + y^2 - 36`
`= (4x^2 - 4xy + y^2) - 36`
`= (2x - y)^2 - 6^2`
`= (2x - y - 6)(2x - y + 6)`
b) `-9x^2 - 6xy - y^2 + 25`
`= -(9x^2 + 6xy + y^2 - 25)`
`= -[(9x^2 + 6xy + y^2) - 25]`
`= -[(3x + y)^2 - 5^2]`
`= -(3x + y - 5)(3x + y + 5)`
BT9:
a) `2(x - 4) - 4x + x^2`
`= 2(x - 4) + x^2 - 4x`
`= 2(x - 4) + x(x - 4)`
`= (x - 4)(x + 2)`
b) `7x^2 - 7xy - 4x + 4y`
`= (7x^2 - 7xy) - (4x - 4y)`
`= 7x(x - y) - 4(x - y)`
`= (x - y)(7x - 4)`
BT10:
TC: `a + b + c = 0`
`=> (a + b + c)^3 = 0`
`=> a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3b^2c + 3bc^2 + 3ac^2 + 3a^2c + 6abc = 0`
`=> a^3 + b^3 + c^3 + (3a^2b + 3ab^2 + 3abc) + (3a^2c + 3ac^2 + 3abc) + (3b^2c + 3bc^2 + 3abc) - 3abc = 0`
`=> a^3 + b^3 + c^3 + 3ab(a + b + c) + 3ac(a + b + c) + 3bc(a + b + c) - 3abc = 0`
Do `a + b + c = 0`
`=> a^3 + b^3 + c^3 - 3abc = 0`
`=> a^3 + b^3 + c^3 = 3abc` (đpcm)
Study well
