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x3 + y3 + z3 - 3xyz
= (x3 + 3x2y + 3xy2 + y3) + z3 - (3x2y + 3xy2 + 3xyz)
= (x + y)3 + z3 - 3xy(x + y + z)
= (x + y + z)[(x + y)2 - z(x + y) + z2] - 3xy(x + y + z)
= ( x + y + z)(x2 + 2xy + y2 - xz - zy + z2 - 3xy)
= (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
b. (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x - 1)(x - 4)(x - 2)(x - 3) - 3
= (x2 - 5x + 4)(x2 - 5x + 6) - 3
Đặt t = x2 - 5x + 4
=> Đa thức
<=> t.(t + 2) - 3
= t2 + 2t - 3
= t2 + 3t - t - 3
= t.(t + 3) - (t + 3)
= (t + 3)(t + 1) (1)
Thay t = x2 - 5x + 4 vào (1):
=> (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x2 - 5x + 4 + 3)(x2 - 5x + 4 + 1)
= (x2 - 5x + 7)(x2 - 5x + 5)
a) \(x^4+2x^3-4x-4=\left[\left(x^2\right)^2-4\right]+\left(2x^3-4x\right)\)
\(=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2+2+2x\right)\left(x^2-2\right)\)
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)=x^2\left(x+1\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)=\left(x^2-2\right)\left(x^2+2x+2\right)\)
b) \(x^2+y^2-x^2y^2+xy-x-y=\left(x^2-x^2y^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)
\(=x^2\left(1-y\right)\left(1+y\right)-y\left(1-y\right)-x\left(1-y\right)=\left(1-y\right)\left(x^2+x^2y-y-x\right)\)
\(=\left(1-y\right)\left[\left(x-1\right)x+y\left(x-1\right)\left(x+1\right)\right]=\left(1-y\right)\left(x-1\right)\left(x+xy+y\right)\)
c) Không phân tích được.
Bạn tải ứng dụng PhotoMath về nha. Ứng dụng này sẽ giải toán số chi tiết
a) \(x^3-4x^2-12x+27\)
\(=\left(x^3+27\right)-\left(4x^2+12x\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
b) \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
a) \(9x^2+6xy+y^2=\left(3x+y\right)^2\)
b) \(6x-9-x^2=-\left(x-3\right)^2\)
a) (x+y)2-(x-y)2
=(x+y)(x-y)
b)(3x+1)2-(x+1)2
=[(3x+1)+(x+1)].[(3x+1)-(x+1)]
=(3x+1+x+1)(3x+1-x-1)
a) \(x^4+5x^3+10x-4\)
\(=\left(x^4+2x^2\right)+\left(5x^3+10x\right)-\left(2x^2+4\right)\)
\(=x^2\left(x^2+2\right)+5x\left(x^2+2\right)-2\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2+5x-2\right)\)
\(=\left(x^2+2\right)\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}-\frac{25}{4}-2\right)\)
\(=\left(x^2+2\right)\left[\left(x+\frac{5}{2}\right)^2-\frac{33}{4}\right]\)
\(=\left(x^2+2\right)\left[\left(x+\frac{5}{2}\right)^2-\left(\frac{\sqrt{33}}{2}\right)^2\right]\)
\(=\left(x^2+2\right)\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x^2+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)
b) \(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+2xy-zx-zy+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-zx-zy\right)\)
a) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=2x.2\left(2x+1\right)=4x\left(2x+1\right)\)
a) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1-x-1\right)\left(3x+1+x+1\right)=2x\left(4x+2\right)=4x\left(2x+1\right)\)
b) \(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+2xy+xz+yz+z^2\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=\frac{\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]}{2}\)
câu a còn thiếu thì phải bạn xem lại xem
a. Đề phải là thế này:
x3 + y3 + z3 - 3xyz
= (x3 + 3x2y + 3xy2 + y3) + z3 - (3x2y + 3xy2 + 3xyz)
= (x + y)3 + z3 - 3xy(x + y + z)
= (x + y + z)[(x + y)2 - z(x + y) + z2] - 3xy(x + y + z)
= ( x + y + z)(x2 + 2xy + y2 - xz - zy + z2 - 3xy)
= (x + y + z)(x2 + y2 + z2 - xy - yz - xz)
b. (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x - 1)(x - 4)(x - 2)(x - 3) - 3
= (x2 - 5x + 4)(x2 - 5x + 6) - 3
Đặt t = x2 - 5x + 4
=> Đa thức
<=> t.(t + 2) - 3
= t2 + 2t - 3
= t2 + 3t - t - 3
= t.(t + 3) - (t + 3)
= (t + 3)(t + 1) (1)
Thay t = x2 - 5x + 4 vào (1):
=> (x - 1)(x - 2)(x - 3)(x - 4) - 3
= (x2 - 5x + 4 + 3)(x2 - 5x + 4 + 1)
= (x2 - 5x + 7)(x2 - 5x + 5)