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a) = (x + 3)2 - y2 = (x + 3 - y)(x + 3 + y)
b) = x2(x - 3) -4(x - 3) = (x - 3)(x2 - 4) = (x - 3)(x - 2)(x + 2)
c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)
d) Nhầm đề. tui sửa lại x3 + y3 + 2x2 - 2xy + 2y2
= x3 + y3 + 2(x2 - xy + y2) = (x + y)(x2 - xy + y2) + 2(x2 - xy + y2) = (x2 - xy + y2)(x + y + 2)
e) = x4 - x3 - x3 + x2 - x2 + x + x - 1 = x3(x - 1) - x2(x - 1) - x(x - 1) + x - 1 = (x - 1)(x3 - x2 - x + 1) = (x - 1)(x - 1)(x2 - 1) = (x - 1)3(x + 1)
f) = x3 - 3x2 - x2 + 3x + 9x - 27 = x2(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x2 - x + 9)
g) chắc là 3xyz
= x2y + xy2 + y2z + yz2 + x2z + xz2 + 3xyz = x2y + xy2 + xyz + y2z + yz2 + xyz + x2z + xz2 + xyz = (x + y + z)(xy + yz + xz)
h) = 23 -(3x)3 = (2 - 3x)(4 + 6x + 9x2)
i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy
k) = (x3 - y3)(x3 + y3) = (x - y)(x2 + xy +y2)(x + y)(x2 - xy +y2).
Bài 1:
a) Ta có: \(x^3+3x^2+3x+2=0\)
\(\Leftrightarrow x^3+2x^2+x^2+2x+x+2=0\)
\(\Leftrightarrow x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên x+2=0
hay x=-2
Vậy: x=-2
b) Ta có: \(x^3-12x^2+48x-72=0\)
\(\Leftrightarrow x^3-6x^2-6x^2+36x+12x-72=0\)
\(\Leftrightarrow x^2\left(x-6\right)-6x\left(x-6\right)+12\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-6x+12\right)=0\)
mà \(x^2-6x+12>0\forall x\)
nên x-6=0
hay x=6
Vậy: x=6
\(\dfrac{8x^3y^2-6x^2y^3}{-2xy}=\dfrac{8x^3y^2}{-2xy}+\dfrac{6x^2y^3}{2xy}=-4x^2y+3xy^2\)
⇒ Chọn A.
a) \(\left(xy+1\right)^2-\left(x+y\right)^2\)
\(=\left(xy+1-x+y\right)\left(xy+1+x-y\right)\)
b) \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(x+y-x+y\right)\left[\left(x^2+2xy+y^2\right)+x^2-y^2+\left(x^2-2xy+y^2\right)\right]\)
\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
c) \(3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=3y^2\left(x^4+x^3+x+1\right)\)
d) \(4\left(x^2-y^2\right)-8\left(x-ay\right)-4\left(a^2-1\right)\)
\(=4\left[\left(x^2-y^2\right)-2\left(x-ay\right)-\left(a^2-1\right)\right]\)
\(=4\left[\left(x^2-y^2\right)-\left(2x-2ay\right)-\left(a^2-1\right)\right]\)
\(=4\left(x^2-y^2-2x+2ay-a^2+1\right)\)
P/s: Ko chắc!
c/
\(=3y^2\left(x^4+x^3+x+1\right)\)
\(=3y^2\left[x^3\left(x+1\right)+x+1\right]\)
\(=3y^2\left(x^3+1\right)\left(x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
d/
\(=\left(4x^2-8x+4\right)-\left(4y^2-8ay+4a^2\right)\)
\(=4\left(x-1\right)^2-4\left(y-a\right)^2\)
\(=4\left[\left(x-1\right)^2-\left(y-a\right)^2\right]\)
\(=4\left(x-1-y+a\right)\left(x-1+y-a\right)\)
Bài 4:
a: \(=7xy\left(2-3-4\right)=-35xy\)
b: \(=\left(x-5\right)\left(x+y\right)\)
c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)
d: \(=\left(x+y\right)^3-\left(x+y\right)\)
=(x+y)(x+y+1)(x+y-1)
e: =x^2+8x-x-8
=(x+8)(x-1)
f: \(=2x^2-4x+x-2=\left(x-2\right)\left(2x+1\right)\)
g: =-5x^2+15x+x-3
=(x-3)(-5x+1)
h: =x^2-3xy+xy-3y^2
=x(x-3y)+y(x-3y)
=(x-3y)*(x+y)
Bài 4:
a: \(=7xy\left(2-3-4\right)=-35xy\)
b: \(=\left(x-5\right)\left(x+y\right)\)
c: \(=10x\left(x-y\right)+8\left(x-y\right)=2\left(x-y\right)\left(5x+4\right)\)
d: \(=\left(x+y\right)^3-\left(x+y\right)\)
=(x+y)(x+y+1)(x+y-1)
e: =x^2+8x-x-8
=(x+8)(x-1)
f: \(=2x^2-4x+x-2=\left(x-2\right)\left(2x+1\right)\)
g: =-5x^2+15x+x-3
=(x-3)(-5x+1)
h: =x^2-3xy+xy-3y^2
=x(x-3y)+y(x-3y)
=(x-3y)*(x+y)
Ta có : x3 - 7x + 6
= x3 - x - 6x + 6
= x(x2 - 1) - 6(x - 1)
= x(x + 1)(x - 1) - 6(x - 1)
= (x - 1) [x(x + 1) - 6]
= (x - 1) (x2 + x - 6) .
CÁC Ý SAU TƯƠNG TỰ
Câu c sai đề rồi kìa :)))
Câu c phải là \(\left(\frac{x}{2}-y\right)^3\) chứ không phải \(\left(\frac{4}{2}-2\right)^3\)