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a: \(A=x^3y-12xy-x^2y\)
\(=xy\cdot x^2-xy\cdot12-xy\cdot x\)
\(=xy\left(x^2-x-12\right)\)
\(=xy\left(x^2-4x+3x-12\right)\)
\(=xy\left[x\left(x-4\right)+3\left(x-4\right)\right]\)
\(=xy\left(x-4\right)\left(x+3\right)\)
c: \(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-120\)
=(x+1)(x+4)(x+2)(x+3)-120
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-120\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)-96\)
\(=\left(x^2+5x+16\right)\left(x^2+5x-6\right)\)
\(=\left(x^2+5x+16\right)\left(x+6\right)\left(x-1\right)\)
d: \(D=x^5-x^4+x^2-1\)
\(=\left(x^5-x^4\right)+\left(x^2-1\right)\)
\(=x^4\left(x-1\right)+\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^4+x+1\right)\)
a: \(x^4+4=\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
b: \(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
c: \(x^8+x^4+1\)
\(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4-x^2+1\right)\cdot\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2+1-x\right)\left(x^2+1+x\right)\)
c) 2x2(x +1) + 4x(x +1); d) 2 x(y - 1) - 2
1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
\(a,=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\\ b,=\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\\ c,=\left[3x-2y-2\left(x+y\right)\right]\left[3x-2y+2\left(x+y\right)\right]\\ =5x\left(x-4y\right)\\ d,=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\\ =\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\\ =\left(x-5y\right)\left(5x-y\right)\\ f,=\left(x+3\right)\left(x^2-3x+9\right)\\ g,=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\\ h,=\left(5x-1\right)\left(25x^2+5x+1\right)\)
\(a)x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)\\ b)x^2-3y^2=\\ c)(3x-2y)^2-4(x+y)^2=(3x-2y)^2-[2(x+y)]^2\\=(3x-2y+2x+2y)(3x-2y-2x-2y)=5x(x-4y)\\ d)9(x-y)^2-4(x+y)^2=[3(x-y)]^2-[2(x+y)]^2=(3x-3y+2x+2y)(3x-3y-2x-2y)\\=(5x-y)(x-5y)\\ f)x^3+27=(x+3)(x^2-3x+9)\\ g)27x^3-0,001=(3x-0,1)(9x+0,3x+0,01)\\ h)125x^3-1=(5x-1)(25x^2+5x+1)\)
a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)
b: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
a: \(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left[\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\right]\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
b: \(x^2-1-xy+y\)
\(=\left(x-1\right)\left(x+1\right)-y\left(x-1\right)\)
\(=\left(x-1\right)\left(x-y+1\right)\)
c: Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)^2\cdot\left(x-2\right)\)
\(=\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3-x-1\right)\)
\(=2\cdot\left(x-1\right)\cdot\left(x-2\right)^2\)
Lời giải:
a. Không phân tích được nữa
b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$
$=x[(x^2+2xy+y^2)-4]=x[(x+y)^2-2^2]=x(x+y-2)(x+y+2)$
a: \(x\left(x^3-4\right)+2x^3-4\)
\(=x^4-4x+2x^3-4\)
\(=\left(x^2-2\right)\left(x^2+2\right)-2x\left(x^2+2\right)\)
\(=\left(x^2+2\right)\left(x^2-2x-2\right)\)
b: \(\left(1+2x\right)\left(1-2x\right)-x\left(x+2\right)\left(x-2\right)\)
\(=1-4x^2-x\left(x^2-4\right)\)
\(=1-4x^2-x^3+4x\)
\(=-\left(x^3-1+4x^2-4x\right)\)
\(=-\left[\left(x-1\right)\left(x^2+x+1\right)+4x\left(x-1\right)\right]\)
\(=-\left(x-1\right)\left(x^2+5x+1\right)\)
a) \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-3x+2y\right)\)
\(=0\cdot0\)
\(=0\)
d) \(9\left(x-y\right)^2-4\left(x+y\right)^2\)
\(=\left(3x-3y\right)^2-\left(2x+2y\right)^2\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)\)
\(=\left(x-5y\right)\left(5x-y\right)\)
e) \(\left(4x^2-4x+1\right)-\left(x+1\right)^2\)
\(=\left(2x-1\right)^2-\left(x+1\right)^2\)
\(=\left(2x-1+x+1\right)\left(2x-1-x-1\right)\)
\(=3x\left(x-2\right)\)
f) \(x^3+27\)
\(=x^3+3^3\)
\(=\left(x+3\right)\left(x^2-3x+9\right)\)
g) \(27x^3-0,001\)
\(=\left(3x\right)^3-\left(0,1\right)^3\)
\(=\left(3x-0,1\right)\left(9x^2+0,3x+0,01\right)\)
h) \(125x^3-1\)
\(=\left(5x\right)^3-1^3\)
\(=\left(5x-1\right)\left(25x^2+5x+1\right)\)
c) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2\)
\(=\left(3x-2y+2x-3y\right)\left(3x-2y-2x+3y\right)\)
\(=\left(5x-5y\right)\left(x+y\right)\)
\(=5\left(x+y\right)\left(x-y\right)\)
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\left(1\right)=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-15=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
Đặt \(t=x^2+5x+4\)
(1) trở thành: \(t\left(t+2\right)-15=t^2+2t+1-16=\left(t+1\right)^2-4^2=\left(t-3\right)\left(t+5\right)\)
Thay t: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-15=\left(x^2+5x+4-3\right)\left(x^2+5x+4+5\right)=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
b) \(\left(2x+5\right)^2-\left(x-9\right)^2=\left(2x+5-x+9\right)\left(2x+5+x-9\right)=\left(x+14\right)\left(3x-4\right)\)
a: Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+3\right)\left(x+4\right)-15\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-15\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24-15\)
\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+9\)
\(=\left(x^2+5x+1\right)\left(x^2+5x+9\right)\)
b: \(\left(2x+5\right)^2-\left(x-9\right)^2\)
\(=\left(2x+5-x+9\right)\left(2x+5+x-9\right)\)
\(=\left(x+15\right)\left(3x-4\right)\)
a) x4 + 4 = (x4 + 4x2 + 4) - 4x2 = (x2 + 2)2 - 4x2 = (x2 + 2x + 2)(x2 - 2x + 2)
b) (x + 2)(x + 3)(x + 4)(x + 5) - 24 = (x + 2)(x + 5)(x + 3)(x + 4) - 24
= (x2 + 7x + 10)(x2 + 7x + 12) - 24
Đặt x2 + 7x + 10 = y => y(y + 2) - 24 = y2 + 2y - 24
= y2 + 6y - 4y - 24 = (y - 4)(y + 6) = (x2 + 7x + 10 - 4)(x2 + 7x + 10 + 6)
= (x2 + 7x + 6)(x2 + 7x + 16) = (x2 + x + 6x + 6)(x2 + 7x + 16) = (x + 1)(x + 6)(x2 + 7x + 16)
ko làm mà đòi có ăn :)