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\(a.\left(8x^4-4x^3+x^2\right):2x^2=4x^2-2x+\frac{1}{2}\)
\(b.\left(2x^4-x^3+3x^2\right):\left(-\frac{1}{3x^2}\right)=-6x^6+3x^5-9x^4\)
\(c.\left(-18x^3y^5+12x^2y^2-6xy^3\right):6xy=-3x^2y^4+2xy-y^2\)
\(d.\left(\frac{3}{4x^3y^6}+\frac{6}{5x^4y^5}-\frac{9}{10x^5y}\right):-\frac{3}{5x^3y}=-\frac{5}{4y^5}-\frac{2}{xy^4}-\frac{3}{2x^2}\)
a) x^2+2xy+y^2-16
=(x+y)2-16
=(x+y-4)(x+y+4)
b) 3x^2+5x-3xy-5y
=(3x2-3xy)+(5x-5y)
=3x(x-y)+5(x-y)
=(x-y)(3x+5)
c) 4x^2-6x^3y-2x^2+8x
ko bik hoặc sai đề
d) x^2-4-2xy+y^2
=(x-y)2-4
=(x-y+2)(x-y-2)
e) x^3-4x^2-12x+27
=sai đề
g) 3x^2-18x+27
=3(x2-6x+9)
=3(x-3)2
h) x^2-y^2-z^2-2yz
=x2-(y2+z2+2yx)
=x2-(y+z)2
=(x-y-z)(x+y+z)
k) 4x^2(x-6)+9y^2(6-x)
=4x2(x-6)-9y2(x-6)
=(x-6)(4x2-9y2)
=(x-6)(2x-3y)(2x+3y)
l)6xy+5x-5y-3x^2-3y^2
=(5x-5y)+(-3x2+6xy-3y2)
=5(x-y)-3(x2-2xy+y2)
=5(x-y)-3(x-y)2
=(x-y)(5-3(x-y))
=(x-y)(5-3x+3y)
a: \(\left(19x^2-14x^3+9-20x+2x^4\right):\left(1+x^2-4x\right)\)
\(=\left(2x^4-14x^3+19x^2-20x+9\right):\left(x^2-4x+1\right)\)
\(=\left(2x^4-8x^3+2x^2-6x^3+24x^2-6x-7x^2+28x-7-42x+16\right)\) :\(\left(x^2-4x+1\right)\)
\(=2x^2-6x-7+\frac{-42x+16}{x^2-4x+1}\)
b: \(\left(3x^4-2x^3-2x^2+4x+8\right):\left(x^2-2\right)\)
\(=\left(3x^4-6x^2-2x^3+4x+4x^2-8+16\right):\left(x^2-2\right)\)
\(=3x^2-2x+4+\frac{16}{x^2-2}\)
c: \(\left(2x^3-26x-24\right):\left(x^2+4x+3\right)\)
\(=\left(2x^3+8x^2+6x-8x^2-32x-24\right):\left(x^2+4x+3\right)\)
=2x-8
\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)
a: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
b: \(=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}\)
\(=\dfrac{y-x}{xy\left(y-x\right)}=\dfrac{1}{xy}\)
c: \(=\dfrac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(1-2x\right)}\)
\(=\dfrac{3\left(1+2x\right)}{2\left(x+4\right)}\)
d: \(=\dfrac{12x}{8x^3}\cdot\dfrac{15y^4}{5y^3}=\dfrac{3}{2x^2}\cdot3y=\dfrac{9y}{2x^2}\)
f: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)
a: \(x^3-x^2-14x+24\)
\(=x^3+x-12-x^2-15x+36\)
=>\(\left(x^3-x^2-14x+24\right):\left(x^3+x-12\right)=1+\frac{-x^2-15x+36}{x^3+x-12}\)
Để dư là 0 thì \(-x^2-15x+36=0\)
=>\(x^2+15x-36=0\) (1)
\(\Delta=15^2-4\cdot1\cdot\left(-36\right)=225+144=369>0\)
Do đó: (1) có hai nghiệm phân biệt là:
\(\left[\begin{array}{l}x=\frac{-15-\sqrt{369}}{2\cdot1}=\frac{-15-3\sqrt{41}}{2}\\ x=\frac{-15+3\sqrt{41}}{2}\end{array}\right.\)
b: \(x^5+4x^3+3x^2-5x+15\)
\(=x^5-x^3+3x^2+5x^3-5x+15=\left(x^3-x+3\right)\left(x^2+5\right)\)
=>\(\frac{x^5+4x^3+3x^2-5x+15}{x^3-x+3}=x^2+5\)
=>Đây là phép chia hết
c: \(2x^4+2x^3+3x^2-5x-20\)
\(=2x^4+2x^3+8x^2-5x^2-5x-20=\left(x^2+x+4\right)\left(2x^2-5\right)\)
=>\(\frac{2x^4+2x^3+3x^2-5x-20}{x^2+x+4}=2x^2-5\)
d: \(2x^4-14x^3+19x^2-20x+9\)
\(=2x^4-8x^3+2x^2-6x^3+24x^2-6x-7x^2+28x-7-42x+16\)
\(=\left(x^2-4x+1\right)\left(2x^2-6x-7\right)-42x+16\)
=>\(\frac{2x^4-14x^3+19x^2-20x+9}{x^2-4x+1}=2x^2-6x-7\) dư -42x+16
để dư bằng 0 thì -42x+16=0
=>-42x=-16
=>\(x=\frac{16}{42}=\frac{8}{21}\)
c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)
\(=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)
d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)
Tương tự với a ; b