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a: \(=\dfrac{5}{3}x^2-x+\dfrac{1}{3}\)
b: \(=-5y-9+xy\)
\(\dfrac{\left(x+y\right)2}{x2+xy}+\dfrac{\left(x-y\right)2}{x2-xy}=-\left(\dfrac{\left(x-y\right)2}{x2-xy}\right)+\dfrac{\left(x-y\right)2}{x2-xy}=0\)
b: \(\dfrac{x^2-4x}{xy-4x-3y+12}+\dfrac{x-2}{y-4}\)
\(=\dfrac{x\left(x-4\right)}{\left(y-4\right)\left(x-3\right)}+\dfrac{x-2}{y-4}\)
\(=\dfrac{x^2-4x+x^2-5x+6}{\left(y-4\right)\left(x-3\right)}=\dfrac{2x^2-9x+6}{\left(y-4\right)\left(x-3\right)}\)
c: \(=\dfrac{y^2}{\left(y-5\right)\left(x+1\right)}+\dfrac{2}{x+1}\)
\(=\dfrac{y^2+2y-10}{\left(y-5\right)\left(x+1\right)}\)
a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)
b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)
1.
\(a,\left(-xy\right)\left(-2x^2y+3xy-7x\right)\)
\(=2x^3y^2-3x^2y^2+7x^2y\)
\(b,\left(\dfrac{1}{6}x^2y^2\right)\left(-0,3x^2y-0,4xy+1\right)\)
\(=-\dfrac{1}{20}x^4y^3-\dfrac{1}{15}x^3y^3+\dfrac{1}{6}x^2y^2\)
\(c,\left(x+y\right)\left(x^2+2xy+y^2\right)\)
\(=\left(x+y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3\)
\(d,\left(x-y\right)\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3\)
\(=x^3-3x^2y+3xy^2-y^3\)
2.
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)\)
\(=x^3-y^3\)
\(b,\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=x^3+y^3\)
\(c,\left(4x-1\right)\left(6y+1\right)-3x\left(8y+\dfrac{4}{3}\right)\)
\(=24xy+4x-6y-1-24xy-4x\)
\(=\left(24xy-24xy\right)+\left(4x-4x\right)-6y-1\)
\(=-6y-1\)
#Toru
\(a,=12x^3y-4x^2y^2+3xy^3\\ b,=x^3+3x^2-5x+3x^2+9x-15-x^3-4x^2+4x\\ =2x^2+8x-15\)
b: Ta có: \(\left(x+3\right)\left(x^2+3x-5\right)-x\left(x-2\right)^2\)
\(=x^3+3x^2-5x+3x^2+9x-15-x^3+4x^2-4x\)
\(=10x^2-15\)
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)
a) \(\left(x^5+4x^3-6x^2\right):4x^2\)
\(=\left(x^5:4x^2\right)+\left(4x^3:4x^2\right)+\left(-6x^2:4x^2\right)\)
\(=\dfrac{1}{4}x^3+x-\dfrac{3}{2}\)
b)
Vậy \(\left(x^3+x^2-12\right):\left(x-2\right)=x^2+3x+6\)
c) (-2x5 : 2x2) + (3x2 : 2x2) + (-4x^3 : 2x^2)
= \(-x^3+\dfrac{3}{2}-2x\)
d) \(\left(x^3-64\right):\left(x^2+4x+16\right)\)
\(=\left(x-4\right)\left(x^2+4x+16\right):\left(x^2+4x+16\right)\)
\(=x-4\)
(dùng hẳng đẳng thức thứ 7)
Bài 2 :
a) 3x(x - 2) - 5x(1 - x) - 8(x2 - 3)
= 3x2 - 6x - 5x + 5x2 - 8x2 + 24
= (3x2 + 5x2 - 8x2) + (-6x - 5x) + 24
= -11x + 24
b) (x - y)(x2 + xy + y2) + 2y3
= x3 - y3 + 2y3
= x3 + y3
c) (x - y)2 + (x + y)2 - 2(x - y)(x + y)
= (x - y)2 - 2(x - y)(x + y) + (x + y)2
= [(x - y) + x + y)2 = [x - y + x + y] = (2x)2 = 4x2
Bài 1 :
a]= \(\frac{1}{4}\)x3 + x - \(\frac{3}{2}\).
b] => [x3 + x2 -12 ] = [ x2 +3 ][x-2] + [-6]
c]= -x3 -2x +\(\frac{3}{2}\).
d] = [ x3 - 64 ] = [ x2 + 4x + 16][ x- 4].
a) \(\left(2x+3y\right)^2=\left(2x\right)^2+2\cdot2x\cdot3y+\left(3y\right)^2=4x^2+12xy+9y^2\)
b) \(\left(x+\dfrac{1}{4}\right)^2=x^2+2\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)
c) \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=\left(x^2\right)^2-\left(\dfrac{2}{5}y\right)^2=x^4-\dfrac{4}{25}y^2\)
d) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y^2+3\cdot2x\cdot\left(y^2\right)^2+\left(y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
e) \(\left(3x^2-2y\right)^2=\left(3x^2\right)^2-2\cdot3x^2\cdot2y+\left(2y\right)^2=9x^4-12x^2y+4y^2\)
f) \(\left(x+4\right)\left(x^2-4x+16\right)=x^3+4^3=x^3+64\)
g) \(\left(x^2-\dfrac{1}{3}\right)\cdot\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3=x^6-\dfrac{1}{27}\)
a ) \(\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)
b ) \(\left(x^2-2xy+y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x-y\right)=\left(x-y\right)^3\)
c ) \(\left(x^2y^2-\dfrac{1}{3}xy+3y\right)\left(x-3y\right)\)
\(=\left(x^2y^2-\dfrac{1}{3}xy+3y\right)x-3y\left(x^2y^2-\dfrac{1}{3}xy+3y\right)\)
\(=x^3y^2-\dfrac{1}{3}x^2y+3xy-3x^2y^3+xy^2-9y^2\)
d ) \(\left(\dfrac{1}{5}x-1\right)\left(x^2-5x+2\right)\)
\(=\dfrac{1}{5}x\left(x^2-5x+2\right)-x^2+5x-2\)
\(=\dfrac{1}{5}x^3-x^2+\dfrac{2}{5}x-x^2+5x-2\)
\(=\dfrac{1}{5}x^3-2x^2+\dfrac{27}{5}x-2\)
\(a,\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-y^3\)