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a) (x + 3y) (2x2y - 6xy2)
= (x + 3y) + 2xy (x - 3y)
= 2xy [(x + 3y) (x - 3y)]
= 2xy (x2 - 3y2)
b) (6x5y2 - 9x4y3 + 15x3y4) : 3x3y2
= (6x5y2 : 3x3y2) + (-9x4y3 : 3x3y2) + (15x3y4 : 3x3y2)
= [(6 : 3) (x5 : x3) (y2 : y2)] + [(-9 : 3) (x4 : x3) (y3 : y2)] + [(15 : 3) (x3 : x3) (y4 : y2)]
= 2x2 + (-3xy) + 5y2
= 2x2 - 3xy + 5y2
a: \(=15x^4-12x^3+9x^2\)
c: \(=5x^3-15x^2-4x^2+12x\)
\(=5x^3-19x^2+12x\)
Bài 3:
3: \(6x\left(x-y\right)-9y^2+9xy\)
\(=6x\left(x-y\right)+9xy-9y^2\)
\(=6x\left(x-y\right)+9y\left(x-y\right)\)
\(=\left(x-y\right)\left(6x+9y\right)\)
\(=3\left(2x+3y\right)\left(x-y\right)\)
Bài 4:
$(x-1)(x^2+x+1)-x^3-6x=11$
Dùng $(x-1)(x^2+x+1)=x^3-1$:
$x^3-1-x^3-6x=11$
$-6x-1=11$
$-6x=12$
$x=-2$
Vậy $x=-2$.
2.$16x^2-(3x-4)^2=0$
$(4x)^2-(3x-4)^2=0$
$(4x-3x+4)(4x+3x-4)=0$
$(x+4)(7x-4)=0$
$x=-4$ hoặc $x=\dfrac47$
Vậy $x=-4,\dfrac47$.
3.$x^3-x^2+3-3x=0$
$=x^2(x-1)-3(x-1)=0$
$=(x-1)(x^2-3)=0$
$x-1=0$ hoặc $x^2-3=0$
$x=1$ hoặc $x=\pm\sqrt3$
Vậy $x=1,\sqrt3,-\sqrt3$.
4.$\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}$
Điều kiện: $x\ne-2,-1$.
$(x-1)(x+1)=(x+2)^2$
$x^2-1=x^2+4x+4$
$-4x=5$
$x=-\dfrac54$
Vậy $x=-\dfrac54$.
5.$\dfrac1{x+2}+\dfrac2{x+1}=0$
Điều kiện: $x\ne-2,-1$.
$\dfrac{x+1+2(x+2)}{(x+2)(x+1)}=0$
$x+1+2x+4=0$
$3x+5=0$
$x=-\dfrac53$
Vậy $x=-\dfrac53$.
6.$\dfrac{9-x^2}{x}:(x-3)=1$
Điều kiện: $x\ne0,3$.
$\dfrac{9-x^2}{x(x-3)}=1$
$9-x^2=x(x-3)$
$9-x^2=x^2-3x$
$2x^2-3x-9=0$
$(2x+3)(x-3)=0$
$x=-\dfrac32$ hoặc $x=3$
Nhưng $x=3$ không thỏa điều kiện.
Vậy $x=-\dfrac32$.
a/ (x-1)2-(4x+3)(2-x)=x2-2x+1-(8x-4x2+6-3x)
=x2-2x+1-8x+4x2-6+3x=5x2-7x-6
b/ (15x3y2 - 6x2y3) : 3x2y2 = 5x - 2y
c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)
sau bạn đăng tách ra cho mn cùng giúp nhé
a, \(\left(-2x^5+3x^2-4x^3\right):2x^2=-x^3+\frac{3}{2}-2x\)
b, \(\left(x^3-2x^2y+3xy^2\right):\left(-\frac{1}{2}x\right)=-\frac{x^2}{2}+xy-\frac{3y^2}{2}\)
c, \(\left(3x^2y^2+6x^3y^3-12xy^2\right):3xy=xy+2x^2y^2-4y\)
d, \(\left(4x^3-3x^2y+5xy^2\right):\frac{1}{2}x=2x^2-\frac{3xy}{2}+\frac{5y^2}{2}\)
e, \(\left(18x^3y^5-9x^2y^2+6xy^2\right):3xy^2=6x^2y^3-3x+2\)
f, \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)=x^2+y^2\)
a: \(=\dfrac{2xy\left(2x^2y-4x+5\right)}{2xy}=2x^2y-4x+5\)
b: \(=\dfrac{x^2y\left(7x^2y-2y-5x^2y^3\right)}{3x^2y}=\dfrac{7}{3}x^2y-\dfrac{2}{3}y-\dfrac{5}{3}x^2y^3\)
a: \(\left(5x-3\right)\left(5x+3\right)=25x^2-9\)
b: \(=2x^2-12x^3+15x^2-18x=-12x^3+17x^2-18x\)
c: \(\dfrac{16x^3y^4}{4x^2y^2}=4xy^2\)