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B = \(\frac{8xy-6x^2}{3y\left(3x-4y\right)}=\frac{2x\left(4y-3x\right)}{-3y\left(4y-3x\right)}=-\frac{2x}{3y}\)
C = \(\frac{2x^3-18x}{x^4-81}=\frac{2x\left(x^2-9\right)}{\left(x^2-9\right)\left(x^2+9\right)}=\frac{2x}{x^2+9}\)
a. 5y(2y-1) - (3y+2)(3-3y)
=10y\(^2\) -5y - (9y-9y\(^2\)+6-6y)
=10y\(^2\) -5y - 9y + 9y\(^2\) - 6 + 6y
=19y\(^2\)- 8y -6
b. (6x+1)2 - 2(6x+1)(6x-1) + (6x-1)2
= (6x+1-6x+1)\(^2\)
= 2\(^2\) = 4
c. (2x+3) - 2(2x+3)(x-2) + (x-2)2
= 2x+3 - 2(2x\(^2\)-4x+3x-6) + (x\(^2\)- 4x + 4)
= 2x +3 - 4x\(^2\) + 8x - 6x +12 + x\(^2\)- 4x + 4
= -3x\(^2\) +19
1)a)3(2x-1)(3x-1)-(2x-3)(9x-1)=0
<=>18x2-15x+1-18x2+29x-3=0
<=>14x-2=0
<=>14x=2
<=>x=1/7
b)4(x+1)2+(2x-1)2-8(x-1)(x+1)=11
<=>4x2+8x+4+4x2-4x+1-8x2+8=11
<=>4x+13=11
<=>4x=11-13
<=>4x=-2
<=>x=-1/2
c)Sai đề phải là dấu - chứ không phải +
(x-3)(x2+3x+9)-x(x-2)(x+2)=1
<=>x3-27-x3+4x=1
<=>4x=1+27
<=>4x=28
<=>x=7
2)a)(2x-3y)(2x+3y)-4(x-y)2-8xy
=4x2-9y2-4x2+8xy-4y2-8xy
=-13y2
b)(x-2)3-x(x+1)(x-1)+6x(x-3)
=x3-6x2+12x+8-x3+x+6x2-18x
=8-5x
c)(x-2)(x2-2x+4)(x+2)(x2+2x+4)
=(x-2)(x2+2x+4)(x+2)(x2-2x+4)
=(x3-8)(x3+8)
=x6-64
\(a,\dfrac{3x\left(1-x\right)}{1\left(x-1\right)}=\dfrac{-3x\left(x-1\right)}{x-1}=-3x\)
\(b,\dfrac{6x^2y^2}{8xy^3}=\dfrac{3x.2xy^2}{4y.2xy^2}=\dfrac{3x}{4y}\)
\(c,\dfrac{3\left(x-y\right)\left(x-z\right)^2}{6\left(x-y\right)\left(x-z\right)}=\dfrac{x-z}{2}\)
6x(x+3y-1)-6x2-8xy
=6x2+18xy-6x-6x2-8xy
=10xy-6x
=2x(5y-3)