\(=8xy-16x^2y+24xy^2\)

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 $x=-2$.

$16x^2-(3x-4)^2=0$

$(4x)^2-(3x-4)^2=0$

$(4x-3x+4)(4x+3x-4)=0$

$(x+4)(7x-4)=0$

$x=-4$ hoặc $x=\dfrac47$

Vậy $x=-4,\dfrac47$.

$x^3-x^2+3-3x=0$

$=x^2(x-1)-3(x-1)=0$

$=(x-1)(x^2-3)=0$

$x-1=0$ hoặc $x^2-3=0$

$x=1$ hoặc $x=\pm\sqrt3$

Vậy $x=1,\sqrt3,-\sqrt3$.

$\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}$

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 $x=-\dfrac54$.

$\dfrac1{x+2}+\dfrac2{x+1}=0$

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 $x=-\dfrac53$.

$\dfrac{9-x^2}{x}:(x-3)=1$

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= 3x² y-11x²-5y.8xy-5+6

=(3-11-5.8-5+6).(x².x².x).(y.y.y)

=-47x⁵y³

(2x-3y)(2x-3y)+(3x-2y)²

=(2x-3y)²-(3x-2y)²

=4x²-12xy+9y²-9x²+12xy-4y²

= -5x²+5y²

=5y²-5x²

=5(y²-x²)

Giúp bạn ấy đi mọi người!

2 tụi mình khác nhau nha!

a) (4x-1)(2-x)-(2x-1)²

= 8x-4x²-2+x-(4x²-4x+1) = -8x²+13x-3

b) (15x⁴y⁵-30x³y⁴+35x³y⁴):(5x³y³)

= 3xy²-6y+7y = 3xy²+y

a: \(=8x-4x^2-2+2x-4x^2+4x-1\)

\(=-8x^2+14x-3\)

Lời giải:

\(\frac{2}{x^2y}-\frac{4xy^2-1}{2x^3y^3}=\frac{4xy^2}{2x^3y^3}-\frac{4xy^2-1}{2x^3y^3}=\frac{4xy^2-(4xy^2-1)}{2x^3y^3}=\frac{1}{2x^3y^3}\)

\(\left(2x+3\right)\left(x-5\right)+2x\left(3-x\right)+x-10\)

\(=\left(2x^2+3x-10x-15\right)+\left(6x-2x^2\right)+x-10\)

\(=-25\)

\(\frac{4xy-5}{10x^3y}-\frac{6y^2-5}{10x^3y}=\frac{\left(4xy-5\right)-\left(6y^2-5\right)}{10x^3y}=\frac{4xy-6y^2}{10x^3y}=\frac{2y\left(2x-3y\right)}{2y.5x^3}=\frac{2x-3y}{5x^3}\)

\(\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\)
\(=\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x+1\right)\left(2x-1\right)}:\frac{4x}{10x-5}\)
\(=\frac{\left(2x+1+2x-1\right)\left(2x+1-2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\times\frac{10x-5}{4x}\)
\(=\frac{4x.2}{\left(2x+1\right)\left(2x-1\right)}\times\frac{5\left(2x-1\right)}{4x}\)
\(=\frac{10}{2x+1}\)

\(a,\frac{4xy-5}{10x^3y}-\frac{6y^2-5}{10x^3y}=\frac{\left(4xy-5\right)-\left(6y^2-5\right)}{10x^3y}=\frac{4xy-5-6y^2+5}{10x^3y}=\frac{4xy-6y^2}{10x^3y}\)

\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\)

\(=\left(\frac{2x+1}{2x-1}+\frac{2x-1}{2x-1}\right):\frac{4x}{10x-5}\)

\(=\frac{2x+1+2x-1}{2x-1}:\frac{4x}{10x-5}\)

\(=\frac{4x}{2x-1}.\frac{10x-5}{4x}\)

\(=\frac{10x-5}{2x-1}\)

\(=\frac{5\left(2x-1\right)}{2x-1}\)

\(=\frac{5}{1}=5\)