a: \(=15x^5y^3-6x^4y^2-6x^3y^3\)

c: \(=2x^4-2x^2-3x^3+3x+x^2-1\)

\(=2x^4-3x^3-x^2+3x-1\)

c: \(=\dfrac{x^3+2x^2+x^2+2x-10x-20}{x+2}\)

\(=x^2+x-10\)

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:

a)\(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4=x^4-y^4\)

b) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+18x-24+8=-16\)

Bài 4:

1: \(\left(x-1\right)\left(x^2+x+1\right)-x^3-6x=11\)

=>\(x^3-1-x^3-6x=11\)

=>-6x-1=11

=>-6x=11+1=12

=>\(x=\dfrac{12}{-6}=-2\)

2: \(16x^2-\left(3x-4\right)^2=0\)

=>\(\left(4x\right)^2-\left(3x-4\right)^2=0\)

=>\(\left(4x-3x+4\right)\left(4x+3x-4\right)=0\)

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

=>\(\left[{}\begin{matrix}x+4=0\\7x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{4}{7}\end{matrix}\right.\)

3: \(x^3-x^2-3x+3=0\)

=>\(\left(x^3-x^2\right)-\left(3x-3\right)=0\)

=>\(x^2\left(x-1\right)-3\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(x^2-3\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x^2=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

4: \(\dfrac{x-1}{x+2}=\dfrac{x+2}{x+1}\)(ĐKXĐ: \(x\notin\left\{-2;-1\right\}\))

=>\(\left(x+2\right)^2=\left(x-1\right)\left(x+1\right)\)

=>\(x^2+4x+4=x^2-1\)

=>4x+4=-1

=>4x=-5

=>\(x=-\dfrac{5}{4}\left(nhận\right)\)

5: ĐKXĐ: \(x\notin\left\{0;-1\right\}\)

\(\dfrac{1}{x}+\dfrac{2}{x+1}=0\)

=>\(\dfrac{x+1+2x}{x\left(x+1\right)}=0\)

=>3x+1=0

=>3x=-1

=>\(x=-\dfrac{1}{3}\left(nhận\right)\)

6: ĐKXĐ: \(x\notin\left\{0;3\right\}\)

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

=>\(\dfrac{-\left(x^2-9\right)}{x\left(x-3\right)}=1\)

=>\(\dfrac{-\left(x-3\right)\left(x+3\right)}{x\left(x-3\right)}=1\)

=>\(\dfrac{-x-3}{x}=1\)

=>-x-3=x

=>-2x=3

=>\(x=-\dfrac{3}{2}\left(nhận\right)\)

ĐKXĐ: \(\left\{{}\begin{matrix}3x\ne-y\\3x\ne y\end{matrix}\right.\)

a. \(\dfrac{x}{3x+y}+\dfrac{x}{3x-y}-\dfrac{2xy}{y^2-9x^2}\)

\(=\dfrac{x.\left(3x-y\right)}{\left(3x+y\right).\left(3x-y\right)}+\dfrac{x.\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}+\dfrac{2xy}{9x^2-y^2}\)

\(=\dfrac{x.\left(3x+y+3x-y\right)+2xy}{\left(3x-y\right).\left(3x+y\right)}\)

\(=\dfrac{6x^2+2xy}{\left(3x-y\right).\left(3x+y\right)}\)

\(=\dfrac{2x\left(3x+y\right)}{\left(3x+y\right).\left(3x-y\right)}\)

\(=\dfrac{2x}{3x-y}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\x\ne-5\end{matrix}\right.\)

b. \(\dfrac{4x+5}{x^2+5x}-\dfrac{3}{x+5}\)

\(=\dfrac{4x+5}{x.\left(x+5\right)}-\dfrac{3x}{x.\left(x+5\right)}\)

\(=\dfrac{x+5}{x.\left(x+5\right)}\)

\(=\dfrac{1}{x}\)

a) \(2xy\left(3x^2-5xy+4y^2\right)=6x^3y-10x^2y^2+8xy^3\)

b) \(\left(x-3\right)^2+\left(x+5\right)\left(5-x\right)=x^2-6x+9+25-x^2=34-6x\)

a: \(2xy\left(3x^2-5xy+4y^2\right)=6x^3y-10x^2y^2+8xy^3\)

b: \(\left(x-3\right)^2+\left(x+5\right)\left(5-x\right)\)

\(=x^2-6x+9+25-x^2\)

=-6x+34

Lời giải:

a.

$(2x-3)^2+(2x+3)(5-2x)=(4x^2-12x+9)-(-4x^2+4x+15)$

$=4x^2-12x+9+4x^2-4x-15$

$=24-8x$
b.

$3(2x-3)+5(x+2)=6x-9+5x+10=11x+1$

c.

$3x(2x-8)+(6x-2)(5-x)=(6x^2-24x)+(-6x^2+32x-10)$

$=6x^2-24x-6x^2-32x+10$

$=8x-10$

d.

$(x-3)(x+3)-(x-5)^2=(x^2-9)-(x^2-10x+25)$

$=x^2-9-x^2+10x-25=10x-34$

e.

$(x-y)^3-(x-y)(x^2+xy+y^2)=(x^3-3x^2y+3xy^2-y^3)-(x^3-y^3)$

$=-3x^2y+3xy^2=3xy(y-x)$

a: ta có: \(\left(2x-3\right)^2+\left(2x+3\right)\left(5-2x\right)\)

\(=4x^2-12x+9+2x-4x^2+15-6x\)

\(=-16x+24\)

b: Ta có: \(3\left(2x-3\right)+5\left(x+2\right)\)

\(=6x-9+5x+10\)

\(=11x+1\)

c: ta có: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)

\(=6x^2-24x+30x-6x^2-10+2x\)

\(=8x-10\)