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,

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

b,

$12x^2y^3z^4+(-7x^2y^3z^4)=12x^2y^3z^4-7x^2y^3z^4=5x^2y^3z^4$

c,

$-6xy^3-(-6xy^3)+6x^3=-6xy^3+6xy^3+6x^3=0+6x^3=6x^3$

d,

$\frac{-x^2}{2}+\frac{7}{2}x^2+x=(\frac{7}{2}-\frac{1}{2})x^2+x$

$=3x^2+x$

e,

$2x^3+3x^3-\frac{1}{3}x^3=(2+3-\frac{1}{3})x^3=\frac{14}{3}x^3$

f,

$5xy^2+\frac{1}{2}xy^2+\frac{1}{4}xy^2=(5+\frac{1}{2}+\frac{1}{4})xy^2$

$=\frac{23}{4}xy^2$

\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)

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

1) Ta có: \(x^2-4xy+4y^2\)

\(=x^2-2.x.2y+\left(2y\right)^2\)

\(=\left(x-2y\right)^2\)

Phép tính trở thành: \(\left(x-2y\right)^2:\left(x-2y\right)=x-2y\)

2) Ta có: \(25x^2+2xy+\dfrac{1}{25}y^2\)

\(=\left(5x\right)^2+2.5x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\)

\(=\left(5x+\dfrac{1}{5}y\right)^2\)

Phép tính trở thành: \(\left(5x+\dfrac{1}{5}y\right)^2:\left(5x+\dfrac{1}{5}y\right)=5x+\dfrac{1}{5}y\)

1) (x² - 4xy + 4y²) : (x - 2y)

= (x - 2y)² : (x - 2y)

= x - 2y

2) (25x² + 2xy + 1/25 y²) : (5x + 1/5 y)

= 5x + 1/5 y)² : (5x + 1/5 y)

= 5x + 1/5 y

1: \(4xy\left(x^2-2xy+3y^2\right)=4x^3y-8x^2y^2+12xy^3\)

2: \(\left(x^2-2\right)\left(x^4+2x^2+4\right)=x^6-8\)

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

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

a, = 3x^2y^2.4x^2y^2 = 12x^4y^4

b, = xy^3.(4+5-6) = 3.xy^3

Tk mk nha

em hông bít làm chị ơi

em nhớ rùi để em nhắn tin nhắn cho

`Answer:`