a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+24\)

\(=24-11x\)

b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)

\(=8x^2y-6y^2-9x^2y+12y^2\)

\(=6y^2-x^2y\)

c) \(3y^2\left[\left(2x-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2\cdot\left(2x-1+y+1\right)-y\cdot\left(1-y-y^2\right)+y\)

\(=6xy^2-3y^2+3y^3+3y^2-y+y^2+y^3+y\)

\(=4y^3+y^2+6xy^2\)

Bài 2:

1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)

=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)

=>(2x-1)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

2: \(9x^3-x=0\)

=>\(x\left(9x^2-1\right)=0\)

=>x(3x-1)(3x+1)=0

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

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

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

=>(2x-3)(2x-3-2)=0

=>(2x-3)(2x-5)=0

=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)

4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)

=>\(2x^2+10x-5x-25-10x+25=0\)

=>\(2x^2-5x=0\)

=>\(x\left(2x-5\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)

Bài 1:

1: \(3x^3y^2-6xy\)

\(=3xy\cdot x^2y-3xy\cdot2\)

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

2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)

\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)

\(=\left(x-2y\right)\left(x+3y-2\right)\)

3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)

\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)

\(=(x-2y)(3x-1+5x)\)

\(=\left(x-2y\right)\left(8x-1\right)\)

4: \(x^2-y^2-6y-9\)

\(=x^2-\left(y^2+6y+9\right)\)

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

\(=\left(x-y-3\right)\left(x+y+3\right)\)

5: \(\left(3x-y\right)^2-4y^2\)

\(=\left(3x-y\right)^2-\left(2y\right)^2\)

\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)

\(=\left(3x-3y\right)\left(3x+y\right)\)

\(=3\left(x-y\right)\left(3x+y\right)\)

6: \(4x^2-9y^2-4x+1\)

\(=\left(4x^2-4x+1\right)-9y^2\)

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

\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)

8: \(x^2y-xy^2-2x+2y\)

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

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

9: \(x^2-y^2-2x+2y\)

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

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

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

a: \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

\(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)

\(=2x^2-4xy+\dfrac{15}{4}y^2\)

b: \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)

\(=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)

\(=2x^2+2x+13-2x^2+2\)

=2x+15

a) \(=x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2=2x^2-4xy+\dfrac{15}{4}y^2\)

b) \(=x^2-4x+4+x^2+6x+9-2x^2+2\)

\(=2x+15\)

a; \(\left(x-2y\right)^2+\left(x-\dfrac{1}{2}y\right)\left(x+\dfrac{1}{2}y\right)\)

= \(x^2-4xy+4y^2+x^2-\dfrac{1}{4}y^2\)

= \(2x^2-4xy+\dfrac{15}{4}y^2\)

b; \(\left(x-2\right)^2+\left(x+3\right)^2-2\left(x-1\right)\left(x+1\right)\)

= \(x^2-4x+4+x^2+6x+9-2x^2+2\)

= \(2x+15\)

a, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+16\)

\(=-11x+16\)

b, \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(=8x^2y-6y^2-\left(9x^2y-12y^2\right)\)

\(=8x^2y-6y^2-9x^2y+12y^2=-x^2y+6y^2\)

c, \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2.3y-y+y^2+y^3+y\)

\(=9y^3+y^2+y^3=10y^3+y^2\)

Chúc bạn học tốt!!!

a, \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+16\)

\(=-11x+16\)

b, \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)

\(=8x^2y-6y^2-9x^2y+12y^2\)

\(=-x^2y+6y^2\)

c, \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2.3y-y\left(1-y-y^2-1\right)\)

\(=9y^3-y\left(-y-y^2\right)\)

\(=9y^3+y^2+y^3=10y^3+y^2\)

a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-2\right)\)

\(=3x^2-6x-5x+5x^2-8x^2+16\)

\(=3x^2+5x^2-8x^2-6x-5x+16\)

\(=-11x+16\)

b) \(\left(4x^2-3y\right)2y-\left(3x^2-4y\right).3y\)

\(=9x^2y-6y^2-\left(9x^2y-12y^2\right)\)

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

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

\(=6y^2\)

c) \(3y^2\left[\left(2y-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)

\(=3y^2\left(2y-1\right)+3y^3+3y^2-y+y^2+y^3+y\)

\(=6y^3-3y^2+3y^3+3y^2-y+y^2+y^3+y\)

\(=6y^3+3y^3+y^3-3y^2+3y^2+y^2-y+y\)

\(=10y^3+y^2\)

MK ĐANG CẦN GẤP Ạ AI NHANH MK SẼ VOTE Ạ

a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)

\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)

\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)

\(=-41x^2+34x-7-2x^3+4x-2\)

\(=-2x^3-41x^2+38x-9\)

b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)

\(=\left(3a+1+3a-1\right)^2\)

\(=36a^2\)

\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)

`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`

`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`

`=(y(y-2x))/3`

`b,(x^2-y^2)/(x^2-y^2+xz-yz)`

`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`

`=(x+y)/(x+y+z)`

`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`

`=(-(x^2-3x+x-3))/((x-1)(x+1))`

`=(-x(x-3)+x-3)/((x-1)(x+1))`

`=((x-3)(1-x))/((x-1)(x+1))`

`=(3-x)/(1+x)`

`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`

`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`

`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`

`=(3x^2-4x+1)/(2x^2+x-3)`

`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`

`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`

`=(3x-1)/(2x+3)`

a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)

\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)

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

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)