48x^2.y^2 - 3y^2 + 6xy - 3x^2

= 3. [16.x^2.y^2-y^2 + 2xy - x^2]

= = 3. { [4.x.y]^2 - [x^2-2xy+y^2 ]}

 = 3. { [4xy]^2 - [x-y]^2] }

= 3. [4xy-x+y]. [4xy + x-y] }

Chọn C

6xy + 48x²y² - 3y² - 3x²

=3(16x²y²-x²+2xy-y²)

=3[(4xy)²-(x-y)²]

=3[(4xy+(x-y)]*[(4xy)-(x-y)]

=3(4xy+x-y)(4xy-x+y)

\(x+2y=5\Rightarrow x+y=5-y\)

\(A=3\left(x^2+2xy+y^2\right)-2\left(x+y\right)-100\)

\(=3\left(x+y\right)^2-2\left(x+y\right)-100\)

\(=3\left(5-y\right)^2-2\left(5-y\right)-100\)

\(=3y^2-28y-35\)

Bạn ghi nhầm đề, ko rút gọn được nữa

Nếu \(x+y=5\) thì rút gọn được

Đề là gì vậy bạn?

\(A=x^3-8-128-x^3=-136\\ B=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)

\(A=\left(x-2\right)\left(x^2+2x+4\right)-\left(128+x^3\right)=x^3-8-128-x^3=-136\)

\(B=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)-\left(3x-2y\right)\left(9x^2+6xy+4y^2\right)=8x^3+27y^3-27x^3+8y^3=-19x^3+35y^3\)

\(A=x^3+2x^2+4x-2x^2-4x-8-128-x^3\)
\(A=-136\)
\(B=\left(2x+3y\right)\left(2x-3y\right)^2-\left(3x-2y\right)\left(3x+2y\right)^2\)
\(B=\left(2x+3y\right)\left(2x-3y\right)\left(2x-3y\right)-\left(3x-2y\right)\left(3x+2y\right)\left(3x+2y\right)\)
\(B=\left(4x^2-9y^2\right)\left(2x-3y\right)-\left(9x^2-4y^2\right)\left(3x+2y\right)\)
\(B=8x^3-12x^2y-18xy^2-27y^3-27x^3-18x^2y+12xy^2+8y^3\)
\(B=-19x^3-30x^2y-6xy^2-19y^3\)
\(B=-19\left(x^3-y^3\right)-6xy\left(5x+y\right)\)

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: \(x^2-y^2-2x+2y\)

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

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

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