X³-6x²y+9xy²

= x(x²-6xy+9xy²)

\(x^3-6x^2y+9xy^2\)

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

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

\(a,M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\\ \Rightarrow M=6x^2+9xy-y^2-5x^2+2xy\\ \Rightarrow M=x^2+11xy-y^2\\ b,\left(25x^2y-13xy^2+y^3\right)-M=11xy^2-2y^3\\ \Rightarrow M=25x^2y-13xy^2+y^3-11xy^2+2y^3\\ \Rightarrow M=25x^2y-24xy^2+3y^3\)

1: \(6x^2y-9xy^2+3xy\)

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

2: \(\left(4-x\right)^2-16\)

\(=\left(4-x-4\right)\left(4-x+4\right)\)

\(=-x\cdot\left(8-x\right)\)

3: \(x^3+9x^2-4x-36\)

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

\(=\left(x+9\right)\left(x-2\right)\left(x+2\right)\)

1) \(6x^2y-9xy^2+3xy=3xy\left(2x-3y+1\right)\)

2) \(\left(4-x\right)^2-16=\left(4-x\right)^2-4^2=\left(4-x-4\right)\left(4-x+4\right)=-x\left(8-x\right)\)

3) \(x^3+9x^2-4x-36\\ =\left(x^3-2x^2\right)+\left(11x^2-22x\right)+\left(18x-36\right)\\ =x^2\left(x-2\right)+11x\left(x-2\right)+18\left(x-2\right)\\ =\left(x^2+11x+18\right)\left(x-2\right)\\ =\left[\left(x^2+2x\right)+\left(9x+18\right)\right]\left(x-2\right)\\ =\left[x\left(x+2\right)+9\left(x+2\right)\right]\left(x-2\right)\\ =\left(x+2\right)\left(x+9\right)\left(x-2\right)\)

\(\begin{cases}x^3-6x^2y+9xy^2-4y^3=0\left(1\right)\\\sqrt{x-y}+\sqrt{x+y}=2\left(2\right)\end{cases}\)

\(\left(1\right)\Leftrightarrow x^3-2x^2y+xy^2-4y^3+8xy^2-4x^2y=0\)

\(\Leftrightarrow x\left(x^2-2xy+y^2\right)-4y\left(x^2-2xy+y^2\right)=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)\left(x-4y\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x-4y\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-y\right)^2=0\\x-4y=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=y\\x=4y\end{array}\right.\)

\(\left(2\right)\Leftrightarrow\sqrt{x-x}+\sqrt{x+x}=2\)

\(\Leftrightarrow\sqrt{2x}=2\Leftrightarrow2x=4\Leftrightarrow x=2\).Mà \(\begin{cases}x=2\\x=y\end{cases}\)\(\Rightarrow x=y=2\)

\(\left(2\right)\Leftrightarrow\sqrt{4y-y}+\sqrt{4y+y}=2\)

\(\Leftrightarrow\sqrt{3y}+\sqrt{5y}=2\)\(\Leftrightarrow\sqrt{y}\left(\sqrt{5}+\sqrt{3}\right)=2\)

\(\Leftrightarrow y\left(\sqrt{15}+4\right)=2\)\(\Leftrightarrow y=\frac{2}{\sqrt{15}+4}\).Mà \(\begin{cases}x=4y\\y=\frac{2}{\sqrt{15}+4}\end{cases}\)\(\Leftrightarrow x=\frac{8}{\sqrt{15}+4}\)

Yêu cầu là gì v c.

`@` `\text {Ans}`

`\downarrow`

`1)`

`x^3 - 2x - 1`

`= x^3 - x^2 - x + x^2 - x - 1`

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

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

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

`2)`

`x^3 +3x^2 -4`

`= x^3 + 4x^2 + 4x - x^2 - 4x - 4`

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

`= x(x^2 + 4x + 4) - (x^2 + 4x + 4)`

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

`3)`

`x^3y^3 + x^2y^2 + 4`

`= x^3 y^3 - x^2y^2 + 2x^2y^2 + 2xy - 2xy + 4`

`= (x^3y^3 - x^2y^2 + 2xy) + (2x^2y^2 - 2xy + 4)`

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

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

`4)`

`x^3 + 5x^2 -9xy^2+5y^3`

Bạn xem lại đề ;-;

`5)`

` x^4 + x^3 + 6x^2 + 5x + 5`

`= x^4 + x^3 + 5x^2 + x^2 + 5x + 5`

`= (x^4 + x^3 + x^2) + (5x^2 + 5x + 5)`

`= x^2(x^2 + x + 1) + 5(x^2 + x + 1)`

`= (x^2 + 5)(x^2 + x + 1)`