2: \(ax+ay+bx+by\)

\(=a\left(x+y\right)+b\left(x+y\right)\)

\(=\left(x+y\right)\left(a+b\right)\)

3: \(x\left(x-2y\right)-x+2y\)

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

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

a) \(x^4+x^3-8x-8\)

\(=x^3\left(x+1\right)-8\left(x+1\right)\)

\(=\left(x^3+8\right)\left(x+1\right)\)

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

a) \(=x^3\left(x+1\right)-8\left(x+1\right)=\left(x+1\right)\left(x^3-8\right)=\left(x+1\right)\left(x-2\right)\left(x^2+2x+4\right)\)

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

c) \(=3\left(x-y\right)-a\left(x-y\right)=\left(x-y\right)\left(3-a\right)\)

bn post nhiều nên mình ghi đáp án thôi nhé phần nào sai đề mình cho qua

b)\(\left(x+1\right)\left(xy+1\right)\)

c)\(\left(a+b\right)\left(x+y\right)\)

d)\(\left(x-a\right)\left(x-b\right)\)

e)\(\left(x+y\right)\left(xy-1\right)\)

f)\(\left(a-b\right)\left(x^2+y\right)\)

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a: \(\dfrac{4x^4y-7x^2y+3y}{-3x^2+2y}\)

\(=\dfrac{4x^4y-4x^2y-3x^2y+3y}{-\left(3x^2-2y\right)}\)

\(=\dfrac{4x^2y\left(x^2-1\right)-3y\left(x^2-1\right)}{-\left(3x^2-2y\right)}\)

\(=\dfrac{y\left(x^2-1\right)\left(4x^2-3\right)}{-\left(3x^2-2y\right)}\)

a) `(4x^4y-7x^2y+3y).(2y-3x^2y)`

`=8x^4y^2-14x^2y^2+6y^2-12x^6y^2+21x^4y^2-9x^2y^2`

`=29x^4y^2-12x^6y^2-23x^2y^2+6y^2`

b) `(x^2+3x-3/2 x^3):2x - x/2 . (1-3/2 x)`

`=(x+3-3/2 x^2):2 - (x/2 - 3/4 x^2)`

`=x/2 + 3/2 - 3/4 x^2 -x/2 +3/4 x^2`

`=3/2`

c) `(-2x^3-x-3+5x^2):(3-2x)`

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

`=x^2-x-1`

1, \(2x^2+4x=2x\left(x+2\right)\)

2, \(15x^3+5x^2-10x=5x\left(3x^2+x-2\right)=5x\left(x-\dfrac{2}{3}\right)\left(x+1\right)\)

3) \(5x^2\left(x-2y\right)+15x\left(x-2y\right)=\left(5x^2+15x\right)\left(x-2y\right)=5x\left(x+3\right)\left(x-2y\right)\)

4) \(3\left(x-y\right)+5x\left(y-x\right)=\left(x-y\right)\left(3-5x\right)\)

5) \(5x^2-10x=5x\left(x-2\right)\)

6) \(3x-6y=3\left(x-2y\right)\)

7) \(25x^2+5x^3+x^2y=x^2\left(25+5x+y\right)\)

8) \(14x^2y-21xy^2+28x^2y^2=7xy\left(2x-3y+4xy\right)\)

9) \(x\left(y-1\right)-y\left(y-1\right)=\left(x-1\right)\left(y-1\right)\)

10) \(10x\left(x-y\right)-8y\left(y-x\right)=\left(10x+8y\right)\left(x-y\right)=2\left(5x+4y\right)\left(x-y\right)\)

\(1,=2x\left(x+2\right)\\ 2,=5x\left(3x^2+x-2\right)\\ 3,=\left(x-2y\right)\left(5x^2+15x\right)=5x\left(x+3\right)\left(x-2y\right)\\ 4,=\left(x-y\right)\left(3-5x\right)\\ 5,=5x\left(x-2\right)\\ 6,=3\left(x-2y\right)\\ 7,=5x^2\left(5+x+y\right)\\ 8,=7xy\left(2x-3y+4xy\right)\\ 9,=\left(y-1\right)\left(x-y\right)\\ 10,=\left(x-y\right)\left(10x+8y\right)=2\left(5x+4y\right)\left(x-y\right)\)

Ta có: \(\left(ax+by\right)^2=\left(a^2+b^2\right)\left(x^2+y^2\right)\)

\(\Leftrightarrow a^2x^2+2abxy+b^2y^2=a^2x^2+a^2y^2+x^2b^2+b^2y^2\)

\(\Leftrightarrow2abxy=a^2y^2+x^2b^2\)

\(\Leftrightarrow\left(ay-xb\right)^2=0\)

\(\Leftrightarrow ay=xb\)

hay \(\dfrac{a}{x}=\dfrac{b}{y}\)

Bài 2: 

a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)

b: \(5x^2+5xy-x-y\)

\(=5x\left(x+y\right)-\left(x+y\right)\)

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

c:\(-6x^2+7x-2\)

\(=-6x^2+3x+4x-2\)

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

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

1.

a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)

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

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

c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)

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

2.

a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

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

c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)

3.

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

c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)

4.

a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)