làm ơn giúp e vs

\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)

Bài 1: 

a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)

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

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

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

b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)

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

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

\(=\left(3x-2\right)\left(3x-6\right)\)

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

Bài 2: 

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

\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)

\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)

\(=\left(a-b\right)\left(2a-4b\right)\)

\(=2\left(a-b\right)\left(a-2b\right)\)

f: Ta có: \(x^2-6xy+9y^2+4x-12y\)

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

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

whatterfice

1.A

2.C

3.B

4.C

a

c

b

c

(x-1)y^2-4(x-1)y

a) x² - 2xy + y² - 4m² + 4mn - n² = (x - y)² - [(2m)² -  2.2m.n + n²] = (x - y)² - (2m - n)²

= [(x - y) - (2m - n)][(x - y) + (2m - n)] = (x - y - 2m + n)(x - y + 2m - n)

b) x² - 4x²y² + y² + 2xy = x² + 2xy + y² - 4x²y² = (x + y)² - (2xy)² = (x + y - 2xy)(x + y + 2xy)

c) x⁶ - y⁶ = (x³)² - (y³)² = (x³ - y³)(x³ + y³) = (x - y)(x² + xy + y²)(x + y)(x² - xy - y²)

d) 25 - a² + 2ab - b² = 25 - (a² - 2ab + b²) = 5² - (a - b)² = (5 - a + b)(5 + a - b)

xin lỗi các bạn, đề mink có vấn đề: ý c phải là: x^6 - y^6

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

1: Đa thức này ko phân tích được nha bạn

2: \(x^2+8x+7\)

\(=x^2+x+7x+7\)

\(=x\left(x+1\right)+7\left(x+1\right)\)

\(=\left(x+1\right)\left(x+7\right)\)

3: \(x^2-6x-16\)

\(=x^2-8x+2x-16\)

\(=x\left(x-8\right)+2\left(x-8\right)\)

\(=\left(x-8\right)\left(x+2\right)\)

4: \(4x^2-8x+3\)

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

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

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

5: \(3x^2-11x+6\)

\(=3x^2-9x-2x+6\)

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

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

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

1. \(=3x\left(2x+5\right)\)

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

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

1, = 3x.(2x + 5)

2. = (3x)² - 1² = (3x - 1).(3x +1 )

3, =(x² + 6x + 9) - y² = (x + 3)² - y² =(x + y -3 ). (x - y +3)

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

\(1)4x^2-25+\left(2x+7\right).\left(5.2x\right)\)

\(=\left(2x\right)^2-5^2-\left(2x+7\right).\left(2x-5\right)\)

\(=\left(2x.5\right)\left(2x+5\right).\left(2x+7\right)\left(2x-5\right)\)

\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)

\(=\left(2x-5\right).12\)

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

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

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

\(3)5x^2-2y^2-10x+10y\)

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

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

\(=\left(x-y\right)[5(x+y)-10]\)

Còn lại bn lm nốt nha! 

bài 1: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x}{x+2}-\dfrac{x}{x-2}\)

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

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

Bài 2:

1: \(x^2y^2-8-1\)

\(=x^2y^2-9\)

\(=\left(xy-3\right)\left(xy+3\right)\)

2: \(x^3y-2x^2y+xy-xy^3\)

\(=xy\cdot x^2-xy\cdot2x+xy\cdot1-xy\cdot y^2\)

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

\(=xy\left[\left(x-1\right)^2-y^2\right]\)

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

3: \(x^3-2x^2y+xy^2\)

\(=x\cdot x^2-x\cdot2xy+x\cdot y^2\)

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

4: \(x^2+2x-y^2+1\)

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

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

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

5: \(x^2+2x-4y^2+1\)

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

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

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

6: \(x^2-6x-y^2+9\)

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

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