a) =(x-y)*(x+y)-(5*(x+y))

=(x+y)*(x-y-5)

Mấy bài còn lại cũng tương tự nha bạn = cách đặt nhân tử chung 

bai nao khong hieu thi pan nhan tin vào nick minh minh se giai đùm ban

a) (x² - y²) - 5(x + y)

= (x - y)(x + y) - 5 (x + y)

= (x + y) (x - y -5)

b) 5x³ - 5x²y - 10x² + 10 xy

= 5[(x³ - x²y) - (2x² - 2 xy)]

=5[x²(x - y) - 2x(x - y)]

=5x(x-y)(x - 2)

c) 2x² - 5x = x(2x - 5)

d) x³ - 3x² +1 - 3x 

= (x³ + 1) - (3x² + 3x)

= (x + 1)(x² - x + 1) - 3x(x + 1)

= (x + 1) [x² - x + 1 - 3x]

= (x + 1)[x² - 4x + 1]

= (x + 1)[x² - 2.x.2 + 2² - 2² + 1]

= (x + 1)[(x - 2)² - 3]

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

e) 3x² - 6xy + 3y² - 12z²

= 3[ x² - 2xy + y² - 4z²]

= 3[ (x - y)² - (2z)²]

= 3(x - y + 2z)(x - y - 2z)

f) 3x² - 7x - 10

= 3x² - 7x - 7 - 3

= (3x² -3) - (7x + 7)

= 3(x² - 1) - 7(x + 1)

= 3 (x + 1)(x - 1) - 7(x + 1)

= (x + 1)[3(x - 1) - 7]

= (x +1)(3x - 8)

g) x⁴ + 1 - 2x² = (x²)² - 2.x² + 1 = (x² - 1)²

= (x + 1)²(x - 1)²

h) 3x² - 3y² - 12x + 12y

= 3(x² - y²) - 12(x - y)

= 3(x - y)(x + y) - 12(x -y)

= (x - y) [3(x + y) - 12]

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

j) x² - 3x + 2 = x² - x - 2x +2

= x(x - 1) - 2(x -1)

=(x - 1)(x - 2)

a) =(x-y)*(x+y)-(5*(x+y))

=(x+y)*(x-y-5)

Mấy bài còn lại cũng tương tự nha bạn = cách đặt nhân tử chung 

a, x^2-9+(x-3)^2 = (x-3)(x+3)+(x-3)^2=(x-3)(x+3+x-3)=2x(x-3)

b,có sai k ạ ! vì mình thấy tự nhiên có ẩn y ở đó , nếu đề bài 2 ẩn thì 1 trong 3 hạng tử chứa ẩn x kia phải có thêm 1 ẩn y

c,đề bài thiếu  ẩn ở hạng tử thứ nhất ạ !

b mình viết đúng rồi mà, c hạng tử 1 là x^3

câu c bạn dùng máy tính bấm nghiệm nhé , dùng mode =>5=> 4 nhé ! 

còn câu b mình chưa ra !

a.16x-5x²-3 = - ( 5x²-16x+3) = -( 5x²-15x-x+3)= -[ 5x(x-3)-(x-3)] = -(5x-1)(x-3) 

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

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

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

c.x^4+8x = \(x\left(x^3+8\right)=x\left(x+2\right)\left(x^2-2x+4\right)\)

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

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

e.5x^2-10xy+5y^2-20z^2\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

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

f.2(x^5)-x^2-5x ( mik ko bik làm)

g.x^3-3x^2-4x+12 = \(x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-2^2\right)\left(x-3\right)\)

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

h.x^4-5x^2+4 \(=\left(x^2\right)^2-4x^2+4-x^2\)

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

a) Ta có: \(5x^2-3x\left(x+2\right)\)

\(=5x^2-3x^2-6x\)

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

b) Ta có: \(3x\left(x-5\right)-5x\left(x+7\right)\)

\(=3x^2-15x-5x^2-35x\)

\(=-2x^2-50x\)

c) Ta có: \(3x^2y\left(2x^2-y\right)-2x^2\left(2x^2y-y^2\right)\)

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

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

d) Ta có: \(3x^2\left(2y-1\right)-\left[2x^2\cdot\left(5y-3\right)-2x\left(x-1\right)\right]\)

\(=6x^2y-3x^2-\left[10x^2y-6x^2-2x^2+2x\right]\)

\(=6x^2y-3x^2-10x^2y+6x^2+2x^2-2x\)

\(=-4x^2y+5x^2-2x\)

e) Ta có: \(4x\left(x^3-4x^2\right)+2x\left(2x^3-x^2+7x\right)\)

\(=4x^4-16x^3+4x^4-2x^3+14x^2\)

\(=8x^4-18x^3+14x^2\)

f) Ta có: \(25x-4\left(3x-1\right)+7x\left(5-2x^2\right)\)

\(=25x-12x+4+35x-14x^3\)

\(=-14x^3+48x+4\)

a) Ta có: \(3x^2+5y-3xy-5x\)

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

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

b) Ta có: \(3y^2-3z^2+3x^2+6xy\)

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

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

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

c) Ta có: \(x^2-25-2xy+y^2\)

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

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

d) Ta có: \(5x^2-10xy+5y^2-20z^2\)

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

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

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

e) Ta có: \(x^2-5x+5y-y^2\)

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

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

f) Ta có: \(3x^2-6xy+3y^2-12z^2\)

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

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

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

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: \(x^2-y^2-2x+2y\)

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

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

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

\(a,-2xy^2\left(x^3y-2x^2y^2+5xy^3\right)\\ =-2x^4y^3+4x^3y^4-10x^2y^5\\ b,\left(-2x\right)\left(x^3-3x^2-x+1\right)\\ =-2x^4+6x^3+2x^2-2x\\ c,\left(-10x^3+\dfrac{2}{5}y-\dfrac{1}{3}z\right)\left(-\dfrac{1}{2}zy\right)\\ =5x^3yz-\dfrac{1}{5}y^2z+\dfrac{1}{6}yz^2\\ d,3x^2\left(2x^3-x+5\right)=6x^5-3x^3+15x^2\\ e,\left(4xy+3y-5x\right)x^2y=4x^3y^2+3x^2y^2-5x^3y\\ f,\left(3x^2y-6xy+9x\right)\left(-\dfrac{4}{3}xy\right)\\ =-4x^3y^2+8x^2y^2-12x^2y\)

1)

a) => 16x² - 8x + 1 - 8(2x² + 3x - 4x - 6) = 15

=> 16x² - 8x + 1 - 8(2x² - x - 6) = 15

=> 16x² - 8x + 1 - 16x² + 8x + 48 = 15

=> 49 = 15 (?) (vô lí)

=> Không tìm được x thoả mãn

b) (5x - 2)(x - 2) - 4(x - 3) = x² + 3

=> 5x² - 10x - 2x + 4 - 4x + 12 = x² + 3

=> 5x² - 16x + 16 = x² + 3

=> 4x² - 16x + 16 = 3

=> (2x)² - 2.2x.4 + 4² = 3

=> (2x - 4)² = 3

=> \(\left[{}\begin{matrix}2x-4=\sqrt{3}\\2x-4=-\sqrt{3}\end{matrix}\right.\)           \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{2}\\x=\dfrac{4-\sqrt{3}}{2}\end{matrix}\right.\)

Mong bạn xem lại đề bài!

2) 

a) 5x² - 10xy + 5y² - 20z²

= 5(x² - 2xy + y² - 4z²)

= 5[(x - y)² - (2z)²]

= 5(x - y - 2z)(x - y + 2z)

b) a³ - ay - a²x + xy

= a(a² - y) - x(a² - y)

= (a - x)(a² - y)

c) 3x² - 6xy + 3y² - 12z²

= 3(x² - 2xy + y² - 4z²)

= 3[(x - y)² - (2z)²]

= 3(x - y - 2z)(x - y + 2z)

d) x² - 2xy + tx - 2ty

= x(x - 2y) + t(x - 2y)

= (x + t)(x - 2y)