a: \(2x^3+x^2-13x+6\)

\(=2x^3-4x^2+5x^2-10x-3x+6\)

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

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

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

b: \(2x^2+y^2-6x+2xy-2y+5=0\)

\(\Leftrightarrow x^2+2xy+y^2+x^2-4x+4-2x-2y+1=0\)

\(\Leftrightarrow\left(x+y\right)^2+\left(x-2\right)^2-2\left(x+y\right)+1=0\)

\(\Leftrightarrow\left(x-2\right)^2+\left(x+y-1\right)^2=0\)

=>x-2=0 và x+y-1=0

=>x=2 và y=-1

a) = (x + 3)² - y² = (x + 3 - y)(x + 3 + y)

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

c) = 3x(x - y) - 5(x - y) = (x - y)(3x - y)

d) Nhầm đề. tui sửa lại x³ + y³ + 2x² - 2xy + 2y²

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

e) = x⁴ - x³ - x³ + x² - x² + x + x - 1 = x³(x - 1) - x²(x - 1) - x(x - 1) + x - 1 = (x - 1)(x³ - x² - x + 1) = (x - 1)(x - 1)(x² - 1) = (x - 1)³(x + 1)

f) = x³ - 3x² - x² + 3x + 9x - 27 = x²(x - 3) - x(x - 3) + 9(x - 3) = (x-3)(x² - x + 9)

g) chắc là 3xyz 

= x²y + xy² + y²z + yz² + x²z + xz² + 3xyz = x²y + xy² + xyz + y²z + yz² + xyz + x²z + xz² + xyz = (x + y + z)(xy + yz + xz)

h) = 2³ -(3x)³ = (2 - 3x)(4 + 6x + 9x²)

i) = (x + y - x + y)(x + y + x - y) = 2y*2x = 4xy

k) = (x³ - y³)(x³ + y³) = (x - y)(x² + xy +y2)(x + y)(x² - xy +y2).

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

\(\left(2x+y\right)^2-4x^2+12x-9=\left(2x+y\right)^2-\left(2x-3\right)^2=\left(y+3\right)\left(4x+y-3\right)\)

\(\left(x+1\right)^2-4\left(x+1\right)\left(y^2+4y^4\right)=\left(x+1\right)\left(x-16y^4-4y^2+1\right)\)

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

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

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

\(b,\left(2x+y\right)^2-4x^2+12x-9=\left(2x+y\right)^2-\left(2x-3\right)^2\)

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

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

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

\(=\left(x+1\right)\left(x+1-4y^2+16y^4\right)\)

Mk sửa lại đề nha:

         x³ + y³ + z³ - 3xyz

= (x + y)³ + z³ - 3x²y - 3xy² - 3xyz

= (x + y + z)[ (x + y)² - z(x + y) + z² ] - 3xy(x + y + z)

= (x + y + z)(x² + 2xy + y² - xz - yz + z² - 3xy)

= (x + y + z)(x² + y² + z² - xy - yz - zx)

a) 1/2(x³+8)=1/2(x+2)(x²-2x+4)

b) x⁴(x-y)+2x³(x-y)=x³(x+2)(x-y)

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

d) xy(x³+y³)=xy(x+y)(x²-xy+y²)

e)3x²(x²-25y²)=3x²(x-5y)(x+5y)

f) 4x⁴+4x²y²+y⁴-4x²y²= (2x²+y²)²-(2xy)²=(2x²-2xy+y²)(2x²+2xy+y²)

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

b) \(x^5-x^4y+2x^4-2x^3y=x^3\left(x^2-xy+2x-2y\right)=x^3\left[x\left(x-y\right)+2\left(x-y\right)\right]=x^2\left(x-y\right)\left(x+2\right)\)

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

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

e) \(3x^4-75x^2y^2=3x^2\left(x^2-25y^2\right)=3x^2\left(x+5y\right)\left(x-5y\right)\).

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

(x + y)³ - 1 - 3xy(x + y - 1)

= x³ + 3x²y + 3xy² + y³ - 1 - 3x²y - 3xy² + 3xy

= x³ - 1 + 3xy

= x(x² + 3y) - 1

k bt lm nx r :v

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

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

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

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

\(a,=\dfrac{1}{2}\left[\left(x^2+y^2\right)^2-4x^2y^2\right]\\ =\dfrac{1}{2}\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\\ =\dfrac{1}{2}\left(x-y\right)^2\left(x+y\right)^2\\ b,=\left(3x-\dfrac{1}{2}y\right)\left(9x^2+\dfrac{3}{2}xy+\dfrac{1}{4}y^2\right)\\ c,=\dfrac{1}{2}\left(x^2+\dfrac{1}{2}x+\dfrac{1}{16}\right)=\dfrac{1}{2}\left(x+\dfrac{1}{4}\right)^2\)

\(\left(2a+3\right)\left(2a+3\right)y+\left(2a+3\right)\)

\(=\left(2a+3\right)[y\left(2a+3\right)+1]\)

\(=\left(2a+3\right)\left(2ay+3y+1\right)\)

\(\left(a-b\right)x+\left(b-a\right)y-\left(a-b\right)\) (Sửa đề)

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

\(=\left(a-b\right)\left(x-y-1\right)\)

\(\left(4x-y\right)\left(a+b\right)+\left(4x-y\right)\left(c-1\right)\)

\(=\left(4x-y\right)[\left(a+b\right)+\left(c-1\right)]\)

\(=\left(4x-y\right)\left(a+b+c-1\right)\)

mấy men or girl bạn nào làm đk giải giúp mk đi mk thanks trước nhá

ta muốn làm nắm nhưng mi bợi thêm năm nữa đi nha đợi ta lên lớp 8 ta giải cho