a.

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

Tại \(x=77;y=22\Rightarrow x\left(x+y+1\right)=77\left(77+22+1\right)=77.100=7700\)

b.

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

\(=\left(53-3\right)^2=50^2=2500\)

c.

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

\(=\left(2001+1999\right)\left(2001-1\right)=4000.2000=8000000\)

a)-(x-y)(x²+xy-1)=-(x³+x²y-x-x²y-xy²+y)

                          =-(x³-xy²-x+y)

                          =-x³+xy²+x-y

b)x²(x-1)-(x³+1)(x-y)=x³-x²-x³+x²y-x+y

                                =-x²+x²y-x+y

c)(3x-2)(2x-1)+(-5x-1)(3x+2)=6x²-3x-4x+2-15x²-10x-3x-2

                                             =-9x²-20x

d) hình như bạn ghi lỗi

Bài 2: C=x(x²-y)-x²(x+y)+y(x²-x)

             =x³-xy-x³-x²y+x²y-xy

             =-2xy

Thay x=1/2,y=-1 vào C, ta có:

        C=-2.1/2.(-1)=1

Vậy C=1 khi x=1/2 và y=-1.

Ta có: 

a) 5x - 20y = 5(x - 4y)

b) 5x(x - 1) - 3x(x - 1) = x(x - 1)2

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

d) x² - 9 = (x - 3)(x + 3)

e) 4² - 25 = (4 - 5)(4 + 5) f) x⁶ - y⁶ = (x⁴)² - (y⁴)² = (x⁴ - y⁴)(x⁴ + y⁴) 

a)  5x - 20y = 5( x - 4y)

b) 5x(x-1) - 3x(x-1) =2x (x-1) 

c) x² - 9 = x² - 3² = ( x-3)(x+3)

e) 4²-25 = 4² - 5² = ( 4-5)(4+5)= -9

f) x⁶ - y⁶ = (x³)² - (y³)² = ( x³ - y³ ) ( x³ + y³)

                                    = (x-y) ( x² + xy + y² ) ( x+y) ( x² -xy +y²)

mình làm thiếu câu c ) x(x+y) - 5x-5y 

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

                                    = (x+y) ( x-5)

Bài 2:

a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

hay \(x=\dfrac{2}{7}\)

b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow x^3=-8\)

hay x=-2

Bài 1: 

a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)

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

\(=xy\)

=1

b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)

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

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

\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)

a) \(x^2-xy+x-y\)

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

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

b) \(x^2+5x+6\)

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

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

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

\(2xy-x^2-y^2+16\)

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

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

a)5x-20y

=5x-4*5y

=5(x-4y)

b)x(x-1)-3x(x-1)

=(x-3x)(x-1)

c)x(x+y)-5x-5y

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

=(x-5)(x+y)

5.(x-4)