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

=(x^2-2x+1)-y^2

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

=(x-1-y).(x-1+y)

hok tốt

\(=\text{1620}\)

\(18^2+36^2=324+1296=1620\)

Oh mammamia, hỏi bài trong giờ thi =)

@Nghệ Mạt

#cua

1.

a)\(\left(2x-5\right)^2-4x\left(x-5\right)=4x^2-20x+25-4x^2+20x=25\)

b)\(\left(9x^4y^3-15x^2y^4\right):3x^2y^2+5xy^2=3x^2y-5xy^2+5xy^2=3x^2y\)

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

MTC: 2(x-1)(x+1)

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

\(=\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}=\frac{x-1}{2\left(x+1\right)}\)

2.

a) 3x^2-3y^2

=3(x^2-y^2)

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

b)x^2-2xy+2x-4y

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

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

c)4x^2-y^2+8y-16=(-y^2+8y+4x^2)-4^2

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

3. a) (2x+3)(x-2)-2x(x-8)=24

=> 2x^2 - 4x + 3x - 6 - 2x^2 +16x = 24

=>  15x-6=24

=>15x=24+6

=>15x=30

=>x=30/15=2

vay x=2

b) 5x(x-3)-x+3=0

=>5x(x-3)-(x-3)=0

=>(5x-1)(x-3)=0

=> 2 TH

*5x-1=0=>5x=1=>x=1/5

*x-3=0=>x=3

Vay x=1/5 hoac x=3

x² + xy + y² = x²y²

=> x²+ 2xy-xy+ y² = x²y²

=> x² + 2xy + y² = x²y² +xy

=> (x+y)²= xy.(xy+1)

Vì xy.(xy+1) là số chính phương mà xy,xy,1 là 2 số nguyên liên tiếp

(x+y)²= xy.(xy+1) 

Biến đổi VP ta có:xy.(xy+1)

=>xy=0

=> 2 TH

 x=0 hoặc y=0 

 (x+y)²= xy.(xy+1)

=> (x+y)²=0

=>x+y=0

=>x=-y

=>x=y=0

Vậy x=0 hoặc y=0

a) (x + 3y) (2x²y - 6xy²)

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

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

= 2xy (x² - 3y²)

b) (6x⁵y² - 9x⁴y³ + 15x³y⁴) : 3x³y²

= (6x⁵y² : 3x³y²) + (-9x⁴y³ : 3x³y²) + (15x³y⁴ : 3x³y²)

= [(6 : 3) (x⁵ : x³) (y² : y²)] + [(-9 : 3) (x⁴ : x³) (y³ : y²)] + [(15 : 3) (x³ : x³) (y⁴ : y²)]

= 2x² + (-3xy) + 5y²

= 2x² - 3xy + 5y²