Trả lời:

\(\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\) \(\left(ĐK:a\ne b\ne c\right)\)

\(=\frac{1}{\left(a-b\right)\left(a-c\right)}-\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{b-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}-\frac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

Bạn thấy ảnh ko ạ ?

* Nguồn : Lazi *

Trả lời:

\(\left(ab-1\right)^2+\left(a+b\right)^2\)

\(=a^2b^2-2ab+1+a^2+2ab+b^2\)

\(=a^2b^2+1+a^2+b^2\)

\(=\left(a^2b^2+a^2\right)+\left(b^2+1\right)\)

\(=a^2\left(b^2+1\right)+\left(b^2+1\right)\)

\(=\left(b^2+1\right)\left(a^1+1\right)\) 

Trả lời:

a, \(27a^2b^2-18ab+3=3\left(9a^2b^2-6ab+1\right)=3\left(3ab-1\right)^2\)

b, \(x^2+2xy+y^2-xz-yz\)

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

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

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

c, \(a^4+a^3-a^2-a\)

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

\(=a^3\left(a+1\right)-a\left(a+1\right)\)

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

\(=a\left(a+1\right)\left(a-1\right)\left(a+1\right)\)

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

d, \(a^3-b^3+2b-2a\)

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

\(=\left(a-b\right)\left(a^2+ab+b^2\right)-2\left(a-b\right)\)

\(=\left(a-b\right)\left(a^2+ab+b^2-2\right)\)

a) 4x² + 4xy + y² 

= (2x + y)²

b) (2x + 1)² - (x - 1)² 

= (2x + 1  + x - 1)(2x + 1 - x + 1)

= 3x(x + 2) 

c) 9 - 6x + x² - y² 

 = (x² - 6x + 9) - y² 

= (x - 3)² - y² 

 = (x - y - 3)(x + y - 3) 

d) (-x - 2) + 3(x² - 4) 

 = -(x + 2) + 3(x - 2)(x + 2) 

 = (x + 2)(3x - 7) 

e) 5x²- 10xy² + 5y⁴

= 5(x² - 2xy² + y⁴) 

 = 5(x - y²)² 

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

g) 49(x - 4)² - 9(x + 2)² 

 = (7x - 28)² - (3x + 6)² 

 = (10x - 22)(4x - 34) 

h) (x² + y² - 5)² - 2(xy + 2)²

= \(\left(x^2+y^2-5\right)^2-\left(\sqrt{2}xy+2\sqrt{2}\right)^2\)

\(=\left(x^2+y^2+2\sqrt{xy}+2\sqrt{2}-5\right)\left(x^2+y^2-\sqrt{2}xy-2\sqrt{2}-5\right)\)

Trả lời:

\(27a^2b^2-18ab+3\)

\(=3\left(9a^2b^2-6ab+1\right)\)

\(=3\left[\left(3ab\right)^2-2.3ab.1+1^2\right]\)

\(=3\left(3ab-1\right)^2\)

a, \(\left(y-2\right)\left(y+2\right)\left(y^2+4\right)-\left(y+3\right)\left(y-3\right)\left(y^2+9\right)\)

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

\(=y^4-16-y^4+81=65\)

b, \(2\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)-2\left(x^6-y^6\right)\)

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

\(=2\left(x^6-y^6\right)-2\left(x^6-y^6\right)=0\)

1) a) 2a + 2b = 2(a + b) ; 

b) 9a - 9b = 9(a - b) ; 

c) 3a - 6b - 9c = 3(a - 2b - 3c) 

d) ab - ac = a(b - c) 

e) 5a - 10ax - 15a = -10a - 10ax = 10a(x + 1) 

f) 3a(ax - 2ay + 4) 

g) 5a²(x - y) + 10a(x - y) 

= 5a(x - y)(a + 2) 

2) ax + ay + bx + by 

 = a(x + y) + b(x + y) 

= (a + b)(x + y) 

b) a² - 49 = (a - 7)(a  + 7) 

c) 9a² - 1 = (3a - 1)(3a + 1) 

d) \(\frac{1}{4}a^2-b^2=\left(\frac{1}{2}a-b\right)\left(\frac{1}{2}a+b\right)\)

e) x² + 14x + 49 = (x + 7)²

f) 4x² + 20x + 25 = (2x + 5)²

g) 4x⁴ + 20x² + 25 = (2x² + 5)²

h) 2x³ + 8x² + 8x = 2x(x²  + 4x + 4) = 2x(x + 2)²

i) 2x³ + 16 = 2(x³ + 8) = 2(x + 2)(x² - 2x + 4) 

3) x² + 4x + 3 = x² + x + 3x + 3 = x(x + 1) + 3(x + 1) = (x + 1)(x + 3) 

x² + 7x + 10 = x² + 2x + 5x + 10 = x(x + 2) + 5(x + 2) = (x + 2)(x + 5) 

fuk                                                  hihi

ta có 2(x^2+x+1)/x^2+1

=2x^2+2x+2/x^2+1

=1+(x+1)^2/x^2+1>=1 với mọi x

dấu bằng xảy ra khi x=-1

bạn tự kết luận nha