a) (x-y)(x⁴+x³y+x²y²+xy³+y⁴) = x(x⁴+x³y+x²y²+xy³+y⁴)-y(x⁴+x³y+x²y²+xy³+y⁴) =(x⁵+x⁴y+x³y²+x²y²+xy⁴)-(x⁴y+x³y²+x²y²+xy⁴+y⁵) = x⁵+x⁴y+x³y²+x²y²+xy⁴-x⁴y-x³y²-x²y²-xy⁴-y⁵ =x⁵-y⁵⇒Điều cần chứng minh

Các câu b d tương tự

a. = \(\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)

= \(x^2\left(x+1\right)+7x\left(x+1\right)+10x\left(x+1\right)\)

= \(\left(x+1\right)\left(x^2+7x+10x\right)\)

= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)

\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{a\left(x-y\right)+b\left(x-y\right)}{2\left(x^2-y^2\right)}\)

\(=\dfrac{2\left(x+y\right)}{\left(a+b\right)^2}.\dfrac{\left(x-y\right)\left(a+b\right)}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{1}{a+b}\)

\(=\dfrac{a+b-c}{\left(a+b\right)^2-c^2}.\dfrac{\left(a+b\right)^2+c\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}\)

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

\(=\dfrac{1}{a-b}\)

\(c,\dfrac{x^3+1}{x^2+2x+1}.\dfrac{x^2-1}{2x^2-2x+2}\)

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

Ta có: \(1+xy=x^2+y^2\ge2xy\Rightarrow xy\le1\)

\(1+xy=x^2+y^2\ge-2xy\Rightarrow xy\ge-\dfrac{1}{3}\)

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

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

Đặt \(a=xy\Rightarrow P=f\left(a\right)=-2a^2+2a+1\)

Xét hàm \(f\left(a\right)=-2a^2+2a+1\) trên \(\left[-\dfrac{1}{3};1\right]\)

\(-\dfrac{b}{2a}=\dfrac{1}{2}\in\left[-\dfrac{1}{3};1\right]\)

\(f\left(-\dfrac{1}{3}\right)=\dfrac{1}{9}\) ; \(f\left(\dfrac{1}{2}\right)=\dfrac{3}{2}\) ; \(f\left(1\right)=1\)

\(\Rightarrow M=\dfrac{3}{2}\) ; \(m=\dfrac{1}{9}\) \(\Rightarrow Mm=\dfrac{1}{6}\)

\(x^2y^2+2x^2+y^2+2\)

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

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

\(a^2-b^2+a-b\)

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

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

\(a,x^2y^2+2x^2+y^2+2\)

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

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

\(b,a^2-b^2+a-b\)

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

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

a.
=y²(x²+1)+2(x²+1)

=(x²+1)(y²+2)

b.
=(a-b)(a+b)+(a-b)
=(a-b)(a+b+1)

T a   c ó : x 2 + y 2 = a 2 + b 2 ⇔ x 2 - a 2 = b 2 - y 2 ⇔ x - a x + a = b - y b + y M à   x + y   = a + b ⇔ x - a = b - y   n ê n   t a   c ó x - a x + a = x - a b + y ⇔ x - a x + a - x - a b + y = 0 ⇔ x - a x + a - b - y = 0 ⇔ x - a = 0 x + a - b - y = 0 ⇔ x = a x - y = b - a

+) Với x = a thay vào x + y = a + b ta có: a + y = a + b

Suy ra y = b

Do đó:   x n + y n = a n + b n

+) Với x - y = b - a suy ra x = b - a + y thay vào x + y = a + b ta có:

 b - a + y  + y = a + b

2y = 2a

y = a

Suy ra x - a = b - a hay x = b

Do đó:  x n + y n = b n + a n = a n + b n

Vậy  x n + y n = a n + b n

Đáp án cần chọn là C

a,x²-y²-2x+2y
= (x+y)(x-y) - 2(x-y)
= (x-y)(x+y-2)
b,2x+2y-x²-xy
= 2(x+y) - x(x+y)
= (x+y)(2-x)
c,3a²-6ab+3b²-12c²
= 3(a² - 2ab + b² - 4c²)
= 3[(a-b)² - 4c²)
= 3(a-b-2c)(a-b+2c)
d,x²-25+y²+2xy
= (x+y)² - 25
= (x+y+5)(x+y-5)

e) a²+2ab+b²-ac-bc

= (a+b)²-c(a+b)

= (a+b)( a+b-c)

f) x²-2x-4x²-4y

= -3x²-2x-4y

= -(3x²+2x+4y)

g)x²y-x³-9y+9x

= x²(y-x)-9(y-x)

= (y-x)(x²-9)

h) x²(x-1)+16(1-x)

= x²(x-1)-16(x-1)

= (x-1)(x²-16)

= (x-1)(x-4)(x+4)

n) 81x²-6yz-9y²-z²

= (9x)²-[(3y)²+6yz+z²]

=(9x)²-(3y+z)²

=(9x+3y+z)(9x-3y-z)

m) xz- yz-x²+2xy-y²

= z(x-y)-(x²-2xy+y²)

= z(x-y)-(x-y)²

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

 p) x² + 8x + 15

= x² + 3x + 5x + 15

= x(x+3) + 5(x+3)

= (x+3)(x+5)

k) x² - x - 12

= x² + 3x - 4x - 12

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

= (x+3)(x-4)