Lời giải:
a. $(x+y)-(x-y)=x+y-x+y=(x-x)+y+y=0+2y=2y$

b. $3x(5x^2-2x-1)-15x^3=15x^3-6x^2-3x-15x^3=-6x^2-3x$
c. $(5x-2y)(x^2-xy+1)+7x^2y=5x^3-5x^2y+5x-2x^2y+2xy^2-2y+7x^2y$

$=5x^3+(-5x^3y-2x^2y+7x^2y)+5x+2xy^2-2y$

$=5x^3+5x+2xy^2-2y$

a; 3\(x\). (5\(x^2\) - 2\(x\) - 1) - 15\(x^3\)

 = 15\(x^3\) - 6\(x^2\) - 3\(x\) - 15\(x^3\)

= (15\(x^3\) - 15\(x^3\)) - 6\(x^2\) - 3\(x\)

= -6\(x^2\) - 3\(x\) 

b; (5\(x\) - 2y).(\(x^2\) - \(xy\) + 1) + 7\(x^2\).y

= 5\(x^3\) - 5\(x^2\)y + 5\(x\) - 2y\(x^2\)y+ 2\(xy^2\) -  2y + 7\(x^2\).y

=  5\(x^3\) - (5\(x^2\)y + 2\(x^2\)y - 7\(x^2\).y) + 5\(x\) + 2\(xy^2\)- 2y 

= 5\(x^3\) - 0 + 5\(x\) + 2\(xy^2\)- 2y

= 5\(x^2\) + 5\(x\) + 2\(xy^2\)- 2y

MP = 5cm.

Ai giúp tui vs

1)

\(\dfrac{x-1}{2014}+\dfrac{x-2}{2013}+\dfrac{x-3}{2012}+...+\dfrac{x-2014}{1}=2014\)

\(\Leftrightarrow\left(\dfrac{x-1}{2014}-1\right)+\left(\dfrac{x-2}{2013}-1\right)+...+\left(\dfrac{x-2014}{1}-1\right)=0\)

\(\Leftrightarrow\dfrac{x-2015}{2014}+\dfrac{x-2015}{2013}+...+\dfrac{x-2015}{1}=0\)

\(\Leftrightarrow\left(x-2025\right)\left(\dfrac{1}{2014}+\dfrac{1}{2013}+...+\dfrac{1}{1}\right)=0\)

\(\Leftrightarrow x=2015\)

Vậy \(S=\left\{2015\right\}\)

a) (x + 3)² - (x - 2)(x + 2) = 1

x² + 6x + 9 - x² + 4 - 1 = 0

6x + 12 = 0

6x = 0 - 12

6x = -12

x = -12/6

x = -2

b) M = x² - 6x

= x² - 6x + 9 - 9

= (x - 3)² - 9

Do (x - 3)² ≥ 0 với mọi x ∈ R

⇒ (x - 3)² - 9 ≥ -9

Vậy giá trị nhỏ nhất của M là -9 khi x = 3

Thiếu đề bài.