ta có:  f(x) + g(x) = ( 7 x^6 - 6x ^5 +5x^4 -4x^3 +3x^2 -2x +1) - ( x - 2x^2 +3x^3 - 4x^4 + 5x^5 - 6x^6)

                          \(=7x^6-6x^5+5x^4-4x^3+3x^2-2x+1-x+2x^2-3x^3+4x^4-5x^5+6x^6\)

                      \(=\left(7x^6+6x^6\right)-\left(6x^5+5x^5\right)+\left(5x^4+4x^4\right)-\left(4x^3+3x^3\right)+\left(3x^2+2x^2\right)-\left(2x+x\right)+1\)

\(=13x^6-11x^5+9x^4-7x^3+5x^2-3x+1\)

Chúc bn học tốt !!!!!!

Uhhhhhhhhhhhhhhhhhhhhhhhhhh😥😥😥😥😥😥😥😥😥😥😥????????????...............

a) 6x² - 5x + 3 = 2x - 3x(2 - x)

<=> 6x² - 5x + 3 = 2x - 6x + 3x²

<=> 6x² - 5x + 3 = -4x + 3x²

<=> 6x² - 5x + 3 + 4x - 3x² = 0

<=> 3x² - x + 3 = 0

=> Pt vô nghiệm

b) 25x² - 9 = (5x + 3)(2x + 1)

<=> 25x² - 9 = 10x² + 5x + 6x + 3

<=> 25x² - 9 = 10x² + 11x + 3

<=> 25x² - 9 - 10x² - 11x - 3 = 0

<=> 15x² - 12 - 11x = 0

<=> 15x² + 9x - 20x - 12 = 0

<=> 3x(5x + 3) - 4(5x + 3) = 0

<=> (5x + 3)(3x - 4) = 0

<=> 5x + 3 = 0 hoặc 3x - 4 = 0

<=> x = -3/5 hoặc x = 4/3

2:

=>x^3-1-2x^3-4x^6+4x^6+4x=6

=>-x^3+4x-7=0

=>x=-2,59

4: =>8x-24x^2+2-6x+24x^2-60x-4x+10=-50

=>-62x+12=-50

=>x=1

a) = 2

b) = 1

c) = 2 

a) \(6x^2-2x-6x^2+13=0\\ -2x=-13\\ x=\dfrac{13}{2}\)

b: =>2x-2x-1=x-6x

=>-5x=-1

hay x=1/5

Lời giải:
a. 

$2x(3x-1)=6x^2-13$

$\Leftrightarrow 6x^2-2x=6x^2-13$

$\Leftrightarrow 2x=13$

$\Leftrightarrow x=\frac{13}{2}$

b. 

$\frac{x}{3}-\frac{2x+1}{6}=\frac{x}{6}-x$

$\Leftrightarrow \frac{2x-(2x+1)}{6}=\frac{-5}{6}x$

$\Leftrightarrow \frac{-1}{6}=\frac{-5}{6}x$
$\Leftrightarrow x=\frac{-1}{6}: \frac{-5}{6}=\frac{1}{5}$

giúp mk đi mà

xy-3x+y=19

<=>x(y-3)+(y-3)=19

<=>(x+1)(y-3)=19

=>