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) ĐKXĐ: x <= 2/3

Pt --> 2 - 3x = 4

<=> 3x = -2

<=> x = -2/3 (thỏa)

b) ĐKXĐ: x >= 2

Pt --> x^2 + 4x + 4 = x^2 - 4x + 4

<=> 8x = 0<=> x = 0(loại)

a: Ta có: \(\sqrt{2-3x}=2\)

\(\Leftrightarrow2-3x=4\)

\(\Leftrightarrow3x=-2\)

hay \(x=-\dfrac{2}{3}\)

b: Ta có: \(\sqrt{x^2+4x+4}=x-2\)

\(\Leftrightarrow\left|x+2\right|=x-2\)

\(\Leftrightarrow x+2=2-x\left(x< -2\right)\)

\(\Leftrightarrow x=0\left(loại\right)\)

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

\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}=3x-1\\2x-\frac{1}{2}=1-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1+\frac{1}{2}\\2x+3x=1+\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)

1. -4x( x + 3 )( x - 4 ) - 3x( x² - x + 1 )

= -4x( x² - x - 12 ) - 3x( x² - x + 1 )

= -4x³ + 4x² + 48x - 3x³ + 3x² - 3x

= -7x³ + 7x² + 45x

2. a) 4x( x - 5 ) - ( x - 1 )( 4x - 3 ) = 5

<=> 4x² - 20x - ( 4x² - 7x + 3 ) = 5

<=> 4x² - 20x - 4x² + 7x - 3 = 5

<=> -13x - 3 = 5

<=> -13x = 8

<=> x = -8/13

b) 6( x - 3 )( x - 4 ) - 6x( x - 2 ) = 4

<=> 6( x² - 7x + 12 ) - 6x² + 12x = 4

<=> 6x² - 42x + 72 - 6x² + 12x = 4

<=> -30x + 72 = 4

<=> -30x = -68

<=> x = 34/15

Bài 1 : 

\(-4x\left(x+3\right)\left(x-4\right)-3x\left(x^2-x+1\right)\)

\(=-7x^3+7x^2+45x\)

Bài 2 : 

a, \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

\(\Leftrightarrow4x^2-20x-\left[4x^2-7x+3\right]=5\)

\(\Leftrightarrow4x^2-20x-4x^2+7x-3=5\)

\(\Leftrightarrow-13x-8=0\Leftrightarrow x=-\frac{8}{13}\)

b, \(6\left(x-3\right)\left(x-4\right)-6x\left(x-2\right)=4\)

\(\Leftrightarrow6x^2-42x+72-6x^2+12x=4\)

\(\Leftrightarrow-30x+68=0\Leftrightarrow x=\frac{34}{15}\)

\(\left(x-1\right)3+3x\left(x-1\right)=0\)

<=>  \(3\left(x-1\right)\left(x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x-1=0\\x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Vậy...

\(Sửa:A=x^4-6x^3+13x^2-12x+2021\\ A=\left(x^4-6x^3+9x^2\right)+4\left(x^2-3x\right)+4+2017\\ A=\left(x^2-3x\right)^2+4\left(x^2-3x\right)+4+2017\\ A=\left(x^2-3x+2\right)^2+2017\ge2017\\ A_{min}=2017\Leftrightarrow x^2-3x+2=0\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a)\(\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x-1\right)=15\)

     \(4x^2-11x-3-\left(4x^2-29x+7\right)=15\)

     \(4x^2-11x-3-4x^2+29x-7=15\)

      \(18x-10=15\)

       \(x=\frac{25}{18}\)

b)\(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)

     \(\left(x+1\right)\left(3x-5-3x+1\right)=x-4\)

       \(\left(x+1\right).\left(-4\right)-x+4=0\)

        \(-4x-4-x+4=0\)

          \(x=0\)

a) \(\left(4x+1\right)\left(x-3\right)-\left(x-7\right)\left(4x+1\right)=15\)

\(\left(4x+1\right)\left(x-3-x+7\right)=15\)

\(\left(4x+1\right)4=15\)

\(4x+1=15:4\)

\(4x+1=3,75\)

\(x=\frac{11}{16}\)

b) \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)

\(\left(x+1\right)\left(3x-5-3x+1\right)=x-4\)

\(\left(x+1\right).-4=x-4\)

\(-4x-4-x+4=0\)

\(-5x=0\)

\(x=0\)

k nhé

a: Ta có: \(x^2-8x+20\)

\(=x^2-8x+16+4\)

\(=\left(x-4\right)^2+4>0\forall x\)

b: Ta có: \(-x^2+6x-19\)

\(=-\left(x^2-6x+19\right)\)

\(=-\left(x^2-6x+9+10\right)\)

\(=-\left(x-3\right)^2-10< 0\forall x\)