C lớn nhất khi (x-3)²+1 bé nhất 

=>x²-9 +1 bé nhất 

x²-8 bé nhất 

=>x² khác 8 và x²-8 bé nhất => x² -8=1

=>x²=9=>x=3

D lớn nhất khi |x-2|+2 bé nhất =>x-2 bé nhất=>x-2=0 =>x=2

\(\frac{x}{4}=\frac{18}{x+1}\)

\(\Leftrightarrow x\left(x+1\right)=72\)

\(\Leftrightarrow x=8\)

P/s tham khảo nha

bn trần hoàng việt ơi, bn có thể giải kĩ hơn đc chứ?

(suy ra )  x(x+1)=72

               x(x+1)=8(7+1)

        Vậy x=8 

mình đảm bảo đúng

\(ĐK:x\ne\pm1\)

\(PT\Leftrightarrow\frac{3x+2}{\left(x-1\right)^2}-\frac{6}{\left(x+1\right)\left(x-1\right)}-\frac{3x-2}{\left(x+1\right)^2}\)

Bạn tự quy đồng rồi rút gọn nhé!!

\(\frac{x-2}{2}-\frac{1+x}{3}=\frac{4-3x}{4}-1\)

\(\Leftrightarrow\frac{3\left(x-2\right)-2\left(1+x\right)}{6}=\frac{4-3x-4}{4}\)

\(\Leftrightarrow\frac{3x-6-2-2x}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow\frac{x-8}{6}=-\frac{3x}{4}\)

\(\Leftrightarrow4x-32=-18x\)

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

\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}+1\)

\(\Leftrightarrow\frac{20}{x+3}-8=8-\frac{18}{x+3}\)

\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=8+8\)

\(\Leftrightarrow\frac{38}{x+3}=16\)

\(\Leftrightarrow x+3=2,375\)

\(\Leftrightarrow x=-0,625\)

\(\left(\frac{5}{x+3}-2\right).4=7-\left(\frac{9}{x+3}+\frac{1}{2}\right).2\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\left(\frac{18}{x+3}+1\right)\)

\(\Leftrightarrow\frac{20}{x+3}-8=7-\frac{18}{x+3}-1\)

\(\Leftrightarrow\frac{20}{x+3}+\frac{18}{x+3}=7-1+8\)

\(\Leftrightarrow\frac{38}{x+3}=14\)

\(\Leftrightarrow\left(x+3\right)14=38\)

\(\Leftrightarrow14x+42=38\)

\(\Leftrightarrow14x=-4\Leftrightarrow x=-\frac{4}{14}=-\frac{2}{7}\)

Vậy \(x=-\frac{2}{7}\)

TRẻ Trâu 

3.(7 + x) = 88 - 8²

3.(7 + x) = 24

     7 + x = 24 : 3

     7 + x = 8

           x = 8 - 7

           x = 1

3 . ( 7 + x ) = 88 - 8²

3 . ( 7 + x ) = 88 - 64

3 . ( 7 + x ) = 24

7 + x = 24 : 3

7 + x = 8

      x = 8 - 7

      x = 1

\(3.\left(7+x\right)=88-8^2\)

\(3\left(x+7\right)=88-64\)

\(3\left(x+7\right)=24\)

\(x+7=8\)

\(x=1\)

vậy \(x=1\)

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😥😥😥😥😥😥😥😥😥😥😥????????????...............

\(\frac{-4}{7}-x=\frac{3}{5}-2x\)

\(\frac{-4}{7}-\frac{-3}{5}=-2x+x\)

                 \(\frac{1}{35}=x.\left(-2+1\right)\)

   \(\frac{1}{35}=-1.x\)

       \(x=\frac{1}{35}:-1\)

       \(x=\frac{-1}{35}\)

Học tốt nha!!!

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. 

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

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

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