a) 13 - 2 ( x + 1 ) = 7

=> 2(x+1)=13-7

=>2(x+1)=6

=> x+1=3

=> x=2

b) 15x - 13x = 12² + 5⁶ . 6

=> 15x-13x=93864

=> x.(15-13)=93864

=> x.2=93864

=> x=46947

a)13-2(x+1)=7

   2(x+1)=6

   x+1=3

      x=3-1=2

b)15x-13x=12²+5⁶.6

   15x-13x=12²+9375⁰

     2x=93894

     x=469847

Thiết nghĩ đề sai bạn ơi

a ) x = 25 28 b ) x = 5 12 .

c) x= 1             d) x = 17

a; \(x\) - \(\dfrac{3}{4}\) = \(\dfrac{1}{7}\)

    \(x\)        = \(\dfrac{1}{7}\) + \(\dfrac{3}{4}\)

    \(x\)       =  \(\dfrac{4}{28}\) + \(\dfrac{21}{28}\)

    \(x\)       =  \(\dfrac{25}{28}\)

Vậy \(x=\dfrac{25}{28}\) 

1) \(f\left(x\right)=6x^2-15x+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}-\dfrac{25}{36}\right)+4\)

\(\Rightarrow f\left(x\right)=6\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)+4-\dfrac{25}{6}\)

\(\Rightarrow f\left(x\right)=6\left(x-\dfrac{5}{6}\right)^2-\dfrac{1}{6}\ge-\dfrac{1}{6}\left(6\left(x-\dfrac{5}{6}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{1}{6}\left(tạix=\dfrac{5}{6}\right)\)

2) \(f\left(x\right)=4x^2-13x+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}-\dfrac{169}{64}\right)+5\)

\(\Rightarrow f\left(x\right)=4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)+5-\dfrac{169}{16}\)

\(\Rightarrow f\left(x\right)=4\left(x-\dfrac{13}{8}\right)^2-\dfrac{89}{16}\ge-\dfrac{89}{16}\left(4\left(x-\dfrac{13}{8}\right)^2\ge0,\forall x\right)\)

\(\Rightarrow GTNN\left(f\left(x\right)\right)=-\dfrac{89}{16}\left(tạix=\dfrac{13}{8}\right)\)

15C 16B

a) x = 6

b) x = 2

c) x = 7

d) x = -1

e) \(E=x^5-15x^4+16x^3-29x^2+13x\) tại x = 14

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

\(E=x^5-x^5-x^4+x^4+2x^3-2x^3-x^2+x^2-x\)

\(E=-x\)

\(E=-14\)

d)  \(D=x^3-30x^2-31+1\) tại x = 31

\(D=31^3-30.31^2-31+1\)

\(D=31^2\left(31-30-1\right)+1\)

\(D=0+1\)

\(D=1\)

\(=x^2+x+6x+6\\ =x\left(x+1\right)+6\left(x+1\right)\\ =\left(x+6\right)\left(x+1\right)\)

\(c,x^2+7x+6=x^2+x+6x+6=\left(x^2+x\right)+\left(6x+6\right)=x.\left(x+1\right)+6.\left(x+1\right)=\left(x+1\right).\left(x+6\right)\)

\(15x-3-x^2+2x+x^2-13x=7\)

\(\Leftrightarrow4x=10\Leftrightarrow x=\dfrac{5}{2}\)

Ta có:

     3(5x - 1) - x(x - 2) + x² - 13x = 7

→15x - 3 - x² + 2x + x² - 13x = 7

→4x - 3 = 7

→4x = 10

→x = \(\dfrac{5}{2}\)