=>9/72<4x/72<16/72

=>9<4x<16

mà x là số nguyên

nên 4x=12

hay x=3

\(\dfrac{1}{8}< \dfrac{x}{18}< \dfrac{2}{9}\\ \Rightarrow\dfrac{9}{4}< x< 4\\ \Rightarrow2,25< x< 4\\ \Rightarrow x=3\)

`C=-2x(x+7)=-2x^2-14x`

`=-(2x^2+14x)`

`=-( (\sqrt2x)^2 + 2.\sqrt2 x . (7\sqrt2)/2 + ((7\sqrt2)/2)^2 )+49/2`

`=-(\sqrt2x+(7\sqrt2)/2)^2+49/2`

`=> C_(max) = 49/2 <=> x=-7/2`

`D=-3x^2+5x-9`

`=-(3x^2-5x+9)`

`=-((\sqrt3x)^2 - 2.\sqrt3x . (5\sqrt3)/6 + ((5\sqrt3)/6)^2)-83/12`

`=-(\sqrt3x-(5\sqrt3)/6)^2-83/12`

`=> D_(max)=-83/12 <=> \sqrt3x - (5\sqrt3)/6=0 <=> x=5/6`

a) Ta có: \(C=-2x\left(x+7\right)\)

\(=-2\left(x^2+7x\right)\)

\(=-2\left(x^2+7x+\dfrac{49}{4}-\dfrac{49}{4}\right)\)

\(=-2\left(x+\dfrac{7}{2}\right)^2+\dfrac{49}{2}\le\dfrac{49}{2}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{7}{2}\)

b) Ta có: \(D=-3x^2+5x-9\)

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

\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{6}+\dfrac{25}{36}+\dfrac{83}{36}\right)\)

\(=-3\cdot\left(x-\dfrac{5}{6}\right)^2-\dfrac{83}{12}\le-\dfrac{83}{12}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{5}{6}\)

`9^(x-1)=9`

`=> 9^(x-1)=9^1`

`=>x-1=1`

`=>x=1+1`

`=>x=2`

`(2x+1)^2=9`

`(2x+1)^2=3^2=(-3)^2`

TH1: `2x+1=3`

`2x=2`

`x=1`

TH2: `2x+1=-3`

`2x=-4`

`x=-2`

(2x + 1)² = 9

(2x + 1)² = 3²

\(\Rightarrow\)Trường hợp 1:                 

  2x + 1 = 3                             

  2x = 3 - 1

  2x = 2

  x = 2 : 2

  x = 1

Trường hợp 2:

  2x + 1 = -3

  2x = -3 - 1

  2x = -4

  x = -4 : 2

  x = -2

x \(\in\) \(\left\{-2;1\right\}\)

Ta có: \(\left(2x+1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

\(\Leftrightarrow9x\left(x+2\right)+9y\left(y-\dfrac{2}{3}\right)=10\\ \Leftrightarrow9x^2+18x+9y^2-6y-10=0\\ \Leftrightarrow\left(9x^2+18x+9\right)+\left(9y^2-6y+1\right)=0\\ \Leftrightarrow9\left(x+1\right)^2+\left(3y-1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=\dfrac{1}{3}\end{matrix}\right.\)

Để B nhỏ nhất thì x-9=1

hay x=10

Ta có: x=9

nên x+1=10

Ta có: \(x^{14}-10x^{13}+10x^{12}-...+10x^2-10x+10\)

\(=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+x+1\)

\(=x^{14}-x^{14}-x^{13}+x^{13}+x^{12}-...+x^3+x^2-x^2-x+x+1\)

=1

b

A 24/3 nhá bạn 

24 : 3=........=..........

a.24/3          b.8          c.9

a, 47,5 x 0,5 x 4 

= 47,5 x 2 

= 95

b, 4 x 6,58 x 2,25

= 4 x 2,25 x 6,58

= 9 x 6,58

= 59,22

c, 1,25 x 2 x 13,8 x 4

= 1,25 x 2 x 4 x 13,8

= 1,25 x 8 x 13,8

= 10 x 13,8

= 138

a,47,5x0.5x4

=47,5 x 2

=95

b,4x6,58x2,25

=4x 2,25x6,58

=9x6,58

=59,22

c,1,25x2x13,8x4

=1,25x4x2x13,8

=5x2x13,8

=10x13,8

=138

a ) 47,5 x 0,5 x 4 = 95

b ) 4 x 6,58 x 2,25 = 59,22

c ) 1,25 x 2 x 13,8 x 4 = 138