Có : \(P=4x+\dfrac{16}{x^2}=2x+2x+\dfrac{16}{x^2}\)

- AD AMGM : \(2x+2x+\dfrac{16}{x^2}\ge3\sqrt[3]{\dfrac{2x.2x.16}{x^2}}=12\)

- Dấu " = " xảy ra \(\Leftrightarrow2x=\dfrac{16}{x^2}\)

\(\Leftrightarrow2x^3=16\)

\(\Leftrightarrow x=2\) ( TM )

Vậy ....

( Chắc đề như vầy :vvv )

Dùng cái này đánh công thức nha bạn

ĐKXĐ: -16/3<=x<=4

\(\)\(x^2-4x+12=4\sqrt{4-x}+\sqrt{3x+16}\)

=>\(x^2-4x+12-\left(-x+12\right)=4\sqrt{4-x}-\left(-\frac43x+8\right)+\sqrt{3x+16}-\left(\frac13x+4\right)\)

=>\(x^2-3x=\frac43\left\lbrack3\sqrt{4-x}-\left(6-x\right)\right\rbrack+\frac13\left\lbrack3\cdot\sqrt{3x+16}-\left(x+12\right)\right\rbrack\)

=>\(x^2-3x=\frac43\cdot\frac{9\left(4-x\right)-\left(6-x\right)^2}{3\sqrt{4-x}+\left(6-x\right)}+\frac13\cdot\frac{9\left(3x+16\right)-\left(x+12\right)^2}{3\sqrt{3x+16}+\left(x+12\right)}\)

=>\(x^2-3x=\frac43\cdot\frac{36-9x-x^2+12x-36}{3\sqrt{4-x}+\left(6-x\right)}+\frac13\cdot\frac{27x+144-x^2-24x-144}{3\sqrt{3x+16}+\left(x+12\right)}\)

=>\(\left(x^2-3x\right)\left(1+\frac43\cdot\frac{1}{3\sqrt{4-x}+6-x}+\frac{1}{3\left(3\sqrt{3x+16}+x+12\right)}\right)=0\)

=>\(x^2-3x=0\)

=>x(x-3)=0

=>x=0(nhận) hoặc x=3(nhận)

Ta có : 

\(x+16⋮x+12\)

\(x+12⋮x+12\)

\(\Rightarrow\left(x+16\right)-\left(x+12\right)⋮x+12\)

\(4⋮x+12\)

\(\Rightarrow x+12\in\text{Ư}\left(12\right)=\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow x\in\left\{-11;-13;-10;-14;-6;-16\right\}\)

cà khịa time x=+
 

cân cặc

10 chữ số 0 nha bạn

click cho mk nhé

Ta có : 

15 = 3.5

20 = 4.5

25 = 5.5

30 = 5.6

35= 5.7

40 = 5.8

45 = 5.9

50 = 5.5.2

Vậy tích trên có tận cùng là 10c/s 0

a: =>x+50=9/4

hay x=-47,75

b: =>4x-16=(4000+12)x2=8024

=>4x=8040

hay x=2010

a) \(\left(x+50\right):4=3^2\)

\(x+50:4=9\)

\(x+50=4.9\)

\(x+50=36\)

\(x=36-50\)

\(x=-14\)

b) \(\left(4x-16\right):2-\left(-12\right)=4000\)

\(4x-16:2-12.2=4000\)

\(4x-16:2-24=4000\)

\(4x-\left(16+24\right);2=4000\)

\(4x-30:2=4000\)

\(4x-30=4000.2\)

\(4x-30=8000\)

a) x 2  – x – 12                              b) x 3  – 64.

c) m 3 n 3   –   m 2 n   +   5 mn 2   –   5          d) 16 x 4  – 1.

giải

a) (x+3)(x-4)=\(x^2\) -4x+3x-12

=\(x^2\) +(-4x+3x)-12

=\(x^2\) -x-12

b) (x-4)(\(x^2\) +4x+16)=\(x^3\) +\(4x^2\) +16x-\(4x^2\) -16x-64

=\(x^3\) +(\(4x^2\) -\(4x^2\) )+(16x-16x)-64

=\(x^3\) -64

\(\Leftrightarrow\sqrt{x-4}\left(4-12\cdot\dfrac{1}{2}+2\cdot2\right)=6\)

=>x-4=9

hay x=13

B

B