\(S=\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y^3}{16\left(x+16\right)}+\dfrac{2021}{2022}\)

\(\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y+16}{100}+\dfrac{16}{80}\ge3\sqrt[3]{\dfrac{x^3\left(y+16\right).16}{16\left(y+16\right).100.80}}=\dfrac{3x}{20}\)

\(tương\) \(tự\Rightarrow\dfrac{y^3}{16\left(x+16\right)}\ge\dfrac{3y}{20}\)

\(\Rightarrow S\ge\dfrac{3x}{20}+\dfrac{3y}{20}-\left(\dfrac{x+16}{100}+\dfrac{y+16}{100}\right)-2.\dfrac{16}{80}+\dfrac{2021}{2022}=\dfrac{3x+3y}{20}-\dfrac{x+y+32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{15x+15y-x-y-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{14\left(x+y\right)-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}\)

\(xy=16\le\dfrac{\left(x+y\right)^2}{4}\Rightarrow x+y\ge8\Rightarrow S\ge\dfrac{14.8-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{2}{5}+\dfrac{2021}{2022}\)

\(\Rightarrow minS=\dfrac{2}{5}+\dfrac{2021}{2022}\Leftrightarrow x=y=4\)

\(\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y+16}{100}+\dfrac{1}{5}\ge3\sqrt[3]{\dfrac{x^3\left(y+16\right)}{16.100.5\left(y+16\right)}}=\dfrac{3x}{20}\)

Tương tự: \(\dfrac{y^3}{16\left(x+16\right)}+\dfrac{x+16}{100}+\dfrac{1}{5}\ge\dfrac{3y}{20}\)

Cộng vế:

\(S+\dfrac{x+y+32}{100}+\dfrac{2}{5}\ge\dfrac{3\left(x+y\right)}{20}+\dfrac{2021}{2022}\)

\(S\ge\dfrac{9}{20}\left(x+y\right)-\dfrac{42}{25}+\dfrac{2021}{2022}\ge\dfrac{9}{20}.2\sqrt{xy}-\dfrac{42}{25}+\dfrac{2021}{2022}=...\)

\(=\dfrac{16\left(48+24+28\right)}{16}=100\)

`[16xx48+16xx24+16xx28]/[325-317+426-418]`

`=[16xx(48+24+28)]/[8+426-418]`

`=[16xx100]/[434-418]=[16xx100]/16=100`

Áp dụng BĐT Bunhiacopski:

Đặt \(A=x\sqrt{16-y}+\sqrt{y\left(16-x^2\right)}\)

\(\Leftrightarrow A^2=\left[x\sqrt{16-y}+\sqrt{y\left(16-x^2\right)}\right]^2\le\left(x^2+16-x^2\right)\left(16-y+y\right)\\ \Leftrightarrow A^2\le16\cdot16=256\\ \Leftrightarrow A\le16\\ A_{max}=16\Leftrightarrow\dfrac{x^2}{16-x^2}=\dfrac{16-y}{y}\Leftrightarrow x^2y=256-16y-16x^2+x^2y\\ \Leftrightarrow16x^2+16y-256=0\\ \Leftrightarrow x^2+y-16=0\\ \Leftrightarrow x^2=16-y\Leftrightarrow x=\sqrt{16-y}\)

1. \(\text{6 + 2.(x - 19) = 16}.\)

\(\Leftrightarrow2.\left(x-19\right)=10.\)

\(\Leftrightarrow x-19=5.\)

\(\Leftrightarrow x=24.\)

Vậy \(x=24.\)

2. \(\text{(-240) : x – 16 = 64}.\)

\(\Leftrightarrow\left(-240\right):x=80.\)

\(\Leftrightarrow x=-3.\)

Vậy \(x=-3.\)

3. \(2x^3=16.\)

\(\Leftrightarrow x^3=8.\)

\(\Leftrightarrow x=2.\)

Vậy \(x=2.\)

Chọn B

B

\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)

\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)

\(X+16=-5\)

\(X=-21\)

327 - 16 + 216 - 27 + 600          16 x 48 + 8 x 48 + 16 x 28

=(327 - 27) + (216 - 16) + 600      =16 x 48 + 16 x 24 + 16 x 28 

=300           + 200           + 600     =16 x (48 + 24 + 28)

=500                               + 600     =16 x 100

=1100                                            =1600

\(\left\{{}\begin{matrix}x-16=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=16\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left(x-16\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-16=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=\dfrac{1}{3}\end{matrix}\right.\)

=>x-16=0 hoặc 3x-1=0

    x     = 16         x   = ⅓

c: C={0;3;6;9;12;15;18}

c,Theo đầu bài ta có:

x⋮3 và x<20 => x thuộc tập hợp 3,6,9,12,15,18

d,Theo đầu bài ta có:

16⋮x và 0<x<16 => x thuộc tập hợp 4,8

C, x={0;3;6;9;12;15;18}

D, x={1;2;4;8}