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

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

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

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

Chọn B

B

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 \(x=2.\)

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}

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

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

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

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

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

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

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

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

Cộng vế:

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

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

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

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

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

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

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

\(X+16=-5\)

\(X=-21\)

a) \(39\times118+29\times82\)

\(=39\times2\times59+29\times2\times41\)

\(=2\times\left(39\times59+29\times41\right)\)

\(=2\times\left(2301+1189\right)\)

\(=2\times3490\)

\(=6980\)

b) \(37\times39+78\times14-13\times85-52\times55\)

\(=37\times3\times13+6\times13\times14+13\times85-4\times13\times55\)

\(=13\times\left(37\times3+6\times14+85-4\times55\right)\)

\(=13\times\left(111+84+85-220\right)\)

\(=13\times\left(280-220\right)\)

\(=13\times60\)

\(=780\)

c) \(368:16+752:16+480:16\)

\(=\left(368+752+480\right):16\)

\(=\left(1120+480\right):16\)

\(=1600:16\)

\(=100\)

d) \(5\times11\times18+9\times31\times10+4\times29\times45\)

\(=5\times11\times9\times2+9\times31\times10+2\times2\times29\times5\times9\)

\(=10\times9\times11+10\times9\times31+10\times29\times2\times9\)

\(=90\times11+90\times31+90\times29\times2\)

\(=90\times\left(11+31+29\times2\right)\)

\(=90\times\left(42+58\right)\)

\(=90\times100\)

\(=9000\)