16 + 3 = ?
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\(P=\dfrac{16^7\cdot5^3\left(5-1\right)}{16^7\cdot\left(25^2-5^3\right)}=\dfrac{5^3\cdot2^2}{5^4-5^3}=\dfrac{5^3\cdot2^2}{5^3\cdot\left(5-1\right)}=1\)
\(S=\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y^3}{16\left(x+16\right)}+\dfrac{2021}{2022}\)
\(\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y+16}{100}+\dfrac{16}{80}\ge3\sqrt[3]{\dfrac{x^3\left(y+16\right).16}{16\left(y+16\right).100.80}}=\dfrac{3x}{20}\)
\(tương\) \(tự\Rightarrow\dfrac{y^3}{16\left(x+16\right)}\ge\dfrac{3y}{20}\)
\(\Rightarrow S\ge\dfrac{3x}{20}+\dfrac{3y}{20}-\left(\dfrac{x+16}{100}+\dfrac{y+16}{100}\right)-2.\dfrac{16}{80}+\dfrac{2021}{2022}=\dfrac{3x+3y}{20}-\dfrac{x+y+32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{15x+15y-x-y-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{14\left(x+y\right)-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}\)
\(xy=16\le\dfrac{\left(x+y\right)^2}{4}\Rightarrow x+y\ge8\Rightarrow S\ge\dfrac{14.8-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{2}{5}+\dfrac{2021}{2022}\)
\(\Rightarrow minS=\dfrac{2}{5}+\dfrac{2021}{2022}\Leftrightarrow x=y=4\)
\(\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y+16}{100}+\dfrac{1}{5}\ge3\sqrt[3]{\dfrac{x^3\left(y+16\right)}{16.100.5\left(y+16\right)}}=\dfrac{3x}{20}\)
Tương tự: \(\dfrac{y^3}{16\left(x+16\right)}+\dfrac{x+16}{100}+\dfrac{1}{5}\ge\dfrac{3y}{20}\)
Cộng vế:
\(S+\dfrac{x+y+32}{100}+\dfrac{2}{5}\ge\dfrac{3\left(x+y\right)}{20}+\dfrac{2021}{2022}\)
\(S\ge\dfrac{9}{20}\left(x+y\right)-\dfrac{42}{25}+\dfrac{2021}{2022}\ge\dfrac{9}{20}.2\sqrt{xy}-\dfrac{42}{25}+\dfrac{2021}{2022}=...\)
a) \(\dfrac{-15+9+11}{16}=\dfrac{5}{16}\)
b) \(\dfrac{2}{3}\left(1,4+1,6-1,2\right)=\dfrac{2}{3}\times\dfrac{9}{5}=\dfrac{6}{5}\)
c) \(3\dfrac{2}{15}\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{31}{15}=\dfrac{47}{15}-\dfrac{31}{15}=\dfrac{16}{15}\)
\(\dfrac{9}{16}-\dfrac{3}{16}:\dfrac{3}{8}\) \(=\dfrac{9}{16}-\dfrac{3}{16}\cdot\dfrac{8}{3}\) \(=\dfrac{9}{16}-\dfrac{1}{2}\) \(=\dfrac{9}{16}-\dfrac{8}{16}=\dfrac{1}{16}\)
bằng 19 nha bé
TL:
16+3=19
HT