ko có số nào bằng dãy trên vì chỗ 4xx7 ko có phép tính nào như vậy

Vi ga co 2 chan nen 100 con ga se co: 2.100=200( chan)

Vi bo co 4 chan nen 200 con bo se co : 4.200=800(chan)

vay co tat ca : 200+800=1000 (chan)

nham ab+bc+ac\(\ge3\sqrt[3]{\left(abc\right)^2}\)

Áp dụng BĐT Schur:

\(a^3+b^3+c^3+3abc\ge ab\left(a+b\right)+bc\left(c+b\right)+ca\left(a+c\right)\Rightarrow a^3+b^3+c^3+3abc+3abc\ge\left(a+b+c\right)\left(ab+bc+ac\right)=3\left(a+b+c\right)\)

Vậy \(A+3abc\ge3\left(a+b+c\right)\)

Cauchy :\(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\Rightarrow a+b+c\ge3\)

\(ab+bc+ac\ge3\sqrt[3]{abc}\Rightarrow1\ge abc\Rightarrow-3\le-3abc\)

A\(\ge\) 3(a+b+c)-3abc\(\ge\)3.3-3=6

Vậy A min=6\(\Leftrightarrow\) a=b=c=1

130, 2

4500 

tick nha

Ta có:\(x+\dfrac{1}{y}\le\dfrac{1}{2}\)\(\Rightarrow\)\(2xy+2\le y\)\(\le\dfrac{y^2+16}{8}\)(Cauchy)

\(\Rightarrow\)\(16xy+16\le y^2+16\Rightarrow16x\le y\)

Ta có:\(\dfrac{x}{y}+\dfrac{y}{x}=\dfrac{x}{y}+\dfrac{y}{256x}+\dfrac{255y}{256x}\)\(\ge2\sqrt{\dfrac{x.y}{y.256x}}+\dfrac{255.16x}{256x}=\dfrac{1}{8}+\dfrac{255}{16}=\dfrac{257}{16}\)

Dấu ''='' xảy ra \(\Leftrightarrow\) y=4;x=\(\dfrac{1}{4}\)

Chuẩn hóa: a+b+c=3k

\(\Rightarrow\)\(\dfrac{a}{k}+\dfrac{b}{k}+\dfrac{c}{k}=3\)

Đặt (\(\dfrac{a}{k};\dfrac{b}{k};\dfrac{c}{k}\))\(\Rightarrow\left(x;y;z\right)\);x+y+z=3

ĐPCM\(\Leftrightarrow\)\(\sum\dfrac{19y^3-x^3}{xy+5y^2}\le3\left(x+y+z\right)\)

Ta CM BĐT:

\(\dfrac{19y^3-x^3}{xy+5y^2}\le4y-x\Leftrightarrow-\dfrac{\left(y-x\right)^2\left(x+y\right)}{xy+5y^2}\le0\)(đúng)

CMTT\(\Rightarrow\)ĐPCM

Bác nào ở đây zậy trời! _ _|