a) \(\frac{a}{9}+\frac{3}{b}=\frac{1}{18}\Leftrightarrow\frac{ab+27}{9b}=\frac{1}{18}\left(1\right)\)

\(\Rightarrow9b=18\Rightarrow b=2\left(2\right)\)

Thay (2) vào (1) ta có 

\(\frac{2a+27}{18}=\frac{1}{18}\Rightarrow2a+27=1\Rightarrow2a=-26\Rightarrow a=-13\)

\(7\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=6\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)+3\ge7\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)\)

\(\Rightarrow\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\le3\)Áp dụng BĐT AM-GM ta có : 

\(A=\frac{1}{\sqrt{a^3+b^3+1}}+\frac{1}{\sqrt{b^3c^3+1+1}}+\frac{4\sqrt{3}}{c^6+1+2a^3+8}\)

\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{4\sqrt{3}}{2c^3+2a^3+8}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+4}\)

\(=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{c^3+a^3+1+1+1+1}\)

\(\le\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{2\sqrt{3}}{6\sqrt{ac}}=\frac{1}{\sqrt{3ab}}+\frac{1}{\sqrt{3bc}}+\frac{1}{\sqrt{3ac}}\)\(=\frac{1}{\sqrt{3}}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{ac}}+\frac{1}{\sqrt{bc}}\right)\)

\(\le\frac{1}{\sqrt{3}}\sqrt{3\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}=\sqrt{\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)}\le\sqrt{3}\) (Bunhiacopxki)

Dấu "=" xảy ra\(\Leftrightarrow a=b=c=1\)

PS : Thánh cx đc phết ha; chế đc bài này tui mới khâm phục :)))

nó ko chém đâu anh nó chép trong toán tuổi thơ đấy,thk này khốn nạn lắm

\(A=1\)

\(B=\)

\(A=\frac{11}{9}-\frac{7}{8}+-\frac{2}{3}-\frac{1}{8}+\frac{25}{9}-\frac{4}{3}\)

\(A=1\)

\(B=1\frac{3}{4}:\frac{3}{5}-\frac{2}{3}x1,75+\left(\frac{1}{2}\right)^2:\frac{1}{7}\)

\(B=3,5\)

bài 1

ÁP dụng AM-GM ta có:

\(\frac{a^3}{b\left(2c+a\right)}+\frac{2c+a}{9}+\frac{b}{3}\ge3\sqrt[3]{\frac{a^3.\left(2c+a\right).b}{b\left(2c+a\right).27}}=a.\)

tương tự ta có:\(\frac{b^3}{c\left(2a+b\right)}+\frac{2a+b}{9}+\frac{c}{3}\ge b,\frac{c^3}{a\left(2b+c\right)}+\frac{2b+c}{9}+\frac{a}{3}\ge c\)

công tất cả lại ta có:

\(P+\frac{2a+b}{9}+\frac{2b+c}{9}+\frac{2c+a}{9}+\frac{a+b+c}{3}\ge a+b+c\)

\(P+\frac{2\left(a+b+c\right)}{3}\ge a+b+c\)

Thay \(a+b+c=3\)vào ta được":

\(P+2\ge3\Leftrightarrow P\ge1\)

Vậy Min là \(1\)

dấu \(=\)xảy ra khi \(a=b=c=1\)

đây là toán đâu phải văn. bạn bị say rượu à