b) 3a = 2b; 7b = 5c

=> a/2 = b/3; b/5 = c/7

=> a/10 = b/15 = c/21

Áp dụng tính chất dãy tỉ số bằng nhau, có:

 \(\frac{a}{10}=\frac{b}{15}=\frac{c}{21}=\frac{a-b+c}{10-15+21}=\frac{32}{16}=2\)

suy ra; a/10 = 2    => a = 10 * 2 = 20

         b/15 = 2       => b = 15 * 2 = 30

       c/21 = 2         => c = 21 * 2 = 42

a: \(=ab\cdot\dfrac{4}{3}a^2b^4\cdot7abc=\dfrac{28}{3}a^4b^6c\)

b: \(a^3b^3\cdot a^2b^2c=a^5b^5c\)

c: \(=\dfrac{2}{3}a^3b\cdot\dfrac{-1}{2}ab\cdot a^2b=\dfrac{-1}{3}a^6b^3\)

d: \(=-\dfrac{7}{3}a^3c^2\cdot\dfrac{1}{7}ac^2\cdot6abc=-2a^5bc^5\)

e: \(=\dfrac{-3}{2}\cdot\dfrac{1}{4}\cdot ab^2\cdot bca^2\cdot b=\dfrac{-3}{8}a^3b^4c\)

Có: a + b = ab \(\le\frac{\left(a+b\right)^2}{4}\)

=> a + b \(\ge4\)

\(\frac{1}{a^2+2a}+\frac{1}{b^2+2b}+\sqrt{\left(1+a^2\right)\left(1+b^2\right)}\)

\(\ge\frac{4}{a^2+b^2+2\left(a+b\right)}+\sqrt{\left(1+ab\right)^2}\)

\(=\frac{4}{a^2+b^2+2ab}+\left(1+a+b\right)=\frac{4}{\left(a+b\right)^2}+\left(a+b\right)+1\)

\(=\frac{4}{\left(a+b\right)^2}+\frac{a+b}{4^2}+\frac{a+b}{4^2}+\frac{7}{8}\left(a+b\right)+1\)

\(\ge3\sqrt[3]{\frac{4}{\left(a+b\right)^2}.\frac{a+b}{4^2}.\frac{a+b}{4^2}}+\frac{7}{8}.4+1=\frac{3}{4}+\frac{7}{2}+1\)

Dấu "=" xảy ra <=> a = b = 2

\(\Leftrightarrow\frac{1}{1+a}+\frac{a}{1+a}+\frac{2b}{21+2b}+\frac{21}{21+2b}\le\frac{4c}{4c+27}+\frac{a}{1+a}+\frac{2b}{21+2b}\)

\(\Leftrightarrow2\le\frac{1}{1+\frac{1}{a}}+\frac{1}{1+\frac{21}{2b}}+\frac{1}{1+\frac{27}{4c}}\)

Đặt \(\left(\frac{1}{a};\frac{21}{2b};\frac{27}{4c}\right)=\left(x;y;z\right)\)

\(\Leftrightarrow\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge2\)

\(\Leftrightarrow\frac{1}{1+x}\ge1-\frac{1}{1+y}+1-\frac{1}{1+z}=\frac{y}{1+y}+\frac{z}{1+z}\ge2\sqrt{\frac{yz}{\left(1+y\right)\left(1+z\right)}}\)

Tương tự: \(\frac{1}{1+y}\ge2\sqrt{\frac{zx}{\left(1+z\right)\left(1+x\right)}}\) ; \(\frac{1}{1+z}\ge2\sqrt{\frac{xy}{\left(1+x\right)\left(1+y\right)}}\)

Nhân vế với vế: \(1\ge8xyz\Rightarrow xyz\le\frac{1}{8}\)

\(\Leftrightarrow\frac{1}{a}.\frac{21}{2b}.\frac{27}{4c}\le\frac{1}{8}\Leftrightarrow abc\ge567\)

Dấu "=" xảy ra khi \(\frac{1}{a}=\frac{21}{2b}=\frac{27}{4c}=\frac{1}{2}\Rightarrow\left(a;b;c\right)=\left(2;21;\frac{27}{2}\right)\)