1) Rút gọn :

   \(B=\frac{\left(a+2b\right)^3-\left(a-2b\right)^3}{\left(2a+b\right)^3-\left(2a-b\right)^3}:\frac{3a^4+7a^2b^2+3b^4}{4a^4+7a^2b^2+3b^4}\)

Cho a,b,c là 3 số thực đôi một phân biệt. CMR: 

\(3+\frac{\left(2a+b\right)\left(2b+c\right)}{\left(a-b\right)\left(b-c\right)}+\frac{\left(2b+c\right)\left(2c+a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(2c+a\right)\left(2a+b\right)}{\left(c-a\right)\left(a-b\right)}=\frac{2a+b}{a-b}+\frac{2b+c}{b-c}+\frac{2c+a}{c-a}\)

Cho \(\left(a+b\right)\left(2b-a-4\right)=\left(a-2b\right)\left(5-a-b\right)\). Tính \(\frac{2a^2-3b^2}{ab+b^2}\)

cho a;b;c là các số thực đôi một khác nhau thỏa mãn 

\(3+\frac{\left(2a+b\right)\left(2b+c\right)}{\left(a-b\right)\left(b-c\right)}+\frac{\left(2b+c\right)\left(2c+a\right)}{\left(b-c\right)\left(c-a\right)}+\frac{\left(2c+a\right)\left(2a+b\right)}{\left(c-a\right)\left(a-b\right)}=\)\(\frac{2a+b}{a-b}+\frac{2b+c}{b-c}+\frac{2c+a}{c-a}\)

Biết a, b , c , d >0

Rút gọn \(\left[\left(\frac{a^2b}{cd^2}\right)^3.\left(\frac{ac^4}{b^2d^3}\right)\right]:\left[\left(\frac{a^2b^2}{cd^3}\right)^4.\left(\frac{c}{b^3d}\right)^3 \right]\)

\(\frac{b^2c^3}{a^2+\left(b+c\right)^3}+\frac{c^2a^3}{b^2+\left(c+a\right)^3}+\frac{a^2b^3}{c^2+\left(a+b\right)^3}\ge\frac{9abc}{4\left(3abc+a^2c+b^2a+c^2b\right)}\)voi a,b,c>0

Cho a,b,c >0, cmr \(\frac{a^3}{b\left(2b+a\right)}+\frac{2b^3}{c\left(2c+b\right)}+\frac{128c^3}{a\left(a+4c\right)}\ge a+2b+4c\)

Rút gọn biểu thức

\(\frac{1}{a^2}\sqrt[3]{a^6+3a^4b^2+3a^2b^4+b^6}-\left[\frac{a^2-\left(a^{\frac{2}{3}}-b^{\frac{2}{3}}\right)^3+2b^2}{a^2+\left(a^{\frac{2}{3}}-b^{\frac{2}{3}}\right)^3+2b^2}\right]\)

Cho a, b, c là các số thực dương. Chứng mịnh rằng:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{2a+b}{a\left(a+2b\right)}+\frac{2b+c}{b\left(2b+c\right)}+\frac{2c+a}{c\left(2c+a\right)}\)