\(a+b+c=0\Rightarrow-a=b+c\Rightarrow a^2=b^2+c^2+2bc\Rightarrow b^2+c^2=a^2-2bc\)

Tương tự như vậy ta được: \(a^2+c^2=b^2-2ac;a^2+b^2=c^2-2ab\)

Suy ra: \(B=\frac{a^2}{a^2-b^2-c^2}+\frac{b^2}{b^2-c^2-a^2}+\frac{c^2}{c^2-b^2-a^2}\)

\(=\frac{a^2}{a^2-\left(a^2-2bc\right)}+\frac{b^2}{b^2-\left(b^2-2ac\right)}+\frac{c^2}{c^2-\left(c^2-2ab\right)}\)

\(=\frac{a^2}{2bc}+\frac{b^2}{2ac}+\frac{c^2}{2ab}=\frac{a^3+b^3+c^3}{2abc}=\frac{\left(a+b+c\right)^3-3\left(a+b\right)\left(b+c\right)\left(c+a\right)}{2abc}\)

Ta lại thấy a+b=-c;b+c=-a;c+a=-b (a+b+c=0)

Vậy \(B=\frac{0^3-3.\left(-c\right)\left(-a\right)\left(-b\right)}{2abc}=\frac{3abc}{2abc}=\frac{3}{2}\)

\(A=2^0+2^3+2^5+...+2^{99}\)

\(A\cdot2^2=2^2+2^5+2^7+...+2^{101}\)

\(A\cdot3=2^2-1+2^{99}\)

\(\text{A}=1+2^3+2^5+....+2^{99}\)

\(4\text{A}=2+2^5+2^8+.....+2^{101}\)

\(4\text{A}-\text{A}=\left(2+2^5+2^8+....+2^{101}\right)-\left(1+2^3+...+2^{99}\right)\)

\(3\text{A}=2^{101}+2^2-2^3+2^0\)

Ta có: A = 2⁰ + 2³ + 2⁵ + ....+ 2⁹⁹

=> 2A = 2¹ + 2⁴ + 2⁶ + ....+ 2¹⁰⁰

=> 2A - A = ( 2¹ + 2⁴ + 2⁶ + ...+2¹⁰⁰ ) - ( 2⁰ + 2³ + ...+ 2⁹⁹)

=> A         = 2 - 2⁹⁹

\(A=29\dfrac{1}{2}\cdot\dfrac{2}{3}+39\dfrac{1}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{2}\cdot\dfrac{2}{3}+\dfrac{118}{3}\cdot\dfrac{3}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{3}+\dfrac{118}{4}+\dfrac{5}{6}\)

\(=\dfrac{59}{3}+\dfrac{59}{2}+\dfrac{5}{6}\)

\(=59\cdot\left(\dfrac{1}{3}+\dfrac{1}{2}\right)+\left(\dfrac{1}{3}+\dfrac{1}{2}\right)\)

\(=\dfrac{5}{6}\cdot\left(59+1\right)=\dfrac{5}{6}\cdot60=50\)