\(A=\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(3^2A=3^2\left(\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\right)-3^2\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(9A=\left(1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(3+\frac{1}{3}+...+\frac{1}{3^{97}}\right)\)

\(9A-A=\left(1-\frac{1}{3^{100}}\right)-\left(3-\frac{1}{3^{99}}\right)\)

\(8A=1-3=-2\)

A=\(\frac{-2}{8}=\frac{-1}{4}\)

\(B=4\left|\frac{-1}{4}\right|+\frac{1}{3^{100}}=1+\frac{1}{3^{100}}=1\)

Vậy B=1

Trl:

          Bạn kia trả lời đúng rồi nhoa : ))

Hok tốt

~ nhé bạn ~

Bạn kia trả lời sai rồi

Áp dụng dãy tỉ số bằng nhau:

\(\frac{a}{3}=\frac{b}{4}=\frac{c}{11}=\frac{b+c-a}{4+11-3}=\frac{b+c-a}{12}=\frac{a+c-b}{3+11-4}=\frac{a+c-b}{10}\)

\(\Rightarrow\frac{b+c-a}{a+c-b}=\frac{12}{10}=\frac{6}{5}\)

mk làm kiểu khác

Đặt \(\frac{a}{3}=\frac{b}{4}=\frac{c}{11}=k\)

\(\Rightarrow a=3k;b=4k;c=11k\)(1)

Thay (1) vào biểu thức A ta được:

\(\frac{4k+11k-3k}{3k+11k-4k}=\frac{12k}{10k}=\frac{6}{5}\)

Vậy..................

\(\begin{array}{l}a)A = (2 - \frac{1}{2} - \frac{1}{8}):(1 - \frac{3}{2} - \frac{3}{4})\\ = (\frac{{16}}{8} - \frac{4}{8} - \frac{1}{8}):(\frac{4}{4} - \frac{6}{4} - \frac{3}{4})\\ = \frac{{11}}{8}:\frac{{ - 5}}{4}\\ = \frac{{11}}{8}.\frac{4}{{ - 5}}\\ = \frac{{ - 11}}{{10}}\\b)B = 5 - \frac{{1 + \frac{1}{3}}}{{1 - \frac{1}{3}}}\\ = 5 - \frac{{\frac{3}{3} + \frac{1}{3}}}{{\frac{3}{3} - \frac{1}{3}}}\\ = 5 - \frac{{\frac{4}{3}}}{{\frac{2}{3}}}\\ = 5 - \frac{4}{3}:\frac{2}{3}\\ = 5 - \frac{4}{3}.\frac{3}{2}\\ = 5 - 2\\ = 3\end{array}\)

Chú ý:

Khi thực hiện phép cộng hai phân số, nếu phân số thu được chưa tối giản thì ta rút gọn thành phân số tối giản.

Ta có: \(x=\sqrt[3]{\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}+\sqrt[3]{\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}}\)

=> \(x^3=\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}+\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}+3\cdot\sqrt[3]{\left(\frac{-b}{2}+\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)\left(\frac{-b}{2}-\sqrt{\frac{b^2}{4}+\frac{a^3}{27}}\right)}\cdot x.\)

    =  \(-b+\sqrt[3]{\frac{b^2}{4}-\left(\frac{b^2}{4}+\frac{a^3}{27}\right)}\cdot x\)

=\(-b+\sqrt[3]{\frac{a^3}{27}}\cdot x=-b+\frac{a}{27}\cdot x\)

=> \(x^3+b=\frac{a}{27}\cdot x\)

Vậy  \(x^3+ax+b=\frac{a}{27}\cdot x+ax=\frac{28a}{27}\cdot x\)