Bài giải

Ta có : \(B=-\frac{1}{3^0}-\frac{1}{3^1}-\frac{1}{3^2}-...-\frac{1}{3^{100}}\)

\(\Rightarrow\text{ }B=-\frac{1}{3^0}-\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)

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

Đặt \(C=\frac{1}{3^1}+\frac{1}{3^2}+..+\frac{1}{3^{100}}\)

\(\Rightarrow\text{ }3C=1+\frac{1}{3^1}+...+\frac{1}{3^{99}}\)

\(\Rightarrow\text{ }3C-C=2C=1-\frac{1}{3^{100}}\)

\(C=\frac{1-\frac{1}{3^{100}}}{2}=\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)

Thay vào biểu thức B ta được :

\(B=-1-\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)

\(B=-\frac{3}{2}-\frac{1}{2\cdot3^{100}}\)

\(B=\frac{\left(-3\right)^{101}}{2\cdot3^{100}}-\frac{1}{2\cdot3^{100}}=\frac{\left(-3\right)^{101}-1}{2\cdot3^{100}}\)

                                                                           Bài giải

Ta có : \(B=-\frac{1}{3^0}-\frac{1}{3^1}-\frac{1}{3^2}-...-\frac{1}{3^{100}}\)

\(\Rightarrow\text{ }B=-\frac{1}{3^0}-\left(\frac{1}{3^1}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)

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

Đặt \(C=\frac{1}{3^1}+\frac{1}{3^2}+..+\frac{1}{3^{100}}\)

\(\Rightarrow\text{ }3C=1+\frac{1}{3^1}+...+\frac{1}{3^{99}}\)

\(\Rightarrow\text{ }3C-C=2C=1-\frac{1}{3^{100}}\)

\(C=\frac{1-\frac{1}{3^{100}}}{2}=\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)

Thay vào biểu thức B ta được :

\(B=-1-\frac{1}{2}-\frac{1}{2\cdot3^{100}}\)

\(B=-\frac{3}{2}-\frac{1}{2\cdot3^{100}}\)

\(B=\frac{\left(-3\right)^{101}}{2\cdot3^{100}}-\frac{1}{2\cdot3^{100}}=\frac{\left(-3\right)^{101}-1}{2\cdot3^{100}}\)

Bổ sung đề:

Cho: \(\frac{a}{b}=\frac{c}{d}\). C/m \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)

Đặt: \(\frac{a}{b}=\frac{c}{d}=k\)\(\left(k\ne0\right)\)

\(\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Khi đó: \(\frac{ac}{bd}=\frac{bk.dk}{bd}=\frac{k^2.\left(bd\right)}{bd}=k^2\)                                                                   \(\left(1\right)\)

Và: \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2.k^2+d^2.k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\)         \(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\)\(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\)\(\left(đpcm\right)\)

tích cho t đi

\(A=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^4}=\frac{2^{30}+2^{20}}{2^{12}+2^8}=\frac{2^{20}\left(2^{10}+1\right)}{2^8\left(2^4+1\right)}=\frac{2^{12}\left(2^{10}+1\right)}{2^4+1}\)