\(E=\frac{1+x^2+x^4+...+x^{98}}{1+x^4+x^8+...+x^{96}}\)

\(=\frac{\left(1+x^4+x^8+...+x^{96}\right)+\left(x^2+x^6+x^{10}...+x^{94}+x^{98}\right)}{1+x^4+x^8+...+x^{96}}\)

\(=\frac{\left(1+x^4+x^8+...+x^{96}\right)+x^2\left(1+x^4+x^8...+x^{96}\right)}{1+x^4+x^8+...+x^{96}}\)

\(=\frac{\left(1+x^4+x^8+...+x^{96}\right)\left(1+x^2\right)}{1+x^4+x^8+...+x^{96}}=1+x^2\)

#alibaba nguyễn 

Cái này có thể gạch đi mà ???

Học tốt ~.~

                                                                           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