\(B=\frac{2^{10}.13+65.2^{10}}{2^8.104}\)

\(B=\frac{2^8.2^2\left(13+65\right)}{2^8.104}\)

\(B=\frac{2^2.3.26}{2^2.26}\)

B = 3

chịu

khó vãi

hok lp 7 rùi toán lp 6 ko bt lm

..........xoxo

Xin lỗi các bạn, mình đánh nhầm đề, mình sẽ gửi lại đề đúng ngay bây giờ đây

a) 

\(P=\left(x^{14}-9x^{13}\right)-\left(x^{13}-9x^{12}\right)+\left(x^{12}-9x^{11}\right)-...+\left(x^2-9x\right)-\left(x-9\right)+1\)

\(=x^{13}\left(x-9\right)-x^{12}\left(x-9\right)+x^{11}\left(x-9\right)+...+x\left(x-9\right)-\left(x-9\right)+1\)

\(P\left(9\right)=1\)

b)

\(Q=\left(x^{15}-7x^{14}\right)-\left(x^{14}-7x^{13}\right)+\left(x^{13}-7x^{12}\right)-...-\left(x^2-7x\right)+\left(x-7\right)+2\)

\(=x^{14}\left(x-7\right)-x^{13}\left(x-7\right)+x^{12}\left(x-7\right)-...-x\left(x-7\right)+\left(x-7\right)+2\)

\(Q\left(7\right)=2\)

E= 7/4x7 + 7/7x10 =7/10x13+...+ 7/301x304

\(=\frac{7}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{301}-\frac{1}{304}\right)\)

\(=\frac{7}{3}\left(\frac{1}{4}-\frac{1}{304}\right)\)

\(=\frac{7}{3}\cdot\frac{75}{304}\)

\(=\frac{175}{304}\)

E = \(\frac{7}{4.7}+\frac{7}{7.10}+\frac{7}{10.13}+...+\frac{7}{301.304}\) 

   =\(\frac{7}{3}.\frac{7-4}{4.7}+\frac{7}{3}.\frac{10-7}{7.10}+\frac{7}{3}.\frac{13-10}{10.13}+...+\frac{7}{3}.\frac{304-301}{301.304}\)

 = \(\frac{7}{3}.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{301}-\frac{1}{304}\right)\)=\(\frac{7}{3}.\left(\frac{1}{4}-\frac{1}{304}\right)=\frac{7}{3}.\frac{75}{304}=\frac{175}{304}\)

A= 1/7 - 1/10 + 1/10 - 1/13 + 1/13 - 1/16 + .... + 1/97 - 1/100

A= 1/7 - 1/100

A= 93/700

\(A=\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}+...+\frac{3}{97.100}\)

\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{37}-\frac{1}{100}\)

\(A=\frac{1}{7}-\frac{1}{100}\)

\(A=\frac{93}{700}\)

Xin lỗi, mình nhầm. Phải bằng 97/700 mới đúng