Question

Tính B,biết:

B=\(\frac{1+3^4+3^8+3^{12}}{1+3^2+3^4+3^6+3^8+3^{10}+3^{12}+3^{14}}\)

Answer 1

\(B=\frac{1+3^4+3^8+3^{12}}{1+3^2+3^4+3^6+3^8+3^{10}+3^{12}+3^{14}}\)

\(B=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+3^2+3^6+3^{10}+3^{14}}\)

Xét mẫu : \(\left(1+3^4+3^8+3^{12}\right)+3^2\left(1+3^4+3^8+3^{12}\right)=\left(3^2+1\right)\left(1+3^4+3^8+3^{12}\right)\)

Ta có : \(\frac{1+3^4+3^8+3^{12}}{\left(3^2+1\right)\left(1+3^4+3^8+3^{12}\right)}=\frac{1}{3^2+1}=\frac{1}{10}\)

Answer 2

(chủ câu hỏi)

sao lại có \(3^2+1\)ở kia

Answer 3

Tính chất phân phối của phép nhân đối với phép cộng đấy bạn ạ

\(1+3^4+3^8+3^{12}+3^2\left(1+3^4+3^8+3^{12}\right)=1\left(1+3^4+3^8+3^{12}\right)+3^2\left(1+3^4+3^8+3^{12}\right)\)

\(=\left(3^2+1\right)\left(1+3^4+3^8+3^{12}\right)\)như nhau cả thôi bạn

Answer 4

B=\(\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)+\left(3^2+3^6+3^{10}+3^{14}\right)}=\frac{1+3^4+3^8+3^{12}}{\left(1+3^4+3^8+3^{12}\right)\left(1+3^2\right)}\)\(=\frac{1}{10}\)