Tính \(\frac{1}{a-b}+\frac{1}{a+b}+\frac{2a}{a^2+b^2}+\frac{4a^2}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
11 tháng 3 2020
Hắc hắc :P Cứ làm từ từ sẽ thành công em ạ :D
\(=\frac{a+b+a-b}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{2a\left(a^2+b^2\right)+2a\left(a^2-b^2\right)}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3\left(a^4+b^4\right)+4a^3\left(a^4-b^4\right)}{a^8-b^8}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{8a^7\left(a^8+b^8\right)+8a^7\left(a^8-b^8\right)}{\left(a^8-b^8\right)\left(a^8+b^8\right)}\)
\(=\frac{16a^{15}}{a^{16}-b^{16}}\)
AH
Akai Haruma
Giáo viên
2 tháng 12 2019
Bài 1:
\(A=\frac{1}{a-b}+\frac{1}{a+b}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{a+b+a-b}{(a-b)(a+b)}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}=\frac{2a}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=(2a).\frac{a^2+b^2+a^2-b^2}{(a^2-b^2)(a^2+b^2)}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=4a^3.\frac{a^4+b^4+a^4-b^4}{(a^4-b^4)(a^4+b^4)}+\frac{8a^7}{a^8+b^8}=\frac{8a^7}{a^8-b^8}+\frac{8a^7}{a^8+b^8}=8a^7.\frac{a^8+b^8+a^8-b^8}{(a^8-b^8)(a^8+b^8)}\)
\(=\frac{16a^{15}}{a^{16}-b^{16}}\)
--------------
\(B=\frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)}+\frac{1}{(a+2)(a+3)}=\frac{(a+1)-a}{a(a+1)}+\frac{(a+2)-(a+1)}{(a+1)(a+2)}+\frac{(a+3)-(a+2)}{(a+2)(a+3)}\)
\(=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+2}-\frac{1}{a+3}\)
\(=\frac{1}{a}-\frac{1}{a+3}=\frac{3}{a(a+3)}\)
AH
Akai Haruma
Giáo viên
2 tháng 12 2019
Bài 2:
Bạn tham khảo lời giải tương tự tại link sau:
Câu hỏi của Law Trafargal - Toán lớp 8 | Học trực tuyến
21 tháng 12 2019
a) \(A=\left(\frac{2}{2a-b}+\frac{6b}{b^2-4a^2}-\frac{4}{2a+b}\right):\left(a+\frac{4a^2+b^2}{4a^2-b^2}\right)\)
\(=\left(\frac{2}{2a-b}+\frac{6b}{\left(b-2a\right)\left(b+2a\right)}-\frac{4}{2a+b}\right):\left(a+\frac{4a^2+b^2}{4a^2-b^2}\right)\)
\(=\left(\frac{-2\left(b+2a\right)}{\left(b-2a\right)\left(b+2a\right)}+\frac{6b}{\left(b-2a\right)\left(b+2a\right)}-\frac{4\left(b-2a\right)}{\left(2a+b\right)\left(b-2a\right)}\right):\left(\frac{a\left(4a^2-b^2\right)}{4a^2-b^2}+\frac{4a^2+b^2}{4a^2-b^2}\right)\)
\(=\frac{-2b-4a+6b-4b+8a}{\left(b-2a\right)\left(b+2a\right)}:\frac{4a^3-ab^2+4a^2+b^2}{4a^2-b^2}\)
\(=\frac{4a}{\left(b-2a\right)\left(b+2a\right)}.\frac{\left(2a-b\right)\left(2a+b\right)}{4a^3-ab^2+4a^2+b^2}\)
\(=\frac{-4a}{\left(2a-b\right)\left(b+2a\right)}.\frac{\left(2a-b\right)\left(2a+b\right)}{4a^3-ab^2+4a^2+b^2}\)
\(=.\frac{-4a}{4a^3-ab^2+4a^2+b^2}\)
b) ĐKXĐ: \(\hept{\begin{cases}2a\ne b\\2a\ne-b\end{cases}}\)
Ta thấy \(a=\frac{1}{3};b=2\)thỏa mãn điều kiện \(\hept{\begin{cases}2a\ne b\\2a\ne-b\end{cases}}\)nên thay vào A ta được:
bạn thay vào tự tính nhé mà cái phần rút gọn bạn vừa làm vừa check giùm bài mik nhé =)) sợ sai
P
0
LM
19 tháng 7 2017
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)
Sửa đề:
\(\frac{1}{a-b}+\frac{1}{a+b}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{a+b+a-b}{\left(a-b\right)\left(a+b\right)}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{2a}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{2a\left(a^2-b^2+a^2+b^2\right)}{\left(a^2-b^2\right)\left(a^2+b^2\right)}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{2a.2a^2}{\left(a^2-b^2\right)\left(a^2+b^2\right)}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3\left(a^4+b^4+a^4-b^4\right)}{a^4-b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3.2a^4}{\left(a^4+b^4\right)\left(a^4-b^4\right)}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{8a^7}{a^8-b^8}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{8a^7\left(a^8+b^8+a^8-b^8\right)}{\left(a^8-b^8\right)\left(a^8+b^8\right)}\)
\(=\frac{16a^{15}}{a^{16}-b^{16}}\)