Rút gọn phân thức sau:
\(\frac{1}{a-b}+\frac{1}{a+b}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
Chu mi ngaaa....
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LH
1
24 tháng 12 2018
Sửa đề:
\(\frac{1}{a-b}+\frac{1}{a+b}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{a+b+a-b}{\left(a-b\right)\left(a+b\right)}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{2a}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{2a\left(a^2-b^2+a^2+b^2\right)}{\left(a^2-b^2\right)\left(a^2+b^2\right)}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{2a.2a^2}{\left(a^2-b^2\right)\left(a^2+b^2\right)}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3\left(a^4+b^4+a^4-b^4\right)}{a^4-b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3.2a^4}{\left(a^4+b^4\right)\left(a^4-b^4\right)}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{8a^7}{a^8-b^8}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{8a^7\left(a^8+b^8+a^8-b^8\right)}{\left(a^8-b^8\right)\left(a^8+b^8\right)}\)
\(=\frac{16a^{15}}{a^{16}-b^{16}}\)
AH
Akai Haruma
Giáo viên
2 tháng 12 2019
Bài 1:
\(A=\frac{1}{a-b}+\frac{1}{a+b}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{a+b+a-b}{(a-b)(a+b)}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}=\frac{2a}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=(2a).\frac{a^2+b^2+a^2-b^2}{(a^2-b^2)(a^2+b^2)}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=4a^3.\frac{a^4+b^4+a^4-b^4}{(a^4-b^4)(a^4+b^4)}+\frac{8a^7}{a^8+b^8}=\frac{8a^7}{a^8-b^8}+\frac{8a^7}{a^8+b^8}=8a^7.\frac{a^8+b^8+a^8-b^8}{(a^8-b^8)(a^8+b^8)}\)
\(=\frac{16a^{15}}{a^{16}-b^{16}}\)
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\(B=\frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)}+\frac{1}{(a+2)(a+3)}=\frac{(a+1)-a}{a(a+1)}+\frac{(a+2)-(a+1)}{(a+1)(a+2)}+\frac{(a+3)-(a+2)}{(a+2)(a+3)}\)
\(=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+2}-\frac{1}{a+3}\)
\(=\frac{1}{a}-\frac{1}{a+3}=\frac{3}{a(a+3)}\)
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
LM
19 tháng 7 2017
a) \(a^4-5a^2+4=\)\(\left(a^4-4a^2\right)-\left(a^2-4\right)=a^2\left(a^2-4\right)-\left(a^2-4\right)=\left(a^2-1\right)\left(a^2-4\right)\)
\(=\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)\)
\(a^4-a^2+4a-4=a^2\left(a^2-1\right)+4\left(a-1\right)=a^2\left(a-1\right)\left(a+1\right)+4\left(a-1\right)\)
\(=\left(a-1\right)\left[a^2\left(a+1\right)+4\right]=\left(a-1\right)\left(a^3+a^2+4\right)\)
\(a^3+a^2+4=\left(a^3+2a^2\right)-\left(a^2+2a\right)+\left(2a+4\right)=a^2\left(a+2\right)-a\left(a+2\right)+2\left(a+2\right)\)
\(=\left(a^2-a+2\right)\left(a+2\right)\)
\(N=\frac{\left(a-1\right)\left(a+1\right)\left(a-2\right)\left(a+2\right)}{\left(a-1\right)\left(a+2\right)\left(a^2-a+2\right)}=\frac{\left(a+1\right)\left(a-2\right)}{a^2-a+2}\)
UI
11 tháng 3 2020
Xet \(M-1=a+\frac{2a+2}{2-b}-\left(\frac{2a-b}{2+b}+1\right)+\frac{4a}{b^2-4}\)
=\(a+\left(2a+2\right)\left(\frac{1}{2-b}-\frac{1}{2+b}\right)+\frac{4a}{b^2-4}\)
=\(\frac{ab^2-4a-4ab-4b+4a}{b^2-4}\)
=\(\frac{ab^2-4ab-4b}{b^2-4}\)
den doan nay em xet rieng tu so \(ab^2-4ab-4b\)
thay b=a/a+1 vao \(\frac{a^3}{\left(a+1\right)^2}-\frac{4a^2}{a+1}-\frac{4}{a+1}\)
=\(\frac{a\left(a+2\right)\left(-3a-2\right)}{\left(a+1\right)^2}\)
xet mau so b^2-4=(a/a+1)^-4
=\(\frac{\left(a+2\right)\left(-3a-2\right)}{\left(a+1\right)^2}\)
den day thay vao la xong nha
LM
19 tháng 7 2017
c)\(P=\)\(\frac{\left(a-b\right)^2-c^2}{\left(a-b+c\right)^2}=\frac{\left(a-b+c\right)\left(a-b-c\right)}{\left(a-b+c\right)^2}=\frac{a-b-c}{a-b+c}\)
LM
19 tháng 7 2017
b)\(M\)\(=\frac{\left(a+2\right)\left(a-1\right)^2}{\left(2a-3\right)\left(a-1\right)^2}=\frac{a+2}{2a-3}\)
12 tháng 3 2020
\(M=a+\frac{\left(2a+b\right)\left(2+b\right)-\left(2a-b\right)\left(2-b\right)}{4-b^2}-\frac{4a}{4-b^2}.\)
\(=a+\frac{4b\left(a+1\right)-4a}{4-b^2}\)
Ta có \(4ab+4b-4a=4\left[\frac{a^2}{a+1}+\frac{a}{a+1}-4a\right]=-12a\)
\(4-b^2=4-\frac{a^2}{\left(a+1\right)^2}=\frac{4\left(a^2+2a+1\right)-a^2}{\left(a+1\right)^2}=\frac{3a^2+8a+4}{\left(a+1\right)^2}\)
\(\Rightarrow M=a+\frac{-12a\left(a+1\right)^2}{3a^2+8a+4}\)
\(=-\frac{9a^3+16a^2+8a}{3a^2+8a+4}\)
12 tháng 3 2020
\(M=a+\frac{2a+b}{2-b}-\frac{2a-b}{2+b}+\frac{4a}{b^2-4}\)
\(=a-\frac{2a+b}{b-2}-\frac{2a-b}{2+b}+\frac{4a}{b^2-4}\)
\(=a-\frac{\left(2a+b\right)\left(2+b\right)+\left(2a-b\right)\left(b-2\right)}{\left(b-2\right)\left(b+2\right)}+\frac{4a}{b^2-4}\)
\(=a-\frac{4b\left(a+1\right)}{b^2-4}+\frac{4a}{b^2-4}\)
\(=a-\frac{4\frac{a}{a+1}\left(a+1\right)}{b^2-4}+\frac{4a}{b^2-4}\)
\(=a-\frac{4a}{b^2-4}+\frac{4a}{b^2-4}\)
\(=a\)
Hắc hắc :P Cứ làm từ từ sẽ thành công em ạ :D
\(=\frac{a+b+a-b}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{2a\left(a^2+b^2\right)+2a\left(a^2-b^2\right)}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3\left(a^4+b^4\right)+4a^3\left(a^4-b^4\right)}{a^8-b^8}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{8a^7\left(a^8+b^8\right)+8a^7\left(a^8-b^8\right)}{\left(a^8-b^8\right)\left(a^8+b^8\right)}\)
\(=\frac{16a^{15}}{a^{16}-b^{16}}\)