Question

Cho 2 số thực a,b thỏa mãn : a^2 khác b^2

Đặt A=\(\frac{a+b}{a-b}+\frac{a-b}{a+b}\) Tính : B=\(\frac{a^4+b^4}{a^4-b^4}+\frac{a^4-b^4}{a^4+b^4}\) theo A.

Chân thành cảm ơn!

Answer 1

Nếu dễ thì tớ đâu cần các cậu giúp!

Answer 2

\(A=\frac{4ab}{a^2-b^2}=\frac{4.\frac{a}{b}}{\left(\frac{a}{b}\right)^2-1}\Leftrightarrow A\left(\frac{a}{b}\right)^2-4\frac{a}{b}-A=0\Leftrightarrow At^2-4t+\frac{4}{A}=A+\frac{4}{A}\)

\(t=\frac{2}{A^2}+-\sqrt{\frac{A^2+4}{A^3}}\)

\(B=\frac{4a^4b^4}{a^8-b^8}=\frac{4t^4}{t^8-1}=..\)

Answer 3

Nhật Minh xem lại đi cậu!

A=\(\frac{2\left(a^2+b^2\right)}{a^2-b^2}\) chứ! 

Answer 4

\(A=\frac{2\left[\left(\frac{a}{b}\right)^2+1\right]}{\left(\frac{a}{b}\right)^2-1}\Leftrightarrow\left(\frac{a}{b}\right)^2=\frac{A+2}{A-2}\)

\(B=\frac{2\left[\left(\frac{a}{b}\right)^8+1\right]}{\left(\frac{a}{b}\right)^8-1}=\frac{2\left[\left(\frac{A+2}{A-2}\right)^4+1\right]}{\left(\frac{A+2}{A-2}\right)^4-1}=\frac{2\left[\left(A+2\right)^4+\left(A-2\right)^4\right]}{\left(A+2\right)^4-\left(A-2\right)^4}=.......\)