Question

[hyun mau]

cho 2 số thực a,b thỏa mãn a² # b²

Đặt A=(a+b) /(a-b) + (a-b)/(a+b). tính B =( a⁴ + b⁴)/(a⁴ - b⁴) + (a⁴ - b⁴) /(a⁴ + b⁴)  theo A

Answer 1

[Trần Thị Loan]

Chia cả tử và mẫu của các phân số cho a khác 0 ta được:

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

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

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

ta có: \(B=\frac{\left(\frac{a}{b}\right)^4+1}{\left(\frac{a}{b}\right)^4-1}+\frac{\left(\frac{a}{b}\right)^4-1}{\left(\frac{a}{b}\right)^4+1}\)

\(\Rightarrow B=\frac{\left(\frac{A+2}{A-2}\right)^2+1}{\left(\frac{A+2}{A-2}\right)^2-1}+\frac{\left(\frac{A+2}{A-2}\right)^2-1}{\left(\frac{A+2}{A-2}\right)^2+1}=\frac{\left(A+2\right)^2+\left(A-2\right)^2}{\left(A+2\right)^2-\left(A-2\right)^2}+\frac{\left(A+2\right)^2-\left(A-2\right)^2}{\left(A+2\right)^2+\left(A-2\right)^2}\)

\(\Rightarrow B=\frac{2.A^2+8}{8.A}+\frac{8.A}{2.A^2+8}=\frac{\left(2A^2+8\right)^2+64.A^2}{8.A\left(2A^2+8\right)}=\frac{\left(A^2+4\right)^2+16.A^2}{4.A\left(A^2+4\right)}\)

Answer 2

[Đinh Cao Sơn]

Chia cả tử và mẫu của các phân số cho a khác 0 ta được:

$A=\frac{a+b}{a-b}+\frac{a-b}{a+b}=\frac{\frac{a}{b}+1}{\frac{a}{b}-1}+\frac{\frac{a}{b}-1}{\frac{a}{b}+1}=\frac{\left(\frac{a}{b}+1\right)^2+\left(\frac{a}{b}-1\right)^2}{\left(\frac{a}{b}-1\right)\left(\frac{a}{b}+1\right)}=\frac{2.\left(\frac{a}{b}\right)^2+2}{\left(\frac{a}{b}\right)^2-1}$A=a+ba−b +a−ba+b =ab +1ab −1 +ab −1ab +1 =(ab +1)2+(ab −1)2(ab −1)(ab +1) =2.(ab )2+2(ab )2−1 

$\Rightarrow A.\left(\frac{a}{b}\right)^2-A=2.\left(\frac{a}{b}\right)^2+2\Rightarrow A.\left(\frac{a}{b}\right)^2-2.\left(\frac{a}{b}\right)^2=A+2$⇒A.(ab )2−A=2.(ab )2+2⇒A.(ab )2−2.(ab )2=A+2

$\Rightarrow\left(A-2\right).\left(\frac{a}{b}\right)^2=A+2\Rightarrow\left(\frac{a}{b}\right)^2=\frac{A+2}{A-2}$⇒(A−2).(ab )2=A+2⇒(ab )2=A+2A−2 

ta có: $B=\frac{\left(\frac{a}{b}\right)^4+1}{\left(\frac{a}{b}\right)^4-1}+\frac{\left(\frac{a}{b}\right)^4-1}{\left(\frac{a}{b}\right)^4+1}$B=(ab )4+1(ab )4−1 +(ab )4−1(ab )4+1 

$\Rightarrow B=\frac{\left(\frac{A+2}{A-2}\right)^2+1}{\left(\frac{A+2}{A-2}\right)^2-1}+\frac{\left(\frac{A+2}{A-2}\right)^2-1}{\left(\frac{A+2}{A-2}\right)^2+1}=\frac{\left(A+2\right)^2+\left(A-2\right)^2}{\left(A+2\right)^2-\left(A-2\right)^2}+\frac{\left(A+2\right)^2-\left(A-2\right)^2}{\left(A+2\right)^2+\left(A-2\right)^2}$⇒B=(A+2A−2 )2+1(A+2A−2 )2−1 +(A+2A−2 )2−1(A+2A−2 )2+1 =(A+2)2+(A−2)2(A+2)2−(A−2)2 +(A+2)2−(A−2)2(A+2)2+(A−2)2 

$\Rightarrow B=\frac{2.A^2+8}{8.A}+\frac{8.A}{2.A^2+8}=\frac{\left(2A^2+8\right)^2+64.A^2}{8.A\left(2A^2+8\right)}=\frac{\left(A^2+4\right)^2+16.A^2}{4.A\left(A^2+4\right)}$⇒B=2.A2+88.A +8.A2.A2+8 =(2A2+8)2+64.A28.A(2A2+8) =(A2+4)2+16.A24.A(A2+4)