Question

\(\frac{4x^2-7x+3}{x^2-1}=\frac{A}{x^2+2x+1}\)Tìm A

Answer 1

4x²-7x+3 = 4x²-3x-4x+3 = x(4x-3)-(4x-3)=(4x-3)(x-1)

=> \(\frac{\left(4x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{A}{\left(x+1\right)^2}\)

=>\(\frac{4x-3}{x+1}=\frac{A}{\left(x+1\right)^2}\)

=> \(A=\left(4x-3\right)\left(x+1\right)^2:\left(x+1\right)=\left(4x-3\right)\left(x+1\right)=4x^2+x-3\).

Answer 2

\(\frac{\left(4x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{A}{\left(x+1\right)^2}\Leftrightarrow A=\left(x+1\right)\left(4x-3\right)=4x^2+x-3\)