1, đk x khác 1/2 ; -1/2
\(A=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{4x^2-1}:\dfrac{4x}{5\left(2x-1\right)}=\dfrac{\left(2x+1-2x+1\right)\left(2x+1+2x-1\right)}{4x^2-1}:\dfrac{4x}{5\left(2x-1\right)}=\dfrac{8x.5}{\left(2x+1\right).4x}=\dfrac{10}{2x+1}\)
2, đk x khác 2;-2
\(A=\dfrac{x+2+x-2+x^2+1}{x^2-4}=\dfrac{x^2+2x+1}{x^2-4}=\dfrac{\left(x+1\right)^2}{\left(x-2\right)\left(x+2\right)}\)
3, đk x khác 1
\(P=x:\left(\dfrac{x^2-1-x^2-x-1+x^2+2}{x^3-1}\right)=x:\left(\dfrac{x^2-x}{x^3-1}\right)=x:\left(\dfrac{x}{x^2+x+1}\right)=x^2+x+1\)
4, đk x khác 0 ; -1
\(P=\dfrac{x+1+x^2}{x\left(x+1\right)}:\dfrac{x}{x\left(x+1\right)}=\dfrac{x^2+x+1}{x}=x+1+\dfrac{1}{x}\)
đk x khác 1/2 ; -1/2 A = ( 2 x + 1 ) 2 − ( 2 x − 1 ) 2 4 x 2 − 1 : 4 x 5 ( 2 x − 1 ) = ( 2 x + 1 − 2 x + 1 ) ( 2 x + 1 + 2 x − 1 ) 4 x 2 − 1 : 4 x 5 ( 2 x − 1 ) = 8 x .5 ( 2 x + 1 ) .4 x = 10 2 x + 1 2, đk x khác 2;-2 3, đk x khác 1 P = x : ( x 2 − 1 − x 2 − x − 1 + x 2 + 2 x 3 − 1 ) = x : ( x 2 − x x 3 − 1 ) = x : ( x x 2 + x + 1 ) = x 2 + x + 1
