\(a.\)

\(\dfrac{16x^2-1}{16x^2-8x+1}\\ =\dfrac{\left(4x\right)^2-1}{\left(4x-1\right)^2}\\ =\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\\ =\dfrac{4x+1}{4x-1}\)

\(b.\)

\(\dfrac{4x^2-4xy+y^2}{-\left(4x^2-y^2\right)}\\ =-\dfrac{\left(2x-y\right)^2}{\left(2x-y\right)\left(2x+y\right)}\\ =\dfrac{-\left(2x-y\right)}{2x+y}\\ =\dfrac{y-2x}{y+2x}\)

a) Ta có: \(\dfrac{16x^2-1}{16x^2-8x+1}\)

\(=\dfrac{\left(4x-1\right)\left(4x+1\right)}{\left(4x-1\right)^2}\)

\(=\dfrac{4x+1}{4x-1}\)

b) Ta có: \(\dfrac{4x^2-4xy+y^2}{y^2-4x^2}\)

\(=\dfrac{\left(2x-y\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)

\(=\dfrac{\left(y-2x\right)^2}{\left(y-2x\right)\left(y+2x\right)}\)

\(=\dfrac{y-2x}{y+2x}\)

\(2x^3y-2xy^3-4xy^2-2xy\)

\(=2xy.\left(x^2-y^2-2y-1\right)\)

\(=2xy.[x^2-\left(y^2+2y+1\right)]\)

\(=2xy.[x^2-\left(y+1\right)^2]\)

\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)

Vậy chọn đáp án A

chọn A

a: \(y^2-9-x^2+6x\)

\(=y^2-\left(x^2-6x+9\right)\)

\(=y^2-\left(x-3\right)^2\)

=(y-x+3)(y+x-3)

b: \(25-4x^2-4xy-y^2\)

\(=25-\left(4x^2+4xy+y^2\right)\)

\(=25-\left(2x+y\right)^2=\left(5-2x-y\right)\left(5+2x+y\right)\)

c: \(x^2-xz+4y^2-2yz+4xy\)

\(=x^2+4xy+4y^2-z\left(x+2y\right)\)

\(=\left(x+2y\right)^2-z\left(x+2y\right)\)

=(x+2y)(x+2y-z)

d: \(3x^2+6xy-48z^2+3y^2\)

\(=3\left\lbrack x^2+2xy+y^2-16z^2\right\rbrack\)

\(=3\left\lbrack\left(x+y\right)^2-\left(4z\right)^2\right\rbrack\)

=3(x+y+4z)(x+y-4z)

e: \(x^2-z^2+4y^2-4t^2-4xy+4zt\)

\(=x^2-4xy+4y^2-z^2+4zt-4t^2\)

\(=\left(x-2y\right)^2-\left(z-2t\right)^2\)

=(x-2y-z+2t)(x-2y+z-2t)

f: \(x^3+2x^2y+xy^2-16x\)

\(=x\left(x^2+2xy+y^2-16\right)\)

\(=x\left\lbrack\left(x+y\right)^2-16\right\rbrack\)

=x(x+y+4)(x+y-4)

\(\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16x}\)

\(=\frac{\left(2x+y\right)^22\left(4x^2-y^2\right)+\left(2x-y\right)^2}{\left(2x-y\right)^2\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)

\(=\frac{16x^2}{16x\left(2x-y\right)^2}=\frac{x}{\left(2x-y\right)^2}\)

\(\left[\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-4^2}+\frac{1}{\left(2x+y\right)^2}\right].\frac{4x^2+4xy+y^2}{16x}\)

\(=\frac{\left(2x+y\right)^22\left(4x^2-y^2\right)+\left(2x-y\right)^2}{\left(2x-y\right)^2\left(2x+y\right)^2}.\frac{\left(2x+y\right)^2}{16x}\)

\(=\frac{16x^2}{16x\left(2x-y\right)^2}=\frac{x}{\left(2x-y\right)^2}\)

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được