Question

[nekkk Ngocc]

làm hộ em đến bài 3 ạ, em cảm ơn ;-;

Answer 1

[Lấp La Lấp Lánh]

Bài 1:

a) \(=\dfrac{\left(2m-2n\right)^2}{5\left(m^2-n^2\right)}=\dfrac{4\left(m-n\right)^2}{5\left(m-n\right)\left(m+n\right)}=\dfrac{4m-4n}{4m+5n}\)

b) \(=\dfrac{\left(x-y\right)\left(x-z\right)}{\left(x+y\right)\left(x-z\right)}=\dfrac{x-y}{x+y}\)

c) \(=\dfrac{\left(x-3\right)\left(y-9\right)}{-\left(x-3\right)}=9-y\)

d) \(=\dfrac{\left(3x+2-x-2\right)\left(3x+2+x+2\right)}{x^2\left(x-1\right)}=\dfrac{8x\left(x+1\right)}{x^2\left(x-1\right)}=\dfrac{8x+8}{x^2-x}\)

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

g) \(=\dfrac{12x\left(1-2x\right)}{24x\left(x-2\right)}=\dfrac{1-2x}{2x-4}\)

Bài 2:

a) \(=\dfrac{3\left(m-2n\right)}{-5\left(m-2n\right)}=-\dfrac{3}{5}\)

b) \(=\dfrac{\left(y+1\right)\left(y^2+4\right)}{\left(y-3\right)\left(y+1\right)}=\dfrac{y^2+4}{y-3}\)

c) \(=\dfrac{y^4\left(y-2\right)+2y^2\left(y-2\right)-3\left(y-2\right)}{\left(y-2\right)\left(y+4\right)}=\dfrac{\left(y-2\right)\left(y^4+2y^2-3\right)}{\left(y-2\right)\left(y+4\right)}=\dfrac{y^4+2y^2-3}{y+4}\)

Bài 3:

\(A=\dfrac{\left(m^2+2mn+n^2\right)+5\left(m+n\right)-6}{\left(m^2+2mn+n^2\right)+6\left(m+n\right)}=\dfrac{\left(m+n\right)^2+5\left(m+n\right)-6}{\left(m+n\right)^2+6\left(m+n\right)}=\dfrac{2013^2+5.2013-6}{2013^2+6.2013}=\dfrac{2012}{2013}\)