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

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

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

1)

\((x+2)(x+3)(x+4)(x+5)-24\\=[(x+2)(x+5)]\cdot[(x+3)(x+4)]-24\\=(x^2+7x+10)(x^2+7x+12)-24\)

Đặt \(x^2+7x+10=y\), khi đó biểu thức trở thành:

\(y(y+2)-24\\=y^2+2y-24\\=y^2+2y+1-25\\=(y+1)^2-5^2\\=(y+1-5)(y+1+5)\\=(y-4)(y+6)\\=(x^2+7x+10-4)(x^2+7x+10+6)\\=(x^2+7x+6)(x^2+7x+16)\)

2) Bạn xem lại đề!

chịu :)))))))))

c) \(\left(x+\dfrac{y}{x}\right)^3\)

\(=\left(\dfrac{x^2}{x}+\dfrac{y}{x}\right)^3\)

\(=\left(\dfrac{x^2+y}{x}\right)^3\)

\(=\dfrac{x^6+3x^4y+3x^2y^3+y^3}{x^3}\)

f) \(\left(x-\dfrac{1}{2}\right)^3\)

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

\(=x^3-\dfrac{3}{2}x^2+\dfrac{3}{4}x-\dfrac{1}{8}\)

h) \(\left(x+\dfrac{y^2}{2}\right)^3\)

\(=\left(\dfrac{2x}{2}+\dfrac{y^2}{2}\right)^3\)

\(=\left(\dfrac{2x+y^2}{2}\right)^3\)

\(=\dfrac{8x^3+12x^2y^2+6xy^4+y^6}{8}\)

k) \(\left(x-\dfrac{1}{3}\right)^3\)

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

\(=x^3-x^2+\dfrac{x}{3}-\dfrac{1}{27}\)

m) \(\left(x+\dfrac{y^2}{3}\right)^3\)

\(=\left(\dfrac{3x}{3}+\dfrac{y^2}{3}\right)^3\)

\(=\left(\dfrac{3x+y^2}{3}\right)^3\)

\(=\dfrac{27x^3+27x^2y^2+9xy^4+y^6}{27}\)

Q) \(2\left(x^2+\dfrac{1}{2}y\right)\left(2x^2-y\right)\)

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

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

\(=4x^4-y^2\)

\(=\left(x-x\right)^2+\left(y-y\right)^2=0^2+0^2=1+1=2\)

\(x^2-x-y^2-y\)

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

\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y-1\right)\)

đề bài là gì thế bn???