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

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

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

1) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x\left(x^2-16\right)\)

\(=x^3-16x-\left(x^2+1\right)\left(x^2-1\right)\)

\(=x^3-16x-x^4+1\)

b) \(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)

\(=28xy-7x^2+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)

\(=28xy-7x^2+4y^2-28xy-4y^2+7x\)

\(=-7x^2+7x\)

c) \(\left(3x-1\right)\left(2x-5\right)-4\left(2x^2-5x+2\right)\)

\(=6x^2-17x+5-4\left(2x^2-5x+2\right)\)

\(=6x^2-17x+5-8x^2+20x-8\)

\(=-2x^2+3x-3\)

a)  x(x+4)(x-4)-(x²+1)(x²-1)

=>x(x²-4²)-(x⁴-1²)

=>x³-16x-x⁴+1

=>-x⁴-x³-15x

b)  7x(4y-x)+4y(y-7x)-2(2y²-3.5x)

=>28xy-7x²+4y²-28xy-4y²+30x

=>-7x²+30x

c)  (3x+1)(2x-5)-4(2x²-5x+2)

=>6x²-15x+2x-5-8x²+20x-8

=>-2x²+7x-13

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)

Đặt \(x^2+7x=t\)

\(\left(t+10\right)\left(t+12\right)-8=t^2+22t+120-8\)

\(=t^2+22t+112=\left(t+8\right)\left(t+14\right)\)

Theo cách đặt \(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)

CHAO BAN

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

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

a) A= 3.(x²-2xy+y²)- 2. (x²+2xy+y²) - x²-y²

A= 3.x²-2xy+y²-2. x²+2xy+y²-x²-y²

a: =(x-y)^2+2(x-y)

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

c: =(x-3)(x+3)+(x-3)^2

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

=2x(x-3)

d: =(x+3)(x^2-3x+9)-4x(x+3)

=(x+3)(x^2-7x+9)

e: =(x^2-8x+7)(x^2-8x+15)-20

=(x^2-8x)^2+22(x^2-8x)+85

=(x^2-8x+17)(x^2-8x+5)

\(\Rightarrow x^2+2x+1-y^2-4y-4-7=0\\ \Leftrightarrow\left(x+1\right)^2-\left(y+2\right)^2=7\\ \Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)^2=16\\\left(y+2\right)^2=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x+1=4\\y+2=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+1=-4\\y+2=-3\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

Bạn làm như thế này là sai rồi nhé bạn dùng HDT số 3 rồi xét các ước của pt=> nghiệm nha