Sửa đề : \(\left(\frac{2x}{2x+y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)

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

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

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

\(\left(\frac{2x}{2x+y}-\frac{4x^2}{4x^2+2xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)

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

\(=\frac{8x^3+4x^2y+2xy^2-8x^3-4x^2y}{\left(2x+y\right)\left(4x^2+2xy+y^2\right)}:\frac{2x-2x-y}{\left(2x-y\right)\left(2x+y\right)}\)

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

\(=\frac{4x^2-2xy^3}{-y\left(4x^2+2xy+y^2\right)}\)

Nếu chuyển \(4x^2+2xy+y^2\)thành \(4x^2-2xy+y^2\)thì nó sẽ dễ tính hơn nhiều == 

2x(3x - 5) - (5 - 3x) = 0

=> 2x(3x - 5) + (3x - 5) = 0

=> (3x - 5)(2x + 1) = 0

=> \(\orbr{\begin{cases}3x-5=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-\frac{1}{2}\end{cases}}\)

Vậy x \(\in\left\{\frac{5}{3};-\frac{1}{2}\right\}\)là giá trị cần tìm

\(P=\frac{1}{x^2-x+1}+1-\frac{x^2+2}{x^3+1}\)

\(\Rightarrow P=\frac{1}{x^2-x+1}+1-\frac{x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\Rightarrow P=\frac{1\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{1\left(x+1\right)\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{x^2+2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\Rightarrow P=\frac{1\left(x+1\right)+1\left(x+1\right)\left(x^2-x+1\right)-x^2-2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\Rightarrow P=\frac{x+1+1\left(x^3+1\right)-x^2-2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\Rightarrow P=\frac{x+1+x^3+1-x^2-2}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\Rightarrow P=\frac{x+x^3-x^2}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x\left(1+x^2-x\right)}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(\Rightarrow P=\frac{x\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x}{x+1}\)

a) x^2 - 2xy + y^2 - xz + yz 

= (x^2 - 2xy + y^2 ) - (xz + yz)

= (x - y)^2 - z(x + y)

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