Ta phân tích mẫu:

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

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

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

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

Thay vào ta có:

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

Vậy ta có điều phải chứng minh

\(a,VT=\dfrac{x^2+2xy+4-3x^2-3xy}{\left(x+y\right)\left(x+2y\right)}=\dfrac{-2x^2-xy+4}{\left(x+y\right)\left(x-2y\right)}=VP\\ b,VP=\dfrac{\left(x+y\right)^2}{\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{x-y}=VT\)

\(Sửa:VP=\dfrac{x^3+2x^2+x}{x^2y+2xy+y}=\dfrac{x\left(x^2+2x+1\right)}{y\left(x^2+2x+1\right)}=\dfrac{x}{y}=VT\)

\(\dfrac{x^3+2x^2+x}{x^2y+xy+y}=\dfrac{x\left(x^2+2x+1\right)}{y\left(x^2+x+1\right)}\) đề sai

Sửa đề: x+y+1/2x+1/2y>=7/2

Đặt \(S=x+y+\frac{1}{2x}+\frac{1}{2y}\)

\(=\left(\frac{x}{2}+\frac{1}{2x}\right)+\left(\frac{y}{2}+\frac{1}{2y}\right)+\frac{x + y}{2}\)

Theo Co-si, ta có: \(\frac{x}{2} + \frac{1}{2x} \ge 2 \sqrt{\frac{x}{2} \cdot \frac{1}{2x}} = 2 \cdot \frac{1}{2} = 1\)

\(\frac{y}{2} + \frac{1}{2y} \ge 2 \sqrt{\frac{y}{2} \cdot \frac{1}{2y}} = 2 \cdot \frac{1}{2} = 1\)

=>\(\left(\frac{x}{2}+\frac{1}{2x}\right)+\left(\frac{y}{2}+\frac{1}{2y}\right)\ge1+1=2\)

=>\(S = \left( \frac{x}{2} + \frac{1}{2x} \right) + \left( \frac{y}{2} + \frac{1}{2y} \right) + \frac{x + y}{2} \ge 2 + \frac{3}{2} = \frac{7}{2}\)

Dấu '=' xảy ra khi x=y=3/2

Sửa đề: \(A=\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y^2-x^2}\right):\dfrac{2y}{x-y}\)

Ta có: \(A=\left(\dfrac{x+y}{2x-2y}-\dfrac{x-y}{2x+2y}-\dfrac{2y^2}{y^2-x^2}\right):\dfrac{2y}{x-y}\)

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

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

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

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

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

\(=1\)

B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé