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

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

\(\left(x-z\right)\left(x+z\right)=x^2+xz-xz-z^2=x^2-z^2\)

\(a)P\left(x\right)+Q\left(x\right)=4x^2y-2xy^2+x^2y-15+2xy+\left(2xy^2+3x^2y-4xy-x^2y^2\right)\)

\(=4x^2y^2-2xy^2+x^2y-15+2xy+2xy^2+3x^2y-4xy-x^2y^2\)

\(=\left(4x^2y^2-x^2y^2\right)+\left(-2xy^2+2xy^2\right)+\left(x^2y+3x^2y\right)-15+\left(2xy-4xy\right)\)

\(=3x^2y^2+4x^2y-2xy-15\)

\(P\left(x\right)-Q\left(x\right)=4x^2y^2-2xy^2+x^2y-15+2xy-\left(2xy^2+3x^2y-4xy-x^2y^2\right)\)

\(=4x^2y^2-2xy^2+x^2y-15+2xy-2xy^2-3x^2y+4xy+x^2y^2\)

\(=\left(4x^2y+x^2y^2\right)+\left(-2xy^2-2xy^2\right)+\left(x^2y-3x^2y\right)-15+\left(2xy+4xy\right)\)

\(=5x^2y-4xy^2-2x^2y+6xy-15\)

Giúp mình với nhé!

\(A\left(x,y\right)=x^2-2xy+y^2+4x^2-4xy+3\)

\(A\left(x,y\right)=5x^2-6xy+y^2+3\)

\(A\left(x,y\right)=2x^2+3x^2-6xy+y^2+3\)

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

Ta thấy: \(2x^2\ge0\forall x\)

             \(\left(3x-y\right)^2\ge0\forall x,y\)

\(\Rightarrow2x^2+\left(3x-y\right)^2+3\ge0\forall x,y\)

KL: Vậy biểu thức A luôn nhận giá trị dương.

\(B\left(x\right)=3x^2-5x+6\)

\(B\left(x\right)=3x^2-5x+\frac{5}{6}+\frac{31}{6}\)

\(B\left(x\right)=3x^2-5x+\left(\frac{5}{6}\right)^2+\frac{5}{36}+\frac{31}{6}\)

\(B\left(x\right)=\left(3x-\frac{5}{6}\right)^2+\frac{5}{36}+\frac{31}{6}\)

Ta thấy: \(\left(3x-\frac{5}{6}\right)^2\ge0\forall x\)

\(\Rightarrow\left(3x-\frac{5}{6}\right)^2+\frac{5}{36}+\frac{31}{6}\ge0\forall x\)

vậy biểu thức B luôn nhận giá trị dương.

\(\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}:\dfrac{4xy}{y^2-x^2}\) \(\left(x,y\ne0;x\ne\pm y\right)\)

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

\(=\dfrac{1}{x^2+2xy+y^2}+\dfrac{1}{4xy}\)

\(=\dfrac{6xy+x^2+y^2}{4xy\left(x+y\right)^2}\)

Ta có: \(\dfrac{1}{x^2+2xy+y^2}-\dfrac{1}{x^2-y^2}:\dfrac{4xy}{y^2-x^2}\)

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

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

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

\(=\dfrac{x^2+6xy+y^2}{4xy\left(x+y\right)^2}\)

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

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

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

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

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