Thay \(x=1;y=-1;z=2\) vào P

\(P=1.\left(-1\right)^2.2-2.1^2.\left(-1\right).2^2+3.\left(-1\right).2+1\\ =5\)

\(=\left(x^2\right)^3-2^3=\left(x^2-2\right)\left(x^4+2x^2+4\right)\\ =>C\)

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

\(R=\left(2x\right)^2-2.2x.y+y^2\\ =\left(2x-y\right)^2\\ =>D\)

\(P=x^2\left(x^2-4\right)=x^2\left(x-2\right)\left(x+2\right)\\ =>A\)

\(\left(x+y-1\right)\left(x+y+1\right)=x^2+xy-x+xy+y^2-y+x+y-1\\ =x^2+\left(xy+xy\right)+\left(-x+x\right)+y^2+\left(-y+y\right)-1\\ =x^2+2xy+y^2-1\\ =>B\)

xem lại câu c 

\(a,\dfrac{x^2-9}{x-2}:\dfrac{x-3}{x}\\ =\dfrac{\left(x-3\right)\left(x+3\right)}{x-2}\times\dfrac{x}{x-3}\\ =\dfrac{x\left(x+3\right)}{\left(x-2\right)}\)

\(b,\dfrac{x}{z^2}.\dfrac{xz}{y^3}:\dfrac{x^3}{yz}\\ =\dfrac{x}{z^2}.\dfrac{xz}{y^3}.\dfrac{yz}{x^3}=\dfrac{x^2yz^2}{z^2y^3x^3}=\dfrac{1}{xy^2}\)

\(c,\dfrac{2}{x}-\dfrac{2}{x}:\dfrac{1}{x}+\dfrac{4}{x}.\dfrac{x^2}{2}\\ =\dfrac{2}{x}-\dfrac{2}{x}\times\dfrac{x}{1}+\dfrac{4x^2}{2x}\\ =\dfrac{2}{x}-\dfrac{2}{1}+2x\\ =\dfrac{2-2x+2x^2}{x}\)

Theo định lý tổng 4 góc trong tứ giác :

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\\ \Leftrightarrow\dfrac{5}{4}\widehat{C}+130^o+\widehat{C}+50^o=360^o\\ \Rightarrow\dfrac{9}{4}\widehat{C}=360^o-130^o-50^o\\ \Rightarrow\dfrac{9}{4}\widehat{C}=180^o\\ \Rightarrow\widehat{C}=80^o\)

\(\Rightarrow\widehat{A}=\dfrac{5}{4}\times80^o=100^o\)

thì trong ngoặc trc mà bạn