\(x^3+y^3=3xy-1\)

\(\Leftrightarrow x^3+y^3-3xy+1=0\)

\(\Leftrightarrow x^3+y^3+3x^2y+3xy^2-3xy-3x^2y-3xy^2+1=0\)

\(\Leftrightarrow\left(x+y\right)^3+1-3xy\left(x+y+1\right)=0\)

\(\Leftrightarrow\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1\right)-3xy\left(x+y+1\right)=0\)

\(\Leftrightarrow\left(x+y+1\right)\left(x^2+2xy+y^2-x-y+1-3xy\right)=0\)

\(\Leftrightarrow\left(x+y+1\right)\left(x^2+y^2-xy-x-y+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+y+1=0\\x^2+y^2-xy-x-y+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x+y=-1\\x^2+y^2-xy-x-y+1=0\end{cases}}\)

Mà x, y dương nên \(x+y=-1\)là vô lí

Vậy \(x^2+y^2-xy-x-y+1=0\)

Đến đây đợi tớ nghĩ tiếp :v

X³ + Y³ =3XY - 1

=> X³ + Y³ + 3X²Y + 3XY² - 3X²Y - 3XY² - 3XY + 1 = 0

=> \(\subset X+Y\supset^3\)+ 1 - 3XY\(\subset X+Y+1\supset\)= 0

=> \(\subset X+Y+1\supset.\)\(\subset\subset X+Y\supset^2-X-Y+1\supset\)-3XY\(\subset X+Y+1\supset=0\)

=>\(\subset X+Y+1\supset.\)\(\subset X^2+Y^2+2XY-X-Y+1-3XY\supset\)=0

=> \(\subset X+Y+1\supset.\subset X^2+Y^2-XY-X-Y+1\)=0

Vì X,Y > 0 =>X+Y+1 > 0

 \(\Rightarrow X^2+Y^2-XY-X-Y+1=0\)

\(\Rightarrow2X^2+2Y^2-2XY-2X-2Y+2=0\)

\(\Rightarrow X^2-2XY+Y^2+X^2-2X+1+Y^2-2Y+1=0\)

\(\Rightarrow\subset X-Y\supset^2+\subset X-1\supset^2+\subset Y-1\supset^2=0\)

Vì \(\subset X-Y\supset^2\ge;\subset X-1\supset^2\ge0;\subset Y-1\supset^2\ge0\)

\(\Rightarrow\hept{\begin{cases}\subset X-Y\supset^2=0\\\subset X-1\supset^2=0\\\subset Y-1\supset^2=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}X-Y=0\\X-1=0\\Y-1=0\end{cases}}\)\(\Rightarrow X=Y=1\) \(\Rightarrow A=1+1=2\)

A đoán bừa

D con ngu ạ

\(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\) (Cô si ngược + Rút gọn)

Tương tự \(\frac{b}{1+c^2}\ge b-\frac{bc}{2};\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)

Cộng theo vế 3 BĐT,ta được: \(VT\ge\left(a+b+c\right)-\left(\frac{ab+bc+ca}{2}\right)=3-\frac{ab+bc+ca}{2}\)

Mặt khác,ta có BĐT \(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\) (bạn tự c/m,không làm được ib)

Thay x = a; y = b ; z = c,ta có: \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=\frac{9}{3}=3\)

Suy ra\(VT\ge3-\frac{ab+bc+ca}{2}\ge3-\frac{3}{2}=\frac{3}{2}^{\left(đpcm\right)}\)

Dấu "=" xảy ra khi a = b = c = 1

\(x^4+4x^3-2x^2-12x+9\)

\(=x^4+3x^3+x^3+3x^2-5x^2-15x+3x+9\)

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

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

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

\(=\left(x+3\right)\left[x^2\left(x+3\right)-2x\left(x+3\right)+\left(x+3\right)\right]\)

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

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