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

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

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

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

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

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

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

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

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

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

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

Áp dụng BĐT Cô si ta có:

\(x^3+8y^3+1\ge3\sqrt[3]{x^3\cdot8y^3\cdot1}=6xy\)

\(\Rightarrow x^3+8y^3+1-6xy\ge0\)

Dấu "=" xảy ra tại \(x=2y=1\Rightarrow x=1;y=\frac{1}{2}\)

Khi đó:

\(A=x^{2018}+\left(y-\frac{1}{2}\right)^{2019}=1^{2018}+0^{2019}=1\)

Ta có : \(3\left(x^2+y^2+z^2\right)=\left(x+y+z\right)^2\)

\(\Leftrightarrow3\left(x^2+y^2+z^2\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)\)

\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

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

\(\Leftrightarrow x=y=z\)

Khi đó : \(3x^{2018}=27^{673}=\left(3^3\right)^{673}=3^{2019}\)

\(\Leftrightarrow x^{2018}=3^{2018}\)

\(\Leftrightarrow\orbr{\begin{cases}x=y=z=3\\x=y=z=-3\end{cases}}\)

Đến đây tự tính A nha!

TÔI CHƯA GIẢI ĐƯỢC

a) Ta có: ab = 132 = 12.11 ( thỏa mãn điều kiện a+b = 23)

 => a² + b² = 12² + 11² = 144 + 121 = 265

Từ \(3\left(x^2+y^2+z^2\right)=\left(x+y+z\right)^2\)

Suy ra: x=y=z

\(\Rightarrow3x^{2018}=3y^{2018}=3z^{2018}=27^{673}=3^{2019}\)

\(\Leftrightarrow x^{2018}=y^{2018}=z^{2018}=3^{2018}\)

\(\Rightarrow x,y,z=3\)

Dễ tính A

Cảm ơn bạn nhé ,,.... 

Câu 4:

5x + 7y = 112

5(x+ y) = 112 - 2y

5(x + y) = 2(56 - y)

\(\begin{cases}x+y=2\\ 56-y=5\end{cases}\)

\(\begin{cases}x+y=2\\ y=56-5\end{cases}\)

\(\begin{cases}x=2-y\\ y=51\end{cases}\)

\(\begin{cases}x=2-51\\ y=51\end{cases}\)

\(\begin{cases}x=-49\\ y=51\end{cases}\)

Vậy (x ; y) = (-49; 51)

1)<=> \(\left(x^2+2x+1\right)-\left(y^2+4x+4\right)-7=0\)

=> \(\left(x+1\right)^2-\left(y+2\right)^2=7\)

\(\Rightarrow\left(x+1-y-2\right)\left(x+1+y+2\right)=7\)

\(\left(x-y-1\right)\left(x+y+3\right)=7\)

vì x;y là số nguyên dương=> \(x;y\ge1\Rightarrow x+y+3\ge1+1+3=5\)

=> \(x+y+3=7\Rightarrow x+y=4\)

\(x-y-1=1\Rightarrow x-y=2\)

cộng hai phương trình ta có:

=> \(\left(x+y\right)+\left(x-y\right)=4+2\)

=> \(2x=6\)

=> x=3

=> y=4-3=1

2)<=> \(\left(x^2+xy+2x\right)+\left(2xy+2y^2+4y\right)+\left(x+y+2\right)=17\)

=> \(x\left(x+y+2\right)+2y\left(x+y+2\right)+1\left(x+y+2\right)=17\)

=> \(\left(x+y+2\right)\left(x+2y+1\right)=17\)