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

\(\Rightarrow P=xy+\dfrac{1}{xy}=xy+\dfrac{1}{16xy}+\dfrac{15}{16xy}\ge2\sqrt{xy.\dfrac{1}{16xy}}+\dfrac{15}{16.\dfrac{1}{4}}=\dfrac{1}{2}+\dfrac{15}{4}=\dfrac{17}{4}\)

\(min_P=\dfrac{17}{4}\Leftrightarrow x=y=\dfrac{1}{2}\)

sorry pạn ( nãy mik làm sai )

Ta có x²+y² / x-y = x²-2xy+y²+2xy / x-y

                            = (x-y)²+2xy / x-y

Mà xy = 1 => 2xy = 2. Thay vào, ta có

(x-y)²+2xy / x-y = (x-y)²+2 / x-y = (x-y)² / x-y + 2 / x-y

                                                  = x-y + 2 / x-y

Áp dụng BĐT Cauchy, ta có

x-y + 2 / x-y ≥ 2.√(x-y).2 / x-y] = 2.√2 = (√2)³

Vậy Min A = (√2)³

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

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

\(P=2x^2-2x+1=\dfrac{1}{2}\left(2x-1\right)^2+\dfrac{1}{2}\ge\dfrac{1}{2}\)

\(P_{min}=\dfrac{1}{2}\) khi \(x=y=\dfrac{1}{2}\)

\(A=2\left(x^2+y^2\right)+\left(8y^2+\dfrac{1}{2}z^2\right)+\left(8x^2+\dfrac{1}{2}z^2\right)\ge2.2\sqrt{x^2y^2}+2\sqrt{8x^2.\dfrac{1}{2}z^2}+2.\sqrt{8x^2.\dfrac{1}{2}z^2}=4\left(xy+yz+zx\right)=4\)

\(A_{min}=4\) khi \(\left(x;y;z\right)=\left(\dfrac{1}{3};\dfrac{1}{3};\dfrac{4}{3}\right)\)

\(a,A=x^2+y^2\\=x^2-2xy+y^2+2xy\\=(x-y)^2+2xy\\=2^2+2\cdot1\\=4+2\\=6\)

\(b,x+y=1\\\Leftrightarrow (x+y)^3=1^3\\\Leftrightarrow x^3+3x^2y+3xy^2+y^3=1\\\Leftrightarrow x^3+3xy(x+y)+y^3=1\\\Leftrightarrow x^3+3xy\cdot1+y^3=1\\\Rightarrow A=1\)

a) Ta có:

\(x-y=2\)

\(\Rightarrow\left(x-y\right)^2=2^2\)

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

Mà: \(xy=1\)

\(\Rightarrow\left(x^2+y^2\right)-2\cdot1=4\)

\(\Rightarrow x^2+y^2=4+2\)

\(\Rightarrow x^2+y^2=6\)

b) Ta có: 

\(x+y=1\)

\(\Rightarrow\left(x+y\right)^3=1^3\)

\(\Rightarrow x^3+3x^2y+3xy+y^3=1\)

\(\Rightarrow x^3+3xy\left(x+y\right)+y^3=1\) 

Mà: x + y = 1

\(\Rightarrow x^3+3xy\cdot1+y^3=1\)

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

b) Theo đề ra, ta có:

 \(3x=5y\Rightarrow\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x}{10}=\frac{y}{6}\)

\(5y=6z\Rightarrow\frac{y}{6}=\frac{z}{5}\)

\(\Rightarrow\frac{x}{10}=\frac{y}{6}=\frac{z}{5}\)

Áp dụng tính chất của dãy tỷ số bằng nhau

\(\frac{x}{10}=\frac{y}{6}=\frac{z}{5}=\frac{x-y}{10-6}=1\)

\(\Rightarrow x=1.10=10\)

\(\Rightarrow y=1.6\)

\(\Rightarrow z=1.5=5\)

\(xy+x-y=x\left(y+1\right)-y=4\)
\(x\left(y+1\right)-y-1=3\)
\(x\left(y+1\right)-\left(y+1\right)=3\)
\(\left(x-1\right)\left(y+1\right)=3\)
\(=>\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.=>\left[{}\begin{matrix}y+1=3\\y+1=-3\\y+1=1\\y+1=-1\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=0\\x=4\\x=-2\end{matrix}\right.=>\left[{}\begin{matrix}y=2\\y=-4\\y=0\\y=-2\end{matrix}\right.\)

\(\text{x(y+1)-y=4}\)

\(\text{x(y+1)-y-1=3}\)

\(\text{x(y+1)-(y+1)=3}\)

\(\text{(x-1)(y+1)=3}\)

=> \(\left(x-1\right);\left(y+1\right)\in\left(Ư\right)3=\left\{1;3;-1;-3\right\}\)

==> bảng:

=>  (x;y) ∈ { (2;2);(4;0);(0;-4);(-2;-2)}