\(P=\dfrac{x^3+y^3}{x^3y^3}=\dfrac{\left(x+y\right)\left(x^2+y^2-xy\right)}{x^3y^3}=\dfrac{x^2y^2\left(x+y\right)}{x^3y^3}=\dfrac{x+y}{xy}=\dfrac{\left(x+y\right)^2}{xy\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2}{x^2+y^2-xy}=\dfrac{4\left(x^2+y^2-xy\right)-3\left(x^2+y^2-2xy\right)}{x^2+y^2-xy}\)

\(=4-\dfrac{3\left(x-y\right)^2}{x^2+y^2-xy}\le4\)

\(P_{max}=4\) khi \(x=y=\dfrac{1}{2}\)

Đặt \(a=\dfrac{1}{x};b=\dfrac{1}{y}\). khi đó gt trở thành:

\(a+b=a^2+b^2-ab\ge\dfrac{1}{4}\left(a+b\right)^2\Leftrightarrow o\le a+b\le4\);

\(A=a^3+b^3=\left(a+b\right)\left(a^2+b^2-ab\right)=\left(a+b\right)^2\le16\)

Đẳng thức xảy ra khi và chỉ khi a=b=2 <=> x=y=1/2

Vậy Max A = 16

\(x^2+y^2\le x+y\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2\le\dfrac{1}{2}\)

Áp dụng BĐT Bunhiacopski:

\(\left[1\cdot\left(x-\dfrac{1}{2}\right)^2+3\left(y-\dfrac{1}{2}\right)^2\right]\le10\left[\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2\right]\le5\)

\(\Leftrightarrow\left(x+3y-2\right)^2\le5\\ \Leftrightarrow x+3y-2\le\sqrt{5}\\ \Leftrightarrow x+3y\le2+\sqrt{5}\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5+\sqrt{5}}{10}\\y=\dfrac{5+3\sqrt{5}}{10}\end{matrix}\right.\)

\(x^2+y^2< x+y\) . Dấu bé hơn hay dấu bé hơn hoặc bằng vậy bạn?

\(x^2+y^2\le x+y\)

\(\Rightarrow x^2-x+y^2-y\le0\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2\le\dfrac{1}{2}\left(1\right)\)

Theo BĐT Bunhiacopxki:

\(\left[\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2\right]\left(1+9\right)\ge\left[1.\left(x-\dfrac{1}{2}\right)+3.\left(y-\dfrac{1}{2}\right)\right]^2=\left(x+3y-2\right)^2\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\dfrac{1}{2}.10\ge\left(x+3y-2\right)^2\)

\(\Rightarrow-\sqrt{5}\le x+3y-2\le\sqrt{5}\)

\(\Rightarrow-\sqrt{5}+2\le P\le\sqrt{5}+2\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{y-\dfrac{1}{2}}{3}\\\left(x-\dfrac{1}{2}\right)^2+\left(y-\dfrac{1}{2}\right)^2=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}+\dfrac{1}{2\sqrt{5}}\\y=\dfrac{1}{2}+\dfrac{3}{2\sqrt{5}}\end{matrix}\right.\) (loại \(\left\{{}\begin{matrix}x=\dfrac{1}{2}-\dfrac{1}{2\sqrt{5}}\\y=\dfrac{1}{2}-\dfrac{3}{2\sqrt{5}}\end{matrix}\right.\) do \(y>0\) )

Và khi dấu bằng xảy ra thì \(P_{max}=\sqrt{5}+2\)

Theo bài ra, ta có: \(x^2-y=y^2-x\Leftrightarrow x^2-y^2=-x+y\)

\(\Leftrightarrow\left(x-y\right)\left(x+y\right)=-\left(x-y\right)\)

\(\Leftrightarrow\left(x+y\right)=-1\)

Ta lại có: \(A=x^2+2xy+y^2-3x-3y=\left(x+y\right)^2-3\left(x+y\right)\)

Thay x+y=-1 vào biểu thức A, ta được: \(A=\left(-1\right)^2-3.\left(-1\right)=1+3=4\)

Vậy A=4

đúng thì like giúp mik nha bạn. Thx bạn

\(a,VP=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\\ =\left(x+2y\right)\left[x^2-x.2y+\left(2y\right)^2\right]\\ =x^3+\left(2y\right)^3=x^3+8y^3=VT\left(đpcm\right)\\ b,VT=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\left(x-y\right)\\ =x^3-y^3-3xy\left(x-y\right)\\ =x^3-3x^2y+3xy^2-y^3\\ =\left(x-y\right)^3=VP\left(đpcm\right)\)

\(c,VT=\left(x-3y\right)\left(x^2+3xy+9y^2\right)-\left(3y+x\right)\left(9y^2-3xy+x^2\right)\\ =\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]-\left(x+3y\right).\left[x^2-x.3y+\left(3y\right)^2\right]\\ =x^3-27y^3-\left(x^3+27y^3\right)\\ =-54y^3=VP\left(đpcm\right)\)