Ta có: \(2x^2+xy+2y^2=\dfrac{3}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x^2+2xy+y^2\right)=\dfrac{3}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x+y\right)^2\)

Theo BĐT Bunhacopxky: \(\left(x^2+y^2\right)\left(1+1\right)\ge\left(x+y\right)^2\Rightarrow\dfrac{3}{2}\left(x^2+y^2\right)\ge\dfrac{3}{4}\left(x+y\right)^2\\ \Rightarrow2x^2+xy+2y^2=\dfrac{3}{2}\left(x^2+y^2\right)+\dfrac{1}{2}\left(x+y\right)^2\ge\dfrac{5}{4}\left(x+y\right)^2\\ \Rightarrow\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)

Chứng minh tương tự:

\(\sqrt{2y^2+yz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)\\ \sqrt{2z^2+xz+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)

Cộng vế theo vế, ta được: \(P\ge\sqrt{5}\left(x+y+z\right)=\sqrt{5}\cdot1=\sqrt{5}\)

Dấu "=" \(\Leftrightarrow x=y=z=\dfrac{1}{3}\) 

Bạn tham khảo nhé

https://hoc24.vn/cau-hoi/cho-cac-so-duong-xyz-thoa-man-xyz1cmrcan2x2xy2y2can2y2yz2z2can2z2zx2x2can5.182722154737

Thay x = -1,76; y = 3/25

⇒ P = 1/2

Đáp án B.

Ta có  4 = 2 x + 2 y ≥ 2 2 x . 2 y = 2 2 x + y

⇔ 4 ≥ 2 x + y ⇔ x + y ≤ 2 .

Suy ra  x y ≤ x + y 2 2 = 1

Khi đó

P = 2 x 3 + y 3 + 4 x 2 y 2 + 10 x y 2 x + y x + y 2 - 3 x y + 2 x y 2 + 10 x y

≤ 4 4 - 3 x y + 4 x 2 y 2 + 10 x y

= 16 + 2 x 2 y 2 + 2 x y x y - 1 ≤ 18

Vậy P_max = 18 khi x = y = 1.

\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)

\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)

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

Xét hàm  trên  ℝ  và đi đến kết quả 

\(A=\left(x-y\right)\left(x^2-xy\right)-x\left(x^2+2y^2\right)\)

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

\(=-2x^2y-xy^2\)

\(=-2\cdot2^2\cdot\left(-3\right)-2\cdot\left(-3\right)^2\)

\(=8\cdot3-2\cdot9\)

=6

Chắc đề bài là \(Q=\dfrac{3}{9x^2+6xy+y^2}+\dfrac{3}{3x^2+6xy+2y^2}\)

Từ giả thiết ta có:

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

\(\Leftrightarrow2x\left(x^2+y^2\right)+y\left(x^2+y^2\right)=2\left(x^2+y^2\right)\)

\(\Leftrightarrow2x+y=2\)

Do đó:

\(Q=3\left(\dfrac{1}{9x^2+6xy+y^2}+\dfrac{1}{3x^2+6xy+2y^2}\right)\)

\(Q\ge\dfrac{3.4}{12x^2+12xy+3y^2}=\dfrac{4}{\left(2x+y\right)^2}=1\)

\(Q_{min}=1\) khi \(\left\{{}\begin{matrix}2x+y=2\\9x^2+6xy+y^2=3x^2+6xy+2y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{6}-2\\y=6-2\sqrt{6}\end{matrix}\right.\)

a) \(2x^2-3xy-2y^2=2\)

\(\Rightarrow2x^2+xy-4xy-2y^2=2\)

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

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

\(\Rightarrow\left(2x+y\right);\left(x-2y\right)\in\left\{-1;1;-2;2\right\}\)

Ta giải các hệ phương trình sau với x;y nguyên 

1) \(\left\{{}\begin{matrix}2x+y=-1\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-2\\x-2y=-2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)

2) \(\left\{{}\begin{matrix}2x+y=1\\x-2y=2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}4x+2y=2\\x-2y=2\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}5x=4\left(loại\right)\\x-2y=-1\end{matrix}\right.\)

3) \(\left\{{}\begin{matrix}2x+y=-2\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}4x+2y=-4\\x-2y=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=-5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=-1\\y=0\end{matrix}\right.\)

4)  \(\left\{{}\begin{matrix}2x+y=2\\x-2y=1\end{matrix}\right.\) \(\left\{{}\begin{matrix}4x+2y=4\\x-2y=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}5x=5\\y=\dfrac{x+1}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(-1;0\right);\left(1;1\right)\right\}\)

b) \(xy-y+x=9\)

\(\Rightarrow y\left(x-1\right)+x-1+1=9\)

\(\Rightarrow\left(x-1\right)\left(y+1\right)=8\)

\(\Rightarrow\left(x-1\right);\left(y+1\right)\in\left\{-1;1;-2;2;-4;4;-8;8\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(0;-9\right);\left(2;7\right);\left(-1;-5\right);\left(3;3\right);\left(-3;-3\right);\left(5;1\right);\left(-7;-2\right);\left(9;0\right)\right\}\)