\(x\ge xy+1\Rightarrow1\ge y+\dfrac{1}{x}\ge2\sqrt{\dfrac{y}{x}}\Rightarrow\dfrac{y}{x}\le\dfrac{1}{4}\)

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

Đặt \(\dfrac{y}{x}=t\le\dfrac{1}{4}\) 

\(Q^2=\dfrac{t^2+2t+1}{t^2-t+3}=\dfrac{t^2+2t+1}{t^2-t+3}-\dfrac{5}{9}+\dfrac{5}{9}\)

\(Q^2=\dfrac{\left(4t-1\right)\left(t+6\right)}{9\left(t^2-t+3\right)}+\dfrac{5}{9}\le\dfrac{5}{9}\)

\(\Rightarrow Q_{max}=\dfrac{\sqrt{5}}{3}\) khi \(t=\dfrac{1}{4}\) hay \(\left(x;y\right)=\left(2;\dfrac{1}{2}\right)\)

từ giả thiết: \(x+y\le xy\le\frac{\left(x+y\right)^2}{4}\)(theo BĐT AM-GM)

\(\Leftrightarrow\left(x+y\right)\left(x+y-4\right)\ge0\)mà x,y dương nên \(x+y\ge4\)

ta có:\(16P\le\left(x+y\right)^2\left(\frac{1}{5x^2+7y^2}+\frac{1}{5y^2+7x^2}\right)\)

Áp dụng BĐT cauchy-schwarz theo chiều ngược lại:

\(\frac{\left(x+y\right)^2}{5x^2+7y^2}\le\frac{x^2}{3\left(x^2+y^2\right)}+\frac{y^2}{2\left(x^2+2y^2\right)}\)

\(\frac{\left(x+y\right)^2}{5y^2+7x^2}\le\frac{y^2}{3\left(x^2+y^2\right)}+\frac{x^2}{2\left(y^2+2x^2\right)}\)

\(\Rightarrow\left(x+y\right)^2\left(\frac{1}{5x^2+7y^2}+\frac{1}{5y^2+7x^2}\right)\le\frac{x^2+y^2}{3\left(x^2+y^2\right)}+\frac{x^2}{2\left(y^2+2x^2\right)}+\frac{y^2}{2\left(x^2+2y^2\right)}\)(*)

xét \(\frac{x^2}{y^2+2x^2}+\frac{y^2}{x^2+2y^2}=2-\frac{x^2+y^2}{y^2+2x^2}-\frac{x^2+y^2}{x^2+2y^2}=2-\left(x^2+y^2\right)\left(\frac{1}{y^2+2x^2}+\frac{1}{x^2+2y^2}\right)\)

Áp dụng BĐT cauchy:\(\frac{1}{y^2+2x^2}+\frac{1}{x^2+2y^2}\ge\frac{4}{3\left(x^2+y^2\right)}\)

do đó \(\frac{x^2}{y^2+2x^2}+\frac{y^2}{x^2+2y^2}\le2-\frac{4}{3}=\frac{2}{3}\)

kết hợp với (*):\(16VT\le\frac{1}{3}+\frac{1}{2}.\frac{2}{3}=\frac{2}{3}\)

\(VT\le\frac{1}{24}\)

Dấu = xảy ra khi x=y=2

tưởng giá trị nhỏ nhất chứ

giá trị lớn nhất nha pn.nếu pit giúp mk vs

\(y\ge1+xy\Rightarrow1\ge\dfrac{1}{y}+x\ge2\sqrt{\dfrac{x}{y}}\Rightarrow\dfrac{x}{y}\le4\Rightarrow\dfrac{y}{x}\ge4\)

\(G=\dfrac{x}{y}+\dfrac{y}{x}=\left(\dfrac{x}{y}+\dfrac{y}{16x}\right)+\dfrac{15}{16}.\dfrac{y}{x}\ge2\sqrt{\dfrac{xy}{16xy}}+\dfrac{15}{16}.4=\dfrac{17}{4}\)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{2};2\right)\)

Thay y = 4 x  vào biểu thức P và biến đổi ta thu được

P = - 9 log 2 2 + 27 log 2 x - 27 .

Do y ≥ 1  nên x ≤ 4 . Suy ra 1 2 ≤ x ≤ 4 . Đặt t = log 2 x , khi đó - 1 ≤ t ≤ 2 .

Xét hàm số f(t0 = - 9 t 2 + 27t - 27;  t ∈ - 1 ; 2

Ta có f ' (t) = -18t + 27; f ' (t) = 0  ⇔ t = 3 2

f (-1) = -63; f (2) = -9;  f 3 2 = 27 4

Vậy

m a x   P   = - 27 4 ⇔ x = 2 2 ; y = 2

Đáp án A

Đặt \(\left\{{}\begin{matrix}\sqrt{2x+3}=a\ge0\\\sqrt{y}=b\ge0\end{matrix}\right.\)

\(\Rightarrow b\left(b^2+1\right)-3a^2=\left(a^2+1\right)a-3b^2\)

\(\Rightarrow a^3-b^3+3a^2-3b^2+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2\right)+\left(a-b\right)\left(3a+3b\right)+a-b=0\)

\(\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+3a+3b+1\right)=0\)

\(\Leftrightarrow a=b\Rightarrow\sqrt{2x+3}=\sqrt{y}\)

\(\Rightarrow y=2x+3\)

\(\Rightarrow M=x\left(2x+3\right)+3\left(2x+3\right)-4x^2-3\) tới đây chắc chỉ cần bấm máy

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

\(\Rightarrow\left(\sqrt{x^2+y^2}-\sqrt{2}\right)\left(\sqrt{x^2+y^2}+3\sqrt{2}\right)\ge0\)

\(\Rightarrow x^2+y^2\ge2\)

\(\Rightarrow-\left(x^2+y^2\right)\le-2\)

\(P=\sqrt{9-x^2}+\sqrt{9-y^2}+\dfrac{x+y}{4}\le\sqrt{2\left(9-x^2+9-y^2\right)}+\dfrac{\sqrt{2\left(x^2+y^2\right)}}{4}\)

\(P\le\sqrt{2\left(18-x^2-y^2\right)}+\dfrac{1}{4}.\sqrt{2\left(x^2+y^2\right)}\)

\(P\le\left(\sqrt{2}-1\right)\sqrt{18-x^2-y^2}+\sqrt[]{2}\sqrt{\dfrac{\left(18-x^2-y^2\right)}{2}}+\dfrac{1}{2}\sqrt{\dfrac{x^2+y^2}{2}}\)

\(P\le\left(\sqrt{2}-1\right).\sqrt{18-2}+\sqrt{\left(2+\dfrac{1}{4}\right)\left(\dfrac{18-x^2-y^2+x^2+y^2}{2}\right)}=\dfrac{1+8\sqrt{2}}{2}\)

Dấu "=" xảy ra khi \(x=y=1\)

\(6xy=x+y\ge2\sqrt[]{xy}\Rightarrow\sqrt{xy}\ge\dfrac{1}{3}\Rightarrow xy\ge\dfrac{1}{9}\Rightarrow\dfrac{1}{xy}\le9\)

\(M=\dfrac{\dfrac{x+1}{xy+1}+\dfrac{xy+x}{1-xy}+1}{1+\dfrac{xy+x}{1-xy}-\dfrac{x+1}{xy+1}}=\dfrac{\dfrac{x+1}{xy+1}+\dfrac{x+1}{1-xy}}{\dfrac{x+1}{1-xy}-\dfrac{x+1}{xy+1}}=\dfrac{\dfrac{1}{1-xy}+\dfrac{1}{1+xy}}{\dfrac{1}{1-xy}-\dfrac{1}{1+xy}}\)

\(M=\dfrac{1+xy+1-xy}{1+xy-1+xy}=\dfrac{2}{2xy}=\dfrac{1}{xy}\le9\)

Dấu "=" xảy ra khi \(x=y=\dfrac{1}{3}\)

Đáp án C.

Ta có

Khi đó, giả thiết trở thành:

log 3 x + y x 2 + y 2 + x y + 2 = x 2 + y 2 + x y + 2 - 3 x + y - 2

⇔ log 3 x + y - log 3 x 2 + y 2 + x y + 2 = x 2 + y 2 + x y + 2 - 3 x + y - 2

⇔ 3 x + y + log 3 3 x + y = x 2 + y 2 + x y + 2 + log 3 x 2 + y 2 + x y + 2

Xét hàm số  f t = t + log 3   t  trên khoảng  0 ; + ∞ , có  f ' t = 1 + 1 t   ln 3 > 0 ; ∀ t > 0 .

Suy ra f(t) là hàm số đồng biến trên 0 ; + ∞  mà f[3(x + y)] = f(x² + y² + xy + 2)