a)\(5x-xy=12\)

\(\Leftrightarrow x\left(4x-y\right)=12\)

<=>x và 4x-y thuộc Ư(12)=...

thay vào làm

b) \(2x+11=y\left(x+3\right)\)

\(\Rightarrow2x+11-xy-3y=0\)

\(\Rightarrow\left(2x-xy\right)+11-3y=0\)

\(\Rightarrow x\left(2-y\right)+6-3y=-5\)

\(\Rightarrow x\left(2-y\right)+3\left(2-y\right)=-5\)

\(\Rightarrow\left(x+3\right)\left(2-y\right)=-5\)

\(\Rightarrow x+3;2-y\inƯ\left(-5\right)=\left\{\pm1;\pm5\right\}\)

Xét \(x+3=1\Rightarrow x=-2\Rightarrow2-y=5\Rightarrow y=-3\)(loại vì \(x,y\in N\))

Xét \(x+3=-1\Rightarrow x=-4\Rightarrow2-y=-5\Rightarrow y=7\)(loại vì \(x,y\in N\))

Xét \(x+3=5\Rightarrow x=2\Rightarrow2-y=1\Rightarrow y=1\) (thỏa mãn)

Xét \(x+3=-5\Rightarrow x=-8\Rightarrow2-y=-1\Rightarrow y=3\)(loại vì \(x,y\in N\))

Vậy pt có nghiệm (x,y)=(2;1) thỏa mãn

+ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge-3\\y\ge-4\end{matrix}\right.\)

\(gt\Rightarrow x+y=6\left(\sqrt{x+3}+\sqrt{4+y}\right)\le6\sqrt{2\left(x+y+7\right)}\)

\(\Rightarrow\left(x+y\right)^2\le72\left(x+y+7\right)\)

\(\Rightarrow\left(x+y\right)^2-72\left(x+y\right)-504\le0\)

\(\Rightarrow\left(x+y-36\right)^2\le1800\Rightarrow P\le36+30\sqrt{2}\)

max \(P=36+30\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+3=y+4\\x+y=36+30\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{37}{2}+15\sqrt{2}\\y=\frac{35}{2}+15\sqrt{2}\end{matrix}\right.\)

+ \(x+y=6\left(\sqrt{x+3}+\sqrt{y+4}\right)\)

\(\Rightarrow\left(x+y\right)^2=36\left(x+y+7+2\sqrt{\left(x+3\right)\left(y+4\right)}\right)\)

\(\Rightarrow\left(x+y\right)^2-36\left(x+y\right)-252=72\sqrt{\left(x+3\right)\left(y+4\right)}\ge0\)

\(\Rightarrow\left(x+y-42\right)\left(x+y+6\right)\ge0\Rightarrow x+y\ge42\)

Min \(P=42\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\left(x+3\right)\left(y+4\right)}=0\\x+y=42\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-3\\y=45\end{matrix}\right.\\\left\{{}\begin{matrix}x=46\\y=-4\end{matrix}\right.\end{matrix}\right.\)

Để thỏa mãn BPT thì:

\(\left\{{}\begin{matrix}m-1>0\\\Delta< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m>1\\\left[{}\begin{matrix}m>\sqrt{2}\\m< -\sqrt{2}\end{matrix}\right.\end{matrix}\right.\)

=> \(m>\sqrt{2}\)

ơ bạn ơi xét a>0 vớiΔ<0 là thỏa mãn mọi x

còn chỉ lấy x>0 như nào😃😃

\(P=\frac{2}{3xy}+\frac{3}{\sqrt{3\left(1+y\right)}}\ge\frac{2}{3y\left(3-y\right)}+\frac{6}{y+4}\)

\(\Rightarrow P\ge2\left(\frac{-9y^2+28y+4}{3\left(-y^3-y^2+12y\right)}\right)=2\left(\frac{2\left(-y^3-y^2+12y\right)+2y^3-7y^2+4y+4}{3\left(-y^3-y^2+12y\right)}\right)\)

\(P\ge2\left(\frac{2}{3}+\frac{\left(y-2\right)^2\left(2y+1\right)}{3y\left(3-y\right)\left(y+4\right)}\right)\ge\frac{4}{3}\)

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

@Nguyễn Việt Lâm duyệt bài giúp em với ạ @Phạm Minh Quang nick đây

BĐT Bu nhi a cốp xki :

\(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)

\(\Rightarrow\left(x.1+y.1\right)^2\le\left(1^2+1^2\right)\left(x^2+y^2\right)\)

\(\Rightarrow\left(x+y\right)^2\le2\left(x^2+y^2\right)\)

\(\Rightarrow x+y\le\sqrt{2\left(x^2+y^2\right)}\)Nguyễn Thị Thanh Trang

\(P=2018xy+2019\left(x+y\right)\le2018.\frac{x^2+y^2}{2}+2019\sqrt{2\left(x^2+y^2\right)}=2018.\frac{1}{2}+2019\sqrt{2.1}=1009+2019\sqrt{2}\)

Vậy GTLN của P là \(1009+2019\sqrt{2}\) . Dấu \("="\) xảy ra khi \(x=y=\frac{1}{\sqrt{2}}\)

Giả sử: 

x² + x + 6 = k² ( k nguyên dương)

\(\text{=> 4x² + 4x + 24 = 4k² }\)

\(\text{=> -(2x+1)² + 4k² = 23 }\)
\(\text{=>(-2k+2x+1)(2k+2x+1) = -23 }\)
Do x, k đều nguyên và k nguyên dương nên 2x + 2k + 1 > 2x +1-2k do đó chỉ xảy ra các trường hợp 
TH1: -2k+2x+1 = -1 và 2k+2x+1 = 23

=> x = 5 và k = 6 
TH2: -2k+2x+1 = -23 và 2k + 2x +1= 1

=> x = - 6 va k = 6 (loại vì \(k\in N\))

Vậy x = 5

\(P\ge\frac{\left(x+y\right)^2}{2\left(2x^2+1\right)\left(2y^2+1\right)}+\frac{1}{xy}=\frac{2}{\left(2x^2+1\right)\left(2y^2+1\right)}+\frac{2}{9xy}+\frac{7}{9xy}\)

\(P\ge\frac{8}{4x^2y^2+2x^2+2y^2+4xy+5xy+1}+\frac{7}{9xy}\)

\(P\ge\frac{8}{4\left(\frac{x+y}{2}\right)^4+2\left(x+y\right)^2+\frac{5}{4}\left(x+y\right)^2+1}+\frac{28}{9\left(x+y\right)^2}=\frac{11}{9}\)