\(\Leftrightarrow\sqrt{x}+\sqrt{y}=7\sqrt{19}\)

đặt \(\sqrt{x}=a.\sqrt{19}\);\(\sqrt{y}=a.\sqrt{19}\left(a+b=7\right)\)

Vì \(a,b\in N\)nên \(a\in\hept{ }0;1;2;3;4;5;6;7\)

xét từng TH rồi được kết quả (x;y) là (0;931),(19;684),(76;475),(171,304),(304;171),(475;76),(684;19),(931;0)

Ai giúp e vs ạ

\(P=\sqrt{x+1}+\sqrt{y+1}\ge\sqrt{x+1+y+1}=\sqrt{x+y+2}=\sqrt{101}\)

GTNN\(P=\sqrt{101}\)

\(P=\sqrt{x+1}+\sqrt{y+1}\)

\(=>\left(\sqrt{x+1}+\sqrt{y+1}\right)^2\le2\left(x+1+y+1\right)=2.101=202\)

GTLN \(P=202\)

Áp dụng bđt AM-GM dạng \(a+b\ge2\sqrt{ab}\)ta có

\(P^2=x+y+2+2\sqrt{\left(x+1\right)\left(y+1\right)}\)

       \(\le x+y+2+\left(x+1\right)+\left(y+1\right)=202\)

\(\Rightarrow P\le\sqrt{202}\)

Dấu "=" xảy ra khi \(x=y=\frac{99}{2}\)

Áp dụng bất đẳng thức bu - nhi - a - cốp - ski cho 2 cặp số ( \(\sqrt{x+1},\sqrt{y+1}\)) và ( 1 , 1 )

\(\sqrt{x+1}+\sqrt{y+1}\le\left(x+1+y+1\right).\left(1+1\right)\)= 2.101 = 202

Dấu bằng xảy ra \(\Leftrightarrow\)\(\hept{\begin{cases}\frac{\sqrt{x+1}}{1}=\frac{\sqrt{y+1}}{1}\\x+y=99\end{cases}}\) 

                       \(\Leftrightarrow\hept{\begin{cases}\sqrt{x+1}=\sqrt{y+1}\\x+y=99\end{cases}}\)

                       \(\Leftrightarrow\hept{\begin{cases}x=\frac{99}{2}\\y=\frac{99}{2}\end{cases}}\)

P² = x + 1 + \(2\sqrt{\left(x+1\right)\left(y+1\right)}\)+ y +1 = 101 + \(2\sqrt{\left(x+1\right)\left(y+1\right)}\ge101\)

P \(\ge\sqrt{101}\)

Dấu bằng xảy ra \(\Leftrightarrow\)x + 1 = 0

                                  y + 1 = 0

Nếu x = -1 thì y = 100

Nếu y = -1 thì x = 100

\(\sqrt{4x+2\sqrt{x}+1}\le\sqrt{4x+\dfrac{1}{2}\left(2^2+x\right)+1}=\sqrt{\dfrac{9x}{2}+3}\)

\(=\dfrac{1}{\sqrt{21}}.\sqrt{21}.\sqrt{\dfrac{9x}{2}+3}\le\dfrac{1}{2\sqrt{21}}\left(21+\dfrac{9x}{2}+3\right)=\dfrac{1}{2\sqrt{21}}\left(\dfrac{9x}{2}+24\right)\)

Tương tự và cộng lại:

\(A\le\dfrac{1}{2\sqrt{21}}\left(\dfrac{9}{2}\left(x+y+z\right)+72\right)=3\sqrt{21}\)

\(A_{max}=3\sqrt{21}\) khi \(x=y=z=4\)

\(A=1\sqrt{4x+2\sqrt{x}+1}+1.\sqrt{4y+2\sqrt{y}+1}+1\sqrt{4z+2\sqrt{z}+1}\)

\(\le\sqrt{\left(1+1+1\right)\left(4\left(x+y+z\right)+2\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3\right)}\)

\(=\sqrt{3.\left[51+\dfrac{4\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)}{2}\right]}\)

\(\le\sqrt{3.\left[51+\dfrac{x+y+z+12}{2}\right]}\)

\(=\sqrt{189}\)

Dấu "=" xảy ra <=> x = y = z = 4

\(y\left(x+1\right)^2=-x^2+2018x-1\)

\(\Leftrightarrow y=\dfrac{-x^2+2018x-1}{\left(x+1\right)^2}=-1+\dfrac{2020x}{\left(x+1\right)^2}\)

\(\Rightarrow\dfrac{2020x}{\left(x+1\right)^2}\in Z\)

Mà x và \(x\left(x+2x\right)+1\) nguyên tố cùng nhau

\(\Rightarrow2020⋮\left(x+1\right)^2\)

Ta có 2020 chia hết cho đúng 2 số chính phương là 1 và 4

\(\Rightarrow\left[{}\begin{matrix}\left(x+1\right)^2=1\\\left(x+1\right)^2=4\end{matrix}\right.\) \(\Rightarrow x=\left\{0;1\right\}\) \(\Rightarrow y\)

b.

Từ pt đầu:

\(x^2+xy-2y^2+2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+2y\right)+2\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x+2y+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-2y-2\end{matrix}\right.\)

Thế xuống dưới ...

\(P=\sqrt{\left(x-3\right)^2+4^2}+\sqrt{\left(y-3\right)^2+4^2}+\sqrt{\left(z-3\right)^2+4^2}\)

\(P\ge\sqrt{\left(x-3+y-3+z-3\right)^2+\left(4+4+4\right)^2}=6\sqrt{5}\)

\(P_{min}=6\sqrt{5}\) khi \(x=y=z=1\)

Mặt khác với mọi \(x\in\left[0;3\right]\) ta có:

\(\sqrt{x^2-6x+25}\le\dfrac{15-x}{3}\)

Thật vậy, BĐT tương đương: \(9\left(x^2-6x+25\right)\le\left(15-x\right)^2\)

\(\Leftrightarrow8x\left(3-x\right)\ge0\) luôn đúng

Tương tự: ...

\(\Rightarrow P\le\dfrac{45-\left(x+y+z\right)}{3}=14\)

\(P_{max}=14\) khi \(\left(x;y;z\right)=\left(0;0;3\right)\) và hoán vị

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