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PT <=> \(x^2-4x\left(y-1\right)+5y^2-8y-12=0\)
Xét \(\Delta'=\left[-2\left(y-1\right)\right]^2-1.\left(5y^2-8y-12\right)\)
= \(4\left(y^2-2y+1\right)-5y^2+8y+12\)
= \(-y^2+16\)
Để PT có nghiệm <=> \(\Delta'\ge0< =>-y^2+16\ge0\)
<=> \(y^2\le16\) <=> \(-4\le y\le4\)
Mà y nguyên
<=> \(y\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)
Đến đây bn thay y vào PT để tìm x nhé
\(5y^2+3y=-2x^2+8x=8-\left(2x^2-8x+8\right)=8-2\left(x-2\right)^2\le8\)<=> \(5y^2+3y-8\le0< =>\left(5y+8\right)\left(y-1\right)\le0< =>\frac{-8}{5}\le y\le1\)
y nguyên => y = -1; 0; 1
y=-1 => \(2x^2+5-8x-3=0< =>x^2-4x+1=0\)(không có nghiệm x nguyên)
y=0 =>\(2x^2+0-8x-0=0< =>2x^2-8x=0< =>\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
y=1 =>\(2x^2+5-8x+3=0< =>x^2-4x+4=0< =>x=2\)
vậy pt có nghiệm (x;y) = (0;0) (4;0) (2;1)
\(9x+5y+18=2xy\)
\(9x+5y+18-2xy=0\)
\(x\left(9-2y\right)=-18-5y\)
mà \(x,y\in Z\)
\(\Rightarrow x=\frac{-18-5y}{9-2y}\)
đến đây tìm để \(x,y\in Z\) là được
\(9x+5y+18=2xy\)
\(\Leftrightarrow18x+10y-4xy=-36\)
\(\Leftrightarrow\left(18x-4xy\right)+10y=-36\)
\(\Leftrightarrow2x\left(9-2y\right)-\left(45-10y\right)=-81\)
\(\Leftrightarrow2x\left(9-2y\right)-5\left(9-2y\right)=-81\)
\(\Leftrightarrow\left(2x-5\right)\left(9-2y\right)=-81\)
\(P=\sqrt[]{x}+\dfrac{3}{\sqrt[]{x}-1}\left(x>1\right)\)
\(P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\)
Áp dụng bất đẳng thức Cauchy cho 2 số \(\sqrt[]{x}-1;\dfrac{3}{\sqrt[]{x}-1}\) ta được :
\(\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{\sqrt[]{x}-1.\dfrac{3}{\sqrt[]{x}-1}}\)
\(\Rightarrow\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{3}\)
\(\Rightarrow P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\ge2\sqrt[]{3}+1\)
\(\Rightarrow Min\left(P\right)=2\sqrt[]{3}+1\)
Consider the first equation:
\(x+5y=7\Leftrightarrow x=7-5y\)
We can see that as long as \(y\) is an integer, \(x\) will also be an integer. This means the given equation has an infinite amount of integer roots of \(\left(x;y\right)\) such that \(x=7-5y\)
Now consider the second equation:
\(2x+5y=10\Leftrightarrow y=\dfrac{10-2x}{5}\) (1)
Because \(y\) is an integer, \(\dfrac{10-2x}{5}\) must also be an integer. Therefore, \(10-2x⋮5\)
Since \(10⋮5\), \(2x⋮5\).
We have \(\left(2,5\right)=1\), so \(x⋮5\). Thus, \(x=5k\) (\(k\) is an integer)
From this, we subtitute that in (1) to get \(y=\dfrac{10-2.5k}{5}=\dfrac{10-10k}{5}=2-2k\)
As long as \(k\) is an integer, \(y\) and \(x\) will also be an integer. Therefore, the given equation has an infinite amount of integer roots such that \(y=-\dfrac{2}{5}x+2\)