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x+y = 2 => y = 2- x
=> x.y = x.(2 - x) = - x2 + 2x
Xét x.y - 1 = - x2 + 2x - 1 = (-x2 + x) + (x - 1) = - x.(x - 1) + (x - 1) = (x - 1).(-x + 1) = -(x-1).(x-1) = -(x-1)2 \(\le\) 0 với mọi x
=> xy - 1 \(\le\) 0 <=> x.y \(\le\) 1
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3.(x-1)-5.(-7)=(-2)3-2x
3.(x-1)-(-35)=(-8)-2x
3(x-1)+35=(-8)-2x
3x-1x+35=(-8)-2x
2x+35=(-8)-2x
tự giải tiếp nhé
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1, xy+2x-2y-5=0
=> x.( y+2)-2.(y+2)=5
=> (y+2).(x-2)=5
Vì x, y thuộc Z => y+2; x-2 thuộc Z
Mà 5=1.5=-1.(-5) và hoán vị của chúng
Ta có bảng sau:
y+2 1 5 -1 -5
x-2 5 1 -5 -1
y -1 3 -3 -7
x 7 3 -3 1
nHỚ K CHO MIK NHÉ
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Do \(x,y,z>0\Rightarrow xyz\ne0\)
\(\Rightarrow\dfrac{xy}{xyz}+\dfrac{yz}{xyz}+\dfrac{zx}{xyz}=1\)
\(\Rightarrow\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{y}=1\Rightarrow\dfrac{1}{x}< 1\Rightarrow x>1\)
Vì \(x\le y\le z\Rightarrow\dfrac{1}{x}\ge\dfrac{1}{y}\ge\dfrac{1}{z}\)
\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{x}=\dfrac{3}{x}\)
\(\Rightarrow1\le\dfrac{3}{x}\Rightarrow x\le3\) Mà \(x>1\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
Nếu \(x=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{1}{2}\Rightarrow\dfrac{1}{y}< \dfrac{1}{2}\Rightarrow y>2\\\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{2}{y}\Rightarrow\dfrac{2}{y}\ge\dfrac{1}{2}\Rightarrow y\le4\end{matrix}\right.\)
Mà \(y>2\Rightarrow\left[{}\begin{matrix}y=3\\y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y=3\Rightarrow z=6\\y=4\Rightarrow z=4\end{matrix}\right.\)
Nếu \(x=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{y}+\dfrac{1}{z}=\dfrac{2}{3}\Rightarrow\dfrac{1}{y}< \dfrac{2}{3}\Rightarrow y>\dfrac{3}{2}\\\dfrac{1}{y}+\dfrac{1}{z}\le\dfrac{2}{y}\Rightarrow\dfrac{2}{y}\ge\dfrac{2}{3}\Rightarrow y\le3\end{matrix}\right.\)
Do \(x\le y\Rightarrow\left\{{}\begin{matrix}y=3\\z=3\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(3;3;3\right);\left(2;3;6\right);\left(2;4;4\right)\)
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Đặt x/2=y/3=k
=>x=2k; y=3k
\(x^2=-19+y^2+xy\)
\(\Leftrightarrow4k^2=-19+9k^2+6k^2\)
\(\Leftrightarrow k^2=\dfrac{19}{11}\)
Trường hợp 1: \(k=\sqrt{\dfrac{19}{11}}\)
=>\(x=2\sqrt{\dfrac{19}{11}};y=\dfrac{3\sqrt{19}}{11}\)
Trường hợp 2: \(k=-\sqrt{\dfrac{19}{11}}\)
=>\(x=-2\sqrt{\dfrac{19}{11}};y=-\dfrac{3\sqrt{19}}{11}\)