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a) \(6xy+4x-9y-7=0\)
\(\Leftrightarrow2x.\left(3y+2\right)-9y-6-1=0\)
\(\Leftrightarrow2x.\left(3y+x\right)-3.\left(3y+2\right)=1\)
\(\Leftrightarrow\left(2x-3\right).\left(3y+2\right)=1\)
Mà \(x,y\in Z\Rightarrow2x-3;3y+2\in Z\)
Tự làm típ
\(A=x^3+y^3+xy\)
\(A=\left(x+y\right)\left(x^2-xy+y^2\right)+xy\)
\(A=x^2-xy+y^2+xy\)( vì \(x+y=1\))
\(A=x^2+y^2\)
Áp dụng bất đẳng thức Bunhiakovxky ta có :
\(\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\left(x\cdot1+y\cdot1\right)^2=\left(x+y\right)^2=1\)
\(\Leftrightarrow2\left(x^2+y^2\right)\ge1\)
\(\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)
Hay \(x^3+y^3+xy\ge\frac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=\frac{1}{2}\)
Đề bài sai, đề đúng thì phân thức đằng sau dấu chia phải là:
\(\dfrac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)
Bài 3:
\(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\ge\dfrac{4}{xy}\)
\(\Leftrightarrow x^2y^2\left(\dfrac{1}{\left(x-y\right)^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\ge\dfrac{4}{xy}.x^2y^2\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2+y^2\ge4xy\)
\(\Leftrightarrow\dfrac{x^2y^2}{\left(x-y\right)^2}+x^2-2xy+y^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2+\left(x-y\right)^2\ge2xy\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}\right)^2-2xy+\left(x-y\right)^2\ge0\)
\(\Leftrightarrow\left(\dfrac{xy}{x-y}-x+y\right)^2=0\) (luôn đúng)
Ta có:
\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\)
\(\Leftrightarrow2=x^2-2+\dfrac{1}{x^2}+x^2-xy+\dfrac{y^2}{4}+xy\)
\(\Leftrightarrow2=\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2+xy\)
Vì : \(\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2\ge0\)
\(\Rightarrow xy\le2\)
Vậy GTLN của xy=2 \(\Leftrightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x-\dfrac{y}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\Rightarrow y=2\\x=-1\Rightarrow y=-2\end{matrix}\right.\)
\(VT=\dfrac{1}{\left(x-y\right)^2}+\dfrac{x^2+y^2}{x^2y^2}=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x-y\right)^2+2xy}{x^2y^2}\)
\(VT=\dfrac{1}{\left(x-y\right)^2}+\dfrac{\left(x-y\right)^2}{x^2y^2}+\dfrac{2}{xy}\ge2\sqrt{\dfrac{\left(x-y\right)^2}{\left(x-y\right)^2x^2y^2}}+\dfrac{2}{xy}=\dfrac{2}{\left|xy\right|}+\dfrac{2}{xy}\ge\dfrac{2}{xy}+\dfrac{2}{xy}=\dfrac{4}{xy}\)
\(\text{Ta có : }2x^2+\frac{1}{x^2}+\frac{y^2}{4}=4\)
\(\Leftrightarrow\left(x^2+2+\frac{1}{x^2}\right)+\left(x^2-xy+\frac{y^2}{4}\right)=2-xy\)
\(\Leftrightarrow\left(x+\frac{1}{x}\right)^2+\left(x-\frac{y}{2}\right)^2=2-xy\)
\(\text{ Lại có : }\left(x+\frac{1}{x}\right)^2+\left(x-\frac{y}{2}\right)^2\ge0\)
\(\Rightarrow2-xy\ge0\)
\(\Rightarrow xy\le2\)
Mà xy có giá trị lớn nhất
\(\Rightarrow xy\in\left\{\left(1;2\right)\left(2;1\right)\left(-1;-2\right)\left(-2;-1\right)\right\}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy})(x^2+y^2+2xy)\geq (1+1+2)^2=16$
$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\geq \frac{16}{(x+y)^2}=16$
Áp dụng BĐT AM-GM:
$xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}$
$\Rightarrow \frac{2}{xy}\geq 8$
Cộng 2 BĐT trên lại:
$P\geq 16+8=24$
Vậy $P_{\min}=24$ khi $x=y=\frac{1}{2}$
Lời giải:
Áp dụng BĐT Bunhiacopxky:
$(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy})(x^2+y^2+2xy)\geq (1+1+2)^2=16$
$\Rightarrow \frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{xy}\geq \frac{16}{(x+y)^2}=16$
Áp dụng BĐT AM-GM:
$xy\leq \frac{(x+y)^2}{4}=\frac{1}{4}$
$\Rightarrow \frac{2}{xy}\geq 8$
Cộng 2 BĐT trên lại:
$P\geq 16+8=24$
Vậy $P_{\min}=24$ khi $x=y=\frac{1}{2}$
Ta có :
\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\)
\(\Rightarrow\left(x^2+\dfrac{1}{x^2}-2\right)+\left(x^2+\dfrac{y^2}{4}-xy\right)=2-xy\)
\(\Rightarrow\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2=2-xy\)
Ta có:
\(\left(x-\dfrac{1}{x}\right)^2\ge0\forall x\)
\(\left(x-\dfrac{y}{2}\right)^2\ge0\forall x,y\)
\(\Rightarrow\left(x-\dfrac{1}{x}\right)^2+\left(x-\dfrac{y}{2}\right)^2\ge0\forall x,y\)
\(\Rightarrow2-xy\ge0\forall x,y\)
\(\Rightarrow xy\le2\)
Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x-\dfrac{y}{2}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{x}\\x=\dfrac{y}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2=1\\y=2x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\\\left[{}\begin{matrix}y=2\\y=-2\end{matrix}\right.\end{matrix}\right.\)
Vậy (x;y) nguyên thỏa mãn là : (1;2);(-1;-2)