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\(P=\sqrt{\left(x-3\right)^2+4^2}+\sqrt{\left(y-3\right)^2+4^2}+\sqrt{\left(z-3\right)^2+4^2}\)
\(P\ge\sqrt{\left(x-3+y-3+z-3\right)^2+\left(4+4+4\right)^2}=6\sqrt{5}\)
\(P_{min}=6\sqrt{5}\) 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
Ta có:
\(1.\sqrt{1+x^2}+1.\sqrt{2x}\le\sqrt{\left(1+1\right)\left(1+x^2+2x\right)}=\sqrt{2}\left(x+1\right)\)
Tương tự:
\(\sqrt{1+y^2}+\sqrt{2y}\le\sqrt{2}\left(y+1\right)\) ; \(\sqrt{1+z^2}+\sqrt{2z}\le\sqrt{2}\left(z+1\right)\)
Cộng vế:
\(P\le\sqrt{2}\left(x+y+z+3\right)+\left(2-\sqrt{2}\right)\left(x+y+z\right)\le\sqrt{2}\left(3+3\right)+\left(2-\sqrt{2}\right).3=6+3\sqrt{2}\)
\(P_{max}=6+3\sqrt{2}\) khi \(x=y=z=1\)
\(3=x+y+xy\le\sqrt{2\left(x^2+y^2\right)}+\dfrac{x^2+y^2}{2}\)
\(\Rightarrow\left(\sqrt{x^2+y^2}-\sqrt{2}\right)\left(\sqrt{x^2+y^2}+3\sqrt{2}\right)\ge0\)
\(\Rightarrow x^2+y^2\ge2\)
\(\Rightarrow-\left(x^2+y^2\right)\le-2\)
\(P=\sqrt{9-x^2}+\sqrt{9-y^2}+\dfrac{x+y}{4}\le\sqrt{2\left(9-x^2+9-y^2\right)}+\dfrac{\sqrt{2\left(x^2+y^2\right)}}{4}\)
\(P\le\sqrt{2\left(18-x^2-y^2\right)}+\dfrac{1}{4}.\sqrt{2\left(x^2+y^2\right)}\)
\(P\le\left(\sqrt{2}-1\right)\sqrt{18-x^2-y^2}+\sqrt[]{2}\sqrt{\dfrac{\left(18-x^2-y^2\right)}{2}}+\dfrac{1}{2}\sqrt{\dfrac{x^2+y^2}{2}}\)
\(P\le\left(\sqrt{2}-1\right).\sqrt{18-2}+\sqrt{\left(2+\dfrac{1}{4}\right)\left(\dfrac{18-x^2-y^2+x^2+y^2}{2}\right)}=\dfrac{1+8\sqrt{2}}{2}\)
Dấu "=" xảy ra khi \(x=y=1\)
\(\left\{{}\begin{matrix}x+2=a^3\\x+1=b^3\\y-3=c^3\\y-4=d^3\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b+c+d=0\\a^3-b^3=1\\c^3-d^3=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+d=-\left(b+c\right)\\a^3+d^3-\left(b^3+c^3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+d=-\left(b+c\right)\\\left(a+d\right)\left(a^2-ad+d^2\right)=\left(b+c\right)\left(b^2-bc+c^2\right)\end{matrix}\right.\) (1)
TH1: \(a+d=-\left(b+c\right)\ne0\)
Chia vế cho vế 2 pt (1) ta được:
\(a^2-ad+d^2=-\left(b^2-bc+c^2\right)\)
\(\Leftrightarrow\left(a-\frac{d}{2}\right)^2+\frac{3d^2}{4}+\left(b-\frac{c}{2}\right)^2+\frac{3c^2}{4}=0\)
\(\Leftrightarrow a=b=c=d=0\) (vô nghiệm)
TH2: \(a+d=-\left(b+c\right)=0\Rightarrow\left\{{}\begin{matrix}a=-d\\b=-c\end{matrix}\right.\)
\(\Rightarrow x+2=4-y\Rightarrow x+y=2\)
\(\Rightarrow A=x^2+y^2\ge\frac{1}{2}\left(x+y\right)^2=2\)
Dấu "=" xảy ra khi \(x=y=1\)