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với x;y>=0 ta có:
\(A^2=\left(\sqrt{2x+1}+\sqrt{2y+1}\right)^2=2x+1+2y+1+2\sqrt{\left(2x+1\right)\left(2y+1\right)}\)
\(=2\left(x+y\right)+2+\sqrt{4xy+2x+2y+1}=2\left(x+y\right)+2+\sqrt{4xy+2\left(x+y\right)+1}\)
\(2=2\left(x^2+y^2\right)=\left(1+1\right)\left(x^2+y^2\right)>=\left(x+y\right)^2\Rightarrow x+y< =\sqrt{2}\)(bđt bunhiacopxki)
\(2xy< =x^2+y^2=1\Rightarrow2\cdot2xy=4xy< =2\cdot1=2\)
\(\Rightarrow A^2=2\left(x+y\right)+2+2\sqrt{4xy+2\left(x+y\right)+1}< =2\sqrt{2}+2+2\sqrt{2+2\sqrt{2}+1}\)
\(=2\sqrt{2}+2+2\sqrt{\left(\sqrt{2}+1\right)^2}=2\sqrt{2}+2+2\left(\sqrt{2}+1\right)4\sqrt{2}+4\)
\(\Rightarrow A< =\sqrt{4\sqrt{2}+4}\)
dấu = xảy ra khi x=y=\(\sqrt{\frac{1}{2}}\)
vậy max A là \(\sqrt{4\sqrt{2}+4}\)khi \(x=y=\sqrt{\frac{1}{2}}\)
\(-2A=2x^2+2y^2-2xy-4x-4y\)
\(=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2-4y+4\right)-8\)
\(=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\ge-8\)
=> \(A\le4\)
"=" xảy ra <=> x=y=2
Vậy max A=4 tại x=y=2
\(a.A=48-y^2-2y=-\left(y^2+2y+1\right)+49=-\left(y+1\right)^2+49\text{≥}49\) ⇒ \(A_{Max}=49."="\) ⇔ \(y=-1\)
\(b.B=4x^2-12x+17=4x^2-12x+9+8=\left(2x-3\right)^2+8\text{≥}8\)
⇒ \(B_{MIN}=8."="\) ⇔ \(x=\dfrac{3}{2}\)
\(a.A=-x^2-4x+15=-\left(x^2+4x+4\right)+19=-\left(x+2\right)^2+19\le19\)\(\Rightarrow A_{MAX}=19."="\Leftrightarrow x=-2\)
\(b.B=-x^2-4x-y^2+2y=-x^2-4x-4-y^2+2y-1+5=-\left(x+2\right)^2-\left(y-1\right)^2+5\ge5\)
\(\Rightarrow B_{MAX}=5."="\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=1\end{matrix}\right.\)