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Answer:
3.
\(x^2+2y^2+2xy+7x+7y+10=0\)
\(\Rightarrow\left(x^2+2xy+y^2\right)+7x+7y+y^2+10=0\)
\(\Rightarrow\left(x+y\right)^2+7.\left(x+y\right)+y^2+10=0\)
\(\Rightarrow4S^2+28S+4y^2+40=0\)
\(\Rightarrow4S^2+28S+49+4y^2-9=0\)
\(\Rightarrow\left(2S+7\right)^2=9-4y^2\le9\left(1\right)\)
\(\Rightarrow-3\le2S+7\le3\)
\(\Rightarrow-10\le2S\le-4\)
\(\Rightarrow-5\le S\le-2\left(2\right)\)
Dấu " = " xảy ra khi: \(\left(1\right)\Rightarrow y=0\)
Vậy giá trị nhỏ nhất của \(S=x+y=-5\Rightarrow\hept{\begin{cases}y=0\\x=-5\end{cases}}\)
Vậy giá trị lớn nhất của \(S=x+y=-2\Rightarrow\hept{\begin{cases}y=0\\x=-2\end{cases}}\)
Ta có: 3x + y = 1 => y = 1 - 3x
a, Thay y = 1 - 3x vào M, ta có:
\(\Rightarrow M=3x^2+\left(1-3x\right)^2=3x^2+1-6x+9x^2=12x^2-6x+1=3\left(4x^2-2x+\frac{1}{3}\right)\)
\(=3\left(4x^2-2x+\frac{1}{4}+\frac{1}{12}\right)=3\left(2x-\frac{1}{2}\right)^2+\frac{3}{12}=3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\)
Vì \(\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow3\left(2x-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}2x-\frac{1}{2}=0\\3x+y=1\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=1-3x=1-3.\frac{1}{4}=\frac{1}{4}\end{cases}}\)\(\Leftrightarrow x=y=\frac{1}{4}\)
Vậy GTNN M = 1/4 khi x = y = 1/4
b, Thay y = 1 - 3x vào N
\(\Rightarrow N=x\left(1-3x\right)=x-3x^2=-3\left(x^2-\frac{x}{3}+\frac{1}{36}-\frac{1}{36}\right)\)
\(=-3\left(x-\frac{1}{6}\right)^2-3.\left(-\frac{1}{36}\right)=-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\)
Vì \(\left(x-\frac{1}{6}\right)^2\ge0\forall x\)
\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2\le0\forall x\)
\(\Rightarrow-3\left(x-\frac{1}{6}\right)^2+\frac{1}{12}\le\frac{1}{12}\forall x\)
Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-\frac{1}{6}=0\\3x+y=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{6}\\y=1-3x=1-3.\frac{1}{6}=\frac{1}{2}\end{cases}}\)
Vậy GTLN N = 1/12 khi x = 1/6 và y = 1/2
Ta có: 3x + y = 1 => y = 1 - 3x
=> M = 3x2 + y2 = 3x2 + (1-3x)2
= 3x2 + 1 - 6x + 9x2
= 12x2 - 6x + 1
= 12.(x2 -\(\frac{1}{2}x\) + \(\frac{1}{12}\))
= 12.((x2 - 2. \(\frac{1}{4}x\)+ \(\frac{1}{16}\)) - \(\frac{1}{16}\)+ \(\frac{1}{12}\))
= 12.((x-\(\frac{1}{4}\))2 + \(\frac{1}{48}\))
= 12. (x-\(\frac{1}{4}\))2 + \(\frac{1}{4}\)
=> M \(\ge\)\(\frac{1}{4}\)
Dấu ''='' xảy ra khi: (x - \(\frac{1}{4}\))2 = 0 => x = \(\frac{1}{4}\)
Vậy Mmin= \(\frac{1}{4}\)khi x= \(\frac{1}{4}\)
Ta có: \(Q=\dfrac{2}{x^2+y^2}+\dfrac{3}{xy}=\dfrac{2}{x^2+y^2}+\dfrac{6}{2xy}=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{4}{2xy}\)
Áp dụng BĐT phụ: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Rightarrow2\left(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\right)\ge2\left(\dfrac{4}{x^2+2xy+y^2}\right)=2\left[\dfrac{4}{\left(x+y\right)^2}\right]=2.\dfrac{4}{4}=2\)
Dấu "=" xảy ra khi x=y=1
Áp dụng BĐT phụ: \(ab\le\dfrac{\left(a+b\right)^2}{4}\)
\(\Rightarrow xy\le\dfrac{\left(x+y\right)^2}{4}=\dfrac{2^2}{4}=1\)
Dấu"=" xảy ra khi x=y=1
\(\Rightarrow2xy\le2.1=2\)
\(\Rightarrow\dfrac{4}{2xy}\ge\dfrac{4}{2}=2\)
\(\Rightarrow Q=\dfrac{2}{x^2+y^2}+\dfrac{2}{2xy}+\dfrac{4}{2xy}=\dfrac{2}{x^2+y^2}+\dfrac{3}{xy}\ge2+2=4\)
Dấu"=" xảy ra khi x=y=1
By Titu's Lemma we easy have:
\(D=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2\)
\(\ge\frac{\left(x+y+\frac{1}{x}+\frac{1}{y}\right)^2}{2}\)
\(\ge\frac{\left(x+y+\frac{4}{x+y}\right)^2}{2}\)
\(=\frac{17}{4}\)
Mk xin b2 nha!
\(P=\frac{1}{x^2+y^2}+\frac{1}{xy}+4xy=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}+4xy\)
\(\ge\frac{\left(1+1\right)^2}{x^2+y^2+2xy}+\left(4xy+\frac{1}{4xy}\right)+\frac{1}{4xy}\)
\(\ge\frac{4}{\left(x+y\right)^2}+2\sqrt{4xy.\frac{1}{4xy}}+\frac{1}{\left(x+y\right)^2}\)
\(\ge\frac{4}{1^2}+2+\frac{1}{1^2}=4+2+1=7\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
a,Từ \(3x+y=1\Rightarrow x=\dfrac{1-y}{3}\)
\(\Rightarrow M=3x^2+y^2=3.\left(\dfrac{1-y}{3}\right)^2+y^2=3.\dfrac{y^2-2y+1}{9}+y^2\)
\(=\dfrac{3y^2+y^2-2y+1}{3}=\dfrac{4y^2-2y+1}{3}\)
Ta có: \(4y^2-2y+1=4y^2-2.2y.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(2y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Bn tự chứng minh \(\left(2y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\left(2y-\dfrac{1}{2}\right)^2=0\Leftrightarrow y=\dfrac{1}{4}\)
\(\Rightarrow M=\dfrac{\left(2y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}{3}\ge\dfrac{\dfrac{3}{4}}{3}=\dfrac{1}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow\) \(y=\dfrac{1}{4}\);\(x=\dfrac{1-\dfrac{1}{4}}{3}=\dfrac{1}{4}\)
Cx như a, \(x=\dfrac{1-y}{3}\) thay vào N đc:
\(N=xy=\dfrac{1-y}{3}.y=\dfrac{y-y^2}{3}=\dfrac{-\left(y^2-2.\dfrac{1}{2}y+\dfrac{1}{4}+\dfrac{3}{4}\right)}{3}\)
\(=\dfrac{-\left(y-\dfrac{1}{2}\right)^2-\dfrac{3}{4}}{3}\)
Bn tự chứng minh \(-\left(y-\dfrac{1}{2}\right)^2-\dfrac{3}{4}\le-\dfrac{3}{4}\). Dấu "=" xảy ra \(\Leftrightarrow\)\(-\left(y-\dfrac{1}{2}\right)=0\Leftrightarrow y=\dfrac{1}{2}\)
\(\Rightarrow N=\dfrac{-\left(y-\dfrac{1}{2}\right)^2-\dfrac{3}{4}}{3}\le\dfrac{\dfrac{-3}{4}}{3}=\dfrac{-1}{4}\)
Vậy MAX N = \(\dfrac{-1}{4}\Leftrightarrow y=\dfrac{1}{2}\)