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Áp dụng bđt \(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
\(\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5^2\)
\(\Rightarrow-5\le2x+3y\le5\)
Dấu bằng xảy ra khi \(\frac{a}{x}=\frac{b}{y}\)hay \(\frac{\sqrt{2}x}{\sqrt{3}y}=\frac{\sqrt{2}}{\sqrt{3}}\Leftrightarrow x=y\)
Vậy \(A\text{ min }=-5\Leftrightarrow x=y=-1\)
\(A\text{ max }=5\Leftrightarrow x=y=1\)
Áp dụng BĐT Bunhiaskopski:
\(A^2=\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5.5=25\)
\(A^2\le25\Rightarrow-5\le A\le5\)
Max:Dấu ''='' xảy ra khi x=y=1
Min:Dấu ''='' xảy ra khi x=y=-1
Hok bít đúng hok nữa, sai thôi nha
\(A=\sqrt{1-x}+\sqrt{x+1}\)
\(A^2=\left(\sqrt{1-x}\cdot1+\sqrt{x+1}\cdot1\right)^2\)
Áp dụng BĐT Bunhiacospki ta có:
\(A^2\le\left(1^2+1^2\right)\left(1-x+1+x\right)\)
\(A^2\le4\)
\(A\le2\)
\(A_{max}=2\Leftrightarrow x=0\)
E ms tìm dc MAX thôi ah
ĐKXĐ: ....
a/ \(A\le\sqrt{2\left(1-x+1+x\right)}=2\Rightarrow A_{max}=2\) khi \(x=0\)
\(A\ge\sqrt{1-x+1+x}=\sqrt{2}\Rightarrow A_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
b/ \(B\le\sqrt{2\left(x-2+6-x\right)}=2\sqrt{2}\Rightarrow B_{max}=2\sqrt{2}\) khi \(x=4\)
\(B\ge\sqrt{x-2+6-x}=2\Rightarrow B_{min}=2\) khi \(\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
c/ \(A^2=\left(2x+3y\right)^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\)
\(\Rightarrow A^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5.5=25\)
\(\Rightarrow-5\le A\le5\)
\(A_{max}=5\) khi \(x=y=1\)
\(A_{min}=-5\) khi \(x=y=-1\)
GTLN:
Áp dụng BĐT \(a^2+b^2\ge2ab\)
\(\Rightarrow x^2+1\ge2x\Rightarrow2x^2\ge4x-2\)
\(y^2+1\ge2y\Rightarrow3y^2\ge6y-3\)
\(\Rightarrow2x^2+3y^2\ge2\left(2x+3y\right)-5\)
mà \(2x^2+3y^2\le5\)
\(\Rightarrow2\left(2x+3y\right)-5\le5\Rightarrow2x+3y\le5\)
Vậy Max A = 5 khi x = y = 1
\(x^4+2x^2y^2+y^4-3x^2-4y^2+4=1\)
\(\Leftrightarrow\left(x^2+y^2\right)^2-4\left(x^2+y^2\right)+4=1-x^2\)
\(\Leftrightarrow\left(x^2+y^2-2\right)^2=1-x^2\)
Do \(1-x^2\le1\) \(\forall x\)
\(\Rightarrow-1\le x^2+y^2-2\le1\)
\(\Rightarrow1\le x^2+y^2\le3\)
\(A_{min}=1\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm1\end{matrix}\right.\)
\(A_{max}=3\) khi \(\left\{{}\begin{matrix}x=0\\y=\pm\sqrt{3}\end{matrix}\right.\)
hướng dẫn thôi tự trình bày lại nhé
pt đầu bài \(\Leftrightarrow\)\(4x^2+9y^2+25+12xy+20x+30y=-3x^2+24x+36y+40\)
\(\Leftrightarrow\)\(\left(2x+3y+5\right)^2-12\left(2x+3y+5\right)+36=-3x^2+16\)
\(\Leftrightarrow\)\(\left(2x+3y-1\right)^2=-3x^2+16\le16\)
\(\Leftrightarrow\)\(-4\le2x+3y-1\le4\)\(\Leftrightarrow\)\(2\le2x+3y+5\le10\)
\(\Rightarrow\)\(\hept{\begin{cases}S_{min}=2\left(x=0;y=-1\right)\\S_{max}=10\left(x=0;y=\frac{5}{3}\right)\end{cases}}\)
\(A^2=\left(\sqrt{2}.\sqrt{2}x+\sqrt{3}.\sqrt{3}y\right)^2\)
\(\Rightarrow A^2\le\left(2+3\right)\left(2x^2+3y^2\right)\le5.5=25\)
\(\Rightarrow-5\le A\le5\)
\(A_{max}=5\) khi \(x=y=1\)
\(A_{min}=-5\) khi \(x=y=-1\)