tìm gtnn của A=| 2x+18 |-35
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b: \(B=\sqrt{x^2-8x+18}-1\)
\(=\sqrt{\left(x-4\right)^2+2}-1\)
(x-4)^2+2>=2
=>\(\sqrt{\left(x-4\right)^2+2}>=\sqrt{2}\)
=>B>=căn 2-1
Dấu = xảy ra khi x=4
a: \(D=3+\sqrt{2x^2-8x+33}\)
\(=3+\sqrt{2\left(x^2-4x+\dfrac{33}{2}\right)}\)
\(=\sqrt{2\left(x^2-4x+4\right)+25}+3\)
\(=\sqrt{2\left(x-2\right)^2+25}+3>=5+3=8\)
Dấu = xảy ra khi x=2
c/ Ta có:\(6a-5b=1\)
\(\Rightarrow5b=6a-1\)
Theo đề thì: \(A=4a^2+\left(6a-1\right)^2=40a^2-12a+1\)
\(=\left(\left(2\sqrt{10}a\right)^2-\frac{2.2.\sqrt{10}.3a}{\sqrt{10}}+\frac{9}{10}\right)+\frac{1}{10}\)
\(=\left(2\sqrt{10}a-\frac{3}{\sqrt{10}}\right)^2+\frac{1}{10}\ge\frac{1}{10}\)
\(A=2x^2+4y^2+4xy+10x+12y+18\)
\(A=x^2+4xy+4y^2+6x+12y+9+x^2+4x+4+5\)
\(A=\left(x+2y\right)^2+2.3\left(x+2y\right)+9+\left(x+2\right)^2+5\)
\(A=\left(x+2y+3\right)^2+\left(x+2\right)^2+5\)
Do : \(\hept{\begin{cases}\left(x+2y+3\right)^2\ge0\forall x\\\left(x+2\right)^2\ge0\forall x\end{cases}}\)
\(\Leftrightarrow\left(x+2y+3\right)^2+\left(x+2\right)^2+5\ge5\)
\("="\Leftrightarrow\hept{\begin{cases}x+2y+3=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{2}\\x=-2\end{cases}}}\)
Vậy \(A_{min}=5\Leftrightarrow\hept{\begin{cases}x=-2\\y=-\frac{1}{2}\end{cases}}\)
Chúc bạn học tốt !!!
\(...P=x^2-8x+16+x^2+2xy+y^2+2y^2-2y+2\)
\(P=\left(x-4\right)^2+\left(x+y\right)^2+2\left(y^2-y+1\right)\left(1\right)\)
Xét \(y^2-y+1=y^2-y+\dfrac{1}{4}-\dfrac{1}{4}+1=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\left(\left(y-\dfrac{1}{2}\right)^2\ge0\right)\)
\(\Rightarrow2\left(y^2-y+1\right)\ge2.\dfrac{3}{4}=\dfrac{3}{2}\)
mà \(\left(x-4\right)^2\ge0;\left(x+y\right)^2\ge0\)
\(\left(1\right)\Rightarrow P\ge\dfrac{3}{2}\Rightarrow Min\left(P\right)=\dfrac{3}{2}\)
\(A=\left|\sqrt{35}+9\right|-\left|9-\sqrt{35}\right|\)
\(=\left|9+\sqrt{35}\right|-\left(9-\sqrt{35}\right)\)
\(=9+\sqrt{35}-9+\sqrt{35}=2\sqrt{35}\)
\(F=2x^2+8xy+11y^2-4x-2y+18\)
\(=2\left[x^2+2x\left(2y-1\right)+\left(2y-1\right)^2\right]+3\left(y^2+2y+1\right)+13\)
\(=2\left(x+2y-1\right)^2+3\left(y+1\right)^2+13\ge13\)
\(minF=13\Leftrightarrow\) \(\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
\(A=2x^2+4y^2+4xy+10x+12y+18\)
\(A=x^2+4xy+4y^2+6x+12y+9+x^2+4x+4+5\)
\(A=\left(x+2y^2\right)+2.3\left(x+2y\right)+9+\left(x+2\right)^2+5\)
\(A=\left(x+2y+3\right)^2+\left(x+2\right)^2+5\)
Do \(\hept{\begin{cases}\left(x+2y+3\right)^2\ge0\forall x\\\left(x+2\right)^2\ge0\forall x\end{cases}}\)
\(\Leftrightarrow\left(x+2y+3\right)^2+\left(x+2\right)^2+5\ge5\)
" = " \(\Leftrightarrow\hept{\begin{cases}x+2y+3=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-\frac{1}{2}\\x=-2\end{cases}}}\)
\(\Rightarrow A_{min}=5\Leftrightarrow\hept{\begin{cases}x=-2\\y=-\frac{1}{2}\end{cases}}\)
Chúc bạn học tốt !!!
Ta thấy
|2x+18|>=0
=>|2x+18|-35>=0-35=-35
=>A>=-35
Dấu = khi |2x+18|=0 <=>x=-9
Vậy Amin=-35 khi x=-9