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`A=(9(x-2)+18)/(2-x)+2/x`
`=-9+18/(2-x)+2/x`
`=-9+2(9/(2-x)+1/x)`
Áp dụng bđt cosi-schwarts ta có:
`9/(2-x)+1/x>=(3+1)^2/(2-x+x)=8`
`=>A>=16-9=7`
Dấu "=" xảy ra khi `3/(2-x)=1/x`
`<=>3x=2-x`
`<=>4x=2<=>x=1/2(tm)`
b
`y=x/(1-x)+5/x`
`=(x-1+1)/(1-x)+5/x`
`=1/(1-x)+5/x-1`
Áp dụng cosi-schwarts ta có:
`1/(1-x)+5/x>=(1+sqrt5)^2/(1-x+x)=(1+sqrt5)^2=6+2sqrt5`
`=>y>=5+2sqrt5`
Dấu "=" xảy ra khi `1/(1-x)=sqrt5/x`
`<=>x=sqrt5-sqrt5x`
`<=>x(1+sqrt5)=sqrt5`
`<=>x=sqrt5/(sqrt5+1)=(sqrt5(sqrt5-1))/(5-1)=(5-sqrt5)/4`
`c)C=2/(1-x)+1/x`
Áp dụng bđt cosi schwarts ta có:
`C>=(sqrt2+1)^2/(1-x+x)=3+2sqrt2`
Dấu "=" xảy ra khi `sqrt2/(1-x)=1/x`
`<=>sqrt2x=1-x`
`<=>x(sqrt2+1)=1`
`<=>x=1/(sqrt2+1)=(sqrt2-1)/(2-1)=sqrt2-1`
a) \(A=x^2-6x+10=\left(x^2-6x+9\right)+1=\left(x-3\right)^2+1\ge1\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=3\). \(min_A=1\)
b) \(B=3x^2+x-2=3\left(x^2+\dfrac{1}{3}x-\dfrac{2}{3}\right)=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{36}-\dfrac{25}{36}\right)=3\left(x+\dfrac{1}{6}\right)^2-\dfrac{25}{12}\ge\dfrac{-25}{12}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{6}\). \(min_B=\dfrac{-25}{12}\)
c) \(C=\dfrac{4}{x^2}-\dfrac{3}{x}-1=\left(\dfrac{4}{x^2}-\dfrac{3}{x}+\dfrac{9}{16}\right)-\dfrac{25}{16}=\left(\dfrac{2}{x}+\dfrac{2}{3}\right)^2-\dfrac{25}{16}\ge\dfrac{-25}{16}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=-3\). \(min_C=\dfrac{-25}{16}\)
d) \(D=x^2+y^2-x+3y+7=\left(x^2-x+\dfrac{1}{4}\right)+\left(y^2+3y+\dfrac{9}{4}\right)+\dfrac{9}{2}=\left(x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{3}{2}\right)^2+\dfrac{9}{2}\ge\dfrac{9}{2}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-3}{2}\end{matrix}\right.\). \(min_D=\dfrac{9}{2}\)
Ta có: x - y = 2 => x = 2 + y
A = x3 - y3 = (X - y)(x2 + xy + y2) = 2(x2 + xy + y2) = 2(x2 - 2xy + y2) + 6xy = 2(x - y)2 + 6xy = 8 + 6xy
A = 8 + 6y(2 + y) = 8 + 12y + 6y2 = 6(y2 + 2y + 1) + 2 = 6(y + 1)2 + 2 \(\ge\)2 \(\forall\)y
Dấu "=" xảy ra <=> y + 1 = 0 <=> y = -1 <=> x = 2 - 1 = 1
Vậy MinA = 2 khi x = 1 và y = -1
x - y = 2 => y = x - 2
Khi đó: B = 2x2 + y2 = 2x2 + (x- 2)2 = 2x2 + x2 - 4x + 4 = 3x2 - 4x + 4 = 3(x2 - 4/3x + 4/9) + 8/3 = 3(x - 2/3)2 + 8/3 \(\ge\)8/3 \(\forall\)x
Dấu "=" xảy ra <=> x - 2/3 = 0 <=> x = 2/3 => y = 2/3 - 2 = -4/3
Vậy MinB = 8/3 khi x = 2/3 và y = -4/3
có: \(\dfrac{1}{x^2+y^2}=\dfrac{1}{\left(x+y\right)^2-2xy}=\dfrac{1}{1-2xy}\)(1)
có \(\dfrac{1}{xy}=\dfrac{2}{2xy}\left(2\right)\)
từ(1)(2)=>A=\(\dfrac{1}{1-2xy}+\dfrac{2}{2xy}\ge\dfrac{\left(1+\sqrt{2}\right)^2}{1}=\left(1+\sqrt{2}\right)^2\)
=>Min A=(1+\(\sqrt{2}\))^2
b,Ap dung bdt cauchy schwarz dang engel ta co
\(B=\frac{x^2}{1}+\frac{y^2}{1}+\frac{z^2}{1}>=\frac{\left(x+y+z\right)^2}{3}=\frac{a^2}{3}\)
xay ra dau = khi x=y=z=a/3
Ta có \(p=x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}=2\). Ta đi tìm GTNN của \(B=p+\dfrac{1}{p}\).
Do \(B=\dfrac{p}{4}+\dfrac{1}{p}+\dfrac{3p}{4}\) \(\ge2\sqrt{\dfrac{p}{4}.\dfrac{1}{p}}+\dfrac{3.2}{4}\) \(=\dfrac{5}{2}\). ĐTXR \(\Leftrightarrow\left\{{}\begin{matrix}x=y\\p=2\end{matrix}\right.\) \(\Leftrightarrow x=y=1\).
Vậy GTNN của B là \(\dfrac{5}{2}\) khi \(x=y=1\)
\(x+y=2\Rightarrow y=2-x\)
\(P=2x^2-\left(2-x\right)^2-5x+\dfrac{1}{x}+2020=x^2-x+\dfrac{1}{x}+2016\)
\(P=x^2+1-x+\dfrac{1}{x}+2015\ge2x-x+\dfrac{1}{x}+2015\)
\(P\ge x+\dfrac{1}{x}+2015\ge2\sqrt{\dfrac{x}{x}}+2015=2017\)
Dấu "=" xảy ra khi \(x=y=1\)