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\(K=\frac{1}{x^2+y^2}+\frac{1}{xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}\)
\(\ge\frac{4}{x^2+y^2+2xy}+\frac{1}{2.\frac{\left(x+y\right)^2}{4}}\)
\(=\frac{4}{1}+\frac{1}{2.\frac{1}{4}}=6\)
Dấu = xảy ra khi \(x=y=\frac{1}{2}\)
Ta có \(\hept{\begin{cases}\left(x+y\right)^2=1\\\left(x-y\right)^2\ge0\end{cases}}\Leftrightarrow x^2+y^2\ge\frac{1}{2}\)
\(xy\le\frac{\left(x^2+^2\right)}{2}\)nên \(K=\frac{1}{x^2+y^2}+\frac{2}{xy}\ge\frac{1}{x^2+y^2}+\frac{2}{x^2+y^2}=\frac{3}{x^2+y^2}\ge\frac{3}{\frac{1}{2}}=6\)
\(K_{min}=6\)dấu "=" khi \(x=y=\frac{1}{2}\)
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`(-5x^2y) . (-3/5xy^2)^2`.
`<=> (-5x^2y) . (9/25x^2y^4)`.
`<=> -9/5 . x^4y^5`.
Bậc: `4 + 5 = 9`.
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a) \(A=\left(x-1\right)^2+\left(y+3\right)^2+2010\)
Vì \(\left(x-1\right)^2\ge0\forall x;\left(y+3\right)^2\ge0\forall y\)
\(\Rightarrow A\ge2010\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}}\)
Vậy Amin = 2010 <=> x = 1; y = -3
b) tương tự
a, Nhận xét : \(\left(x-1\right)^2\ge0\text{ với}\forall x\)
\(\left(y+3\right)^2\ge0\text{ với}\forall y\)
=> \(\left(x-1\right)^2+\left(y+3\right)^2\ge0\text{ với}\forall x,y\)
=> \(\left(x-1\right)^2+\left(y+3\right)^2+2010\ge2010\)
=> Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
=> \(A_{min}=2010\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)
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a: \(C\left(x\right)=A\left(x\right)+B\left(x\right)\)
\(=3x^4-4x^3+5x^2-4x-3-3x^4+4x^3-5x^2+2x+6\)
=-2x+3
b: Đặt C(x)=0
=>-2x+3=0
hay x=3/2
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a: P(x)=6x^3-4x^2+4x-2
Q(x)=-5x^3-10x^2+6x+11
M(x)=x^3-14x^2+10x+9
b: \(C\left(x\right)=7x^4-4x^3-6x+9+3x^4-7x^3-5x^2-9x+12\)
=10x^4-11x^3-5x^2-15x+21